Superbimatrices and Their Generalizations

SUPERBIMATRICES AND THEIR GENERALIZATIONS

W. B. Vasantha Kandasamy e-mail: vasanthakandasamy@gmail.com web: http://mat.iitm.ac.in/~wbv www.vasantha.in

Florentin Smarandache e-mail: smarand@unm.edu

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Chapter Four SUPER n-MATRICES AND THEIR PROPERTIES 349

FURTHER READING 399 INDEX 403 ABOUT THE AUTHOR 408 S

PREFACE

Thesystematicstudyofsupermatricesandsuperlinear algebrahasbeencarriedoutin2008.Thesenewalgebraic structuresfindtheirapplicationsinfuzzymodels,Leontief economicmodelsanddata-storageincomputers.

Inthisbooktheauthorsintroducethenewnotion ofsuperbimatricesandgeneralizeittosupertrimatrices andsupern-matrices.Studyofthesestructuresisnotonly interestingandinnovativebutisalsobestsuitedforthe computerizedworld.

Themaindifferencebetweensimplebimatrices andsuperbimatricesisthatincaseofsimplebimatrices wehaveonlyonetypeofproductdefinedonthem, whereasincaseofsuperbimatriceswehavedifferenttypes productscalledminorandmajordefinedusingthem.

Thisbookhasfourchapters.Chapteronedescribes thebasicsconceptstomakethisbookaselfcontainedone. Superbimatrices,semisuperbimatrices,symmetric superbimatricesareintroducedinchaptertwo.Chapter threeintroducesthenotionofsupertrimatricesandthe productsdefinedusingthem.Chapterfourgivesthemost generalizedformofsuperbimatrix,viz.supern-matrix.

Thisbookhasgivenseveralexamplessoasto makethereaderunderstandthisnewconcept.Further minorandmajorbyproductdefinedusingthesenew conceptsareillustratedbyexamples.Thesealgebraic structuresarebestsuitedindatastorageincomputers.

Theyarealsousefulinconstructingmultiexpertsuper models

Finallyitisanimmensepleasuretothank Dr.K.Kandasamyforproof-readingandKama,Meenaand Rahulwithoutwhosehelpthebookwouldhavebeenan impossibility.

WededicatethisbooktothemillionsofTamil childreninSriLankawhohavediedorbecomedisabled anddisplacedduetotherecentSriLankanwar.

W.B.VASANTHA KANDASAMY FLORENTIN SMARANDACHE

BASIC CONCEPTS

Weshallcallthemassimplematrices[19].Byasimple matrixwemeanamatrixeachofwhoseelementsarejustan ordinarynumberoraletterthatstandsforanumber.Inother words,theelementsofasimplematrixarescalarsorscalar quantities.

Asupermatrixontheotherhandisonewhoseelementsare themselvesmatriceswithelementsthatcanbeeitherscalarsor othermatrices.Ingeneralthekindofsupermatricesweshall dealwithinthisbook,thematrixelementswhichhaveany scalarfortheirelements.Supposewehavethefourmatrices; a11= 24 01,a12= 040 2112 a21=

31 57 29 anda22=

412 176 311 .

Onecanobservethechangeinnotationaijdenotesamatrixand notascalarofamatrix(1 < i,j<2). Let a= 1112 2122

aa aa ; wecanwriteoutthematrixaintermsoftheoriginalmatrix elementsi.e., a=

24040 012112 31412 57176 29311

.

Heretheelementsaredividedverticallyandhorizontallybythin lines.Ifthelineswerenotusedthematrixawouldbereadasa simplematrix.

Thusfarwehavereferredtotheelementsinasupermatrix asmatricesaselements.Itisperhapsmoreusualtocallthe elementsofasupermatrixassubmatrices.Wespeakofthe submatriceswithinasupermatrix.Nowweproceedontodefine theorderofasupermatrix.

Theorderofasupermatrixisdefinedinthesamewayas thatofasimplematrix.Theheightofasupermatrixisthe numberofrowsofsubmatricesinit.Thewidthofasupermatrix isthenumberofcolumnsofsubmatricesinit.

Allsubmatriceswithinagivenrowmusthavethesame numberofrows.Likewiseallsubmatriceswithinagiven columnmusthavethesamenumberofcolumns.

Adiagrammaticrepresentationisgivenbythefollowing figure.

Inthefirstrowofrectangleswehaveonerowofasquare foreachrectangle;inthesecondrowofrectangleswehavefour rowsofsquaresforeachrectangleandinthethirdrowof rectangleswehavetworowsofsquaresforeachrectangle. Similarlyforthefirstcolumnofrectanglesthreecolumnsof squaresforeachrectangle.Forthesecondcolumnofrectangles wehavetwocolumnofsquaresforeachrectangle,andforthe thirdcolumnofrectangleswehavefivecolumnsofsquaresfor eachrectangle.

Thuswehaveforthissupermatrix3rowsand3columns.

Onethingshouldnowbeclearfromthedefinitionofa supermatrix.Thesuperorderofasupermatrixtellsusnothing aboutthesimpleorderofthematrixfromwhichitwasobtained

bypartitioning.Furthermore,theorderofsupermatrixtellsus nothingabouttheordersofthesubmatriceswithinthat supermatrix.

Nowweillustratethenumberofrowsandcolumnsofa supermatrix.

Example 1.1.1: Let a=

33014 12116 03456 17890 21234

aisasupermatrixwithtworowsandtwocolumns.

Nowweproceedontodefinethenotionofpartitionedmatrices. Itisalwayspossibletoconstructasupermatrixfromanysimple matrixthatisnotascalarquantity.

Thesupermatrixcanbeconstructedfromasimplematrix thisprocessofconstructingsupermatrixiscalledthe partitioning.

Asimplematrixcanbepartitionedbydividingorseparating thematrixbetweencertainspecifiedrows,ortheproceduremay bereversed.Thedivisionmaybemadefirstbetweenrowsand thenbetweencolumns.

Weillustratethisbyasimpleexample.

Example 1.1.2: Let A=

301120 100352 516784 091201 252346 161239

isa6×6simplematrixwithrealnumbersaselements.

A1=

301120 100352 516784 091201 252346 161239

.

Nowletusdrawathinlinebetweenthe2ndand3rdcolumns. ThisgivesusthematrixA1.ActuallyA1mayberegardedas asupermatrixwithtwomatrixelementsformingonerowand twocolumns. Nowconsider A2=

301120 100352 516784 091201 252346 161239

Drawathinlinebetweentherows4and5whichgivesusthe newmatrixA2.A2isasupermatrixwithtworowsandone column. Nowconsiderthematrix

A3=

301120 100352 516784 091201 252346 161239

,

A3isnowasecondordersupermatrixwithtworowsandtwo columns.WecansimplywriteA3as

aa aa where a11=

1112 2122

30 10 51 09

, a12=

1120 0352 6784 1201

, a21= 25 16 anda22= 2346 1239 .

Theelementsnowarethesubmatricesdefinedasa11,a12,a21and a22andthereforeA3isintermsofletters.

Accordingtothemethodswehaveillustratedasimple matrixcanbepartitionedtoobtainasupermatrixinanyway thathappenstosuitourpurposes.

Thenaturalorderofasupermatrixisusuallydeterminedby thenaturalorderofthecorrespondingsimplematrix.Further morewearenotusuallyconcernedwithnaturalorderofthe submatriceswithinasupermatrix.

Nowweproceedontorecallthenotionofsymmetric partition,formoreinformationabouttheseconceptspleaserefer [19].Byasymmetricpartitioningofamatrixwemeanthatthe rowsandcolumnsarepartitionedinexactlythesameway.Ifthe matrixispartitionedbetweenthefirstandsecondcolumnand betweenthethirdandfourthcolumn,thentobesymmetrically partitioning,itmustalsobepartitionedbetweenthefirstand secondrowsandthirdandfourthrows.Accordingtothisruleof symmetricpartitioningonlysquaresimplematrixcanbe

12

symmetricallypartitioned.Wegiveanexampleofa symmetricallypartitionedmatrixas,

Example 1.1.3: Let as=

2341 5692 0619 5115

.

Hereweseethatthematrixhasbeenpartitionedbetween columnsoneandtwoandthreeandfour.Ithasalsobeen partitionedbetweenrowsoneandtwoandrowsthreeandfour.

Nowwejustrecallfrom[19]themethodofsymmetric partitioningofasymmetricsimplematrix.

Example 1.1.4: Letustakeafourthordersymmetricmatrixand partitionitbetweenthesecondandthirdrowsandalsobetween thesecondandthirdcolumns. a=

4327 3614 2152 7427

.

Wecanrepresentthismatrixasasupermatrixwithletter elements. a11= 43 36,a12= 27 14 a21= 21 74 anda22= 52 27, sothat

a= 1112 2122

aa aa .

Thediagonalelementsofthesupermatrixaarea11anda22.We alsoobservethematricesa11anda22arealsosymmetric matrices.

Thenondiagonalelementsofthissupermatrixaarethe matricesa12anda21.Clearlya21isthetransposeofa12.

Thesimpleruleaboutthematrixelementofa symmetricallypartitionedsymmetricsimplematrixare(1)The diagonalsubmatricesofthesupermatrixareallsymmetric matrices.(2)Thematrixelementsbelowthediagonalarethe transposesofthecorrespondingelementsabovethediagonal.

Theforthordersupermatrixobtainedfromasymmetric partitioningofasymmetricsimplematrixaisasfollows. a=

aaaa ' aaaa '' aaaa ''' aaaa

11121314 12222324 13233334 14243444

.

Howtoexpressthatasymmetricmatrixhasbeensymmetrically partitioned(i)a11andat11areequal.(ii)atij(ij);taij=ajiand t aji=aij.Thusthegeneralexpressionforasymmetrically partitionedsymmetricmatrix; a=

aa...a a'a...a a'a'...a

11121n 12222n 1n2nnn

.

Ifwewanttoindicateasymmetricallypartitionedsimple diagonalmatrixwewouldwrite

D=

1 2 n

D0...0 0D...0 00...D

0'onlyrepresentstheorderisreversedortransformed.We denotetaij=a'ijjustthe'meansthetranspose. Dwillbereferredtoasthesuperdiagonalmatrix.The identitymatrix I= s t r

I00 0I0 00I s,tandrdenotethenumberofrowsandcolumnsofthefirst secondandthirdidentitymatricesrespectively(zerosdenote matriceswithzeroasallentries).

Example 1.1.5: Wejustillustrateageneralsuperdiagonal matrixd; d=

31200 56000 00025 00013 000910 i.e.,d=1 2

m0 0m .

Anexampleofasuperdiagonalmatrixwithvectorelementsis given,whichcanbeusefulinexperimentaldesigns.

Example 1.1.6: Let 1000 1000 1000 0100 0100 0010 0010 0010 0010 0001 0001 0001 0001

.

Herethediagonalelementsareonlycolumnunitvectors.In caseofsupermatrix[19]hasdefinedthenotionofpartial triangularmatrixasasupermatrix.

Example 1.1.7: Let u= 21132 05211 00102 uisapartialuppertriangularsupermatrix.

Example 1.1.8: Let L=

50000 72000 12300 45670 12526 12345 01010

;

Lispartialuppertriangularmatrixpartitionedasasupermatrix.

ThusT= T a whereTisthelowertriangularsubmatrix,with T=

50000 72000 12300 45670 12526

anda'= 12345 01010 .

Weproceedontodefinethenotionofsupervectorsi.e.,TypeI columnsupervector.Asimplevectorisavectoreachofwhose elementsisascalar.Itisnicetoseethenumberofdifferent typesofsupervectorsgivenby[19].

Example

whereeachviisacolumnsubvectorsofthecolumnvectorv.

TypeIrowsupervectorisgivenbythefollowingexample.

Example 1.1.10: v1=[231|5784]isatypeIrow supervector.i.e.,v'=[v'1,v'2,…,v'n]whereeachv'iisarow subvector;1in.

NextwerecallthedefinitionoftypeIIsupervectors. TypeIIcolumnsupervectors.

DEFINITION 1.1.1: Let a =

aaa aaa aaa a1 1 = [a11 … a1m] a2 1 = [a21 … a2m] an 1 = [an1 … anm] i.e., a =

a a a

11 21 1n

, a 2 =

... ... ............ ...

11121 21222 12

m m nnnm

1 1 1 2 1 n m

a a a is defined to be the type II column supervector. Similarly if a 1 =

a a a

12 22 2 n

, …, a m =

a a a

m m nm

1 2

.

Hence now a = [a1 a 2 … am] n is defined to be the type II row supervector.

Clearly

a =

a a a

1 1 1 2 1 n m

= [a1 a 2 … am]n the equality of supermatrices. Example 1.1.11: Let A=

36045 21630 11121 01010 20121 beasimplematrix.LetaandbthesupermatrixmadefromA. a=

36045 21630 11121 01010 20121 where a11=

360 216 111 ,a12=

45 30 21 , a21= 010 201 anda22= 10 21 . i.e.,a=1112 2122

aa aa .

b=

36045 21630 11121 01010 20121

bb bb where b11=

3604 2163 1112 0101

= 1112 2122

,b12=

5 0 1 0

, b21=[2012]andb22=[1]. a=

36045 21630 11121 01010 20121 and b=

36045 21630 11121 01010 20121

.

Weseethatthecorrespondingscalarelementsformatrixaand matrixbareidentical.Thustwosupermatricesareequalifand onlyiftheircorrespondingsimpleformsareequal.

NowwegiveexamplesoftypeIIIsupervectorformore refer[19].

Example 1.1.12: a= 32178 02169 00512 =[T'|a'] and b=

200 940 836 529 473

= T b aretypeIIIsupervectors.

OneinterestingandcommonexampleofatypeIIIsupervector isapredictiondatamatrixhavingbothpredictorandcriterion attributes. Thenextinterestingnotionaboutsupermatrixisits transpose.Firstweillustratethisbyanexamplebeforewegive thegeneralcase.

Example 1.1.13: Let a=

21356 02011 11102 22011 56101 20004 10115 = 1112 2122 3132

aa aa aa

where a11=

56 11 02 , a21= 220 561,a22= 11 01, a31= 200 101 anda32= 04 15

Thetransposeofa at=a'=

213 020 111 ,a12=

2012521 1212600 3010101 5101001 6121145

Letusconsiderthetransposesofa11,a12,a21,a22,a31anda32 a'11= t 11

201 a121 301 a'12= t 12 510 a 612 a'21= t 21

25 a26 01

a'31= t 31

21 a00 01 a'22= t 22 10 a 11 a'32= t 32 01 a 45 a'=112131 122232

aaa aaa

Nowwedescribethegeneralcase.Let a=

11121m 21222m n1n2nm

aaa aaa aaa bean×msupermatrix.Thetransposeofthesupermatrixa denotedby a'=

aaa aaa aaa

1121n1 1222n2 1m2mnm

.

a'isambynsupermatrixobtainedbytakingthetransposeof eachelementi.e.,thesubmatricesofa.

Nowwewillfindthetransposeofasymmetricallypartitioned symmetricsimplematrix.Letabethesymmetricallypartitioned symmetricsimplematrix.

Letabeam×msymmetricsupermatrixi.e., a=

aaa aaa aaa

1121m1 1222m2 1m2mmm

thetransposeofthesupermatrixisgivenbya' a'= 11121m 12222m 1m2mmm

a(a)(a) aa'(a) aaa

Thediagonalmatrixa11aresymmetricmatricessoareunaltered bytransposition.Hence a'11=a11,a'22=a22,…,a'mm=amm.

Recallalsothetransposeofatransposeistheoriginalmatrix. Therefore (a'12)'=a12,(a'13)'=a13,…,(a'ij)'=aij

Thusthetransposeofsupermatrixconstructedby symmetricallypartitionedsymmetricsimplematrixaofa'is givenby a'= 11121m 21222m 1m2mmm

aaa aaa aaa

.

1 2 n

d d d d'1=d1,d'2=d2etc.ThusD=D'. NowweseethetransposeofatypeIsupervector. Example 1.1.14: Let

V= 3

d d d D'= 1 2 n 1 2 4 5 7 5 1

v v v where v1=

V= 1 2 3

4 5 7 andv3= 5 1 V'=[v'1v'2v'3]. Thusif V=

3 1 2 ,v2=

1 2 n

v v v then V'=[v'1v'2…v'n].

Example 1.1.15: Let t= 301152 420135 101016 =[T|a].Thetransposeoft i.e.,t'=

341 020 101 110 531 256

= T a .

Theadditionofsupermatricesmaynotbealwaysbedefined.

Example 1.1.16: Forinstancelet a= 1112 2122

aa aa and b=1112 2122

bb bb where a11= 30 12,a12= 1 7 a21=[43],a22=[6]. b11=[2],b12=[13] b21= 5 2 andb22= 41 02 .

Itisclearbothaandbaresecondordersquaresupermatrices butherewecannotaddtogetherthecorrespondingmatrix elementsofaandbbecausethesubmatricesdonothavethe sameorder. Nowweproceedontorecallthedefinitionofminorproduct oftwosupervectors.

Suppose va=

v v v

a a a

1 2 n

andvb=

v v v

b b b

1 2 n

.

Theminorproductofthesetwosupervectorsvaandvbisgiven by =

1 2 12n n

b b abaaa b

v v vvvvv v = 1122nn ababab vvvvvv.

Weillustratethisbythefollowingexample.

Example 1.1.17: LetVaandVbbetwotypeIsupervectors where Va= 1 2 3

v v v with 1a

0 v1 2 ,2 a

4 0 v 1 1

a a a

and 3 a 1 v 2 . Let Vb= 1 2 3

v v v where 1b

1 v1 0 ,2 b

4 1 v 2 0

b b b

and 3b 1 v 1 .

v VVvvvv v = 112233 ababab vvvvvv =

1 1232 3

b abaaab b

4 1 11 0121401112 21 0 0 =–1+(–16+2)+(–1+2) =–1–16+2–1+2 =–14. ItiseasilyprovedV'aVb=V'bVa. Nowweproceedontorecallthedefinitionofmajorproduct oftypeIsupervectors. Suppose Va=

v v v

a a a

1 2 n

andVb=

b b b

1 2 m

v v v beanytwosupervectorsoftypeI.Themajorproductisdefined as VaV'b=

v v v

a a a

1 2 n

12m bbb vvv

=

vvvvvv vvvvvv vvvvvv

ababab ababab ababab

11121m 21222m n1n2nm

Nowweillustratethisbythefollowingexample.

Example 1.1.18: Let Va= 1 2 3

.

v v v andVb=

a a a

v v v v where 1a v=[2],2 a 1 v 1 and 3 a

b b b b

1 2 3 4

1 v2 0 and 1b

3 v1 2 ,2 b 1 v 2,3 b

3 4 v 1 0

and 4b v=[5]. VaV'b=

2 1 1 3121234105 1 2 0

=

2312[2]12[2]3410[2]5 1111 3121234105 1111 1111 23122122341025 0000 =

62424642010 3121234105 3121234105 3121234105 62424682010 0000000000

. Weleaveitforthereadertoverifythat(VaV'b)'=VbV'a. Example 1.1.19: Wejustrecallif v= 3 4 7 isacolumnvectorandv'thetransposeofvisarowvectorthen wehave v'v= 3 3474 7 =32+42+72=74. Thusif V'x=[x1x2…xn]

x x x = 222 12n xxx. Also [11…1] 1 2 n

V'xVx=[x1x2…xn] 1 2 n

x x x

=[x1+x2+…+xn] and [x1x2…xn]

1 1 1

=[x1+x2+…+xn]; i.e.,1'vx=v'x1=i x where vx=

1 2 n

x x x and xi=x1+x2+…+xn.

Wehavethefollowingtypesofproductsdefined.

Example 1.1.20: Wehave [0100]

0 1 0 0

=1, [0100]

1 0 0 0

=0, [0100]

1 1 1 1

=1 and 0 1 0 0

[100]=

000 100 000 000

. Recall a= 11121m 21222m n1n2nm

aaa aaa aaa wehave a=

a a a

1 1 1 2 1 nm

(1) and a=[a1 a 2 …a m]n.(2)

Nowtransposeof a=

1 1 1 2 1 nm

a a a isgivenbytheequation a'=111 12n m (a)(a)(a) a'=

(a) (a) (a)

1 2 m n

Thematrix b= 11121s 21222s t1t2ts

bbb bbb bbb

rowsupervectorofbis b=[b1b2…bs]t=[b1b2…bs]t.

Columnsupervectorofbis b=

Transposeofb;

b b b

1 1 1 2 1 ts

.

b'=

b b b

1 1 1 2 1 st

b'=[b1b2…bt]s.

TheproductoftwomatricesasaminorproductoftypeII supervector. ab=[a1 a 2 …a m]n

1 1 1 2 1 ts

b b b = 111 1122mt ns ababab.

Howevertomakethispointclearwegiveanexample. Example 1.1.21: Let 11 12aa 21 a35 61 and b= 1 2 12b 31b . ab= 21 312531 61

35

= 2431 36155 61231 = 55 1811 913 .

ItiseasilyverifiedthatifthemajorproductofthetypeII supervectoriscomputedbetweenaandb,thenthemajor productcoincideswiththeminorproduct.Fromtheabove example. ab=

12 2121 31 12 3535 31 12 6161 31 = 21132211 31533251 61136211 = 55 1811 913 .

Wecanfindtheminorandmajorproductofsupervectorsby reversingtheorderofthefactors.Sincethetheoryof multiplicationofsupermatricesinvolveslotsofnotationswe haveresolvedtoexplaintheseconceptsbyworkingoutthese conceptswithnumericalillustrations,whichwefeeliseasyfor

36

thegraspofthereader.Nowwegivethenumericalillustration oftheminorproductofTypeIIIvectors.

Example 1.1.22: Let X= 234222 111101 002400 and Y=

20 11 21 53 11 02 betwotypeIIIsupervectors.TofindtheproductXY. XY= 234222 111101 002400

20 11 21 53 11 02 = 23422253 20 1112110111 11 00240002 = 7384128 112155 00422012

= 2715 67 1610 .

210 310 412 214 200 210 = 212102 412 013101

YtXt = 212510 011312

214 510 200 312 210 = 71082412520 3104128512 = 27616 15710 .

Fromthisexampleitisveryclear. (XY)t=YtXt .

NowweillustratetheminorproductmomentoftypeIIIrow supervectorbyanexample.

Example 1.1.23: Let X= 2343450 1411116 2120211 .

38

whereYisthecolumnsupervector

Example 1.1.26: Let X=

131251 211120 151112 410221 321011 101101 421311 and Y=

110131212 201012001 110230114 101121120 120110112 011011021

.

XY=

131251 211120 151112 410221 321011 101101 421311

× 110131212 201012001 110230114 101121120 120110112 011011021

201012001 312312312 110230114 111111111 101121120 511511511 2010120 102102102 1102301 210210210 1011211 011011011

01 14 20 201012001 213110221330121314 101121120 +

511201511015112 200110201102021 121212 211201211012112 110110111101121 010101 120110112 111111 011011021 =

110131212 220262424 110131212 4404124848 330393636 110131212 4404124848

+

9154108357 412363235 121631011239 403254241 512254116 211351234 8155117476 +

51115615711 240220224 142132154 251231245 131121133 011011021 131121133 =

15136101910101320 87271458713 1468516145915 10948209121214 97361688715 112231043 1386102512131417

. WenowillustrateminorproductmomentoftypeIVrowvector

Example 1.1.27: Let X=

111101 212212 011311 101132 510213 110124 211502

.

=

3522624 59341137 2321112 2412513 6111526611 2311623 47231136

+ 1231215 25755410 371067515 15610575 25755410 1457455 51015510525

1212342 2424684 1212342 2424684 36369126 484812168 2424684 =

+

596511711 9181213221521 612139111019 513916161612 11221116402227 7151016222316 11211912271635

. TheminorproductmomentoftypeIVcolumn vectorisillustratedforthesameXjustgivenincaseofrow product.

Example 1.1.28: Let X=

111101 212212 011311 101132 510213 110124 211502

. XtX=

1201512 1110111 1211001 1231215 0113120 1212342

111101 212212 011311 101132 510213 110124 211502 =

12 11 12111101 12212212 01 12 +

0151 1011011311 1100101132 3121510213 1312110124 1234

+

2 1 1 5 0 2

[211|50|2]

=

121212 111101 111111 212212 121212 121111210121 012120121012 111101 121212 212212

+ 011311 015101510151 101132 101110111011 510213 110011001100 110124 01131 312110131211331 1312510131221 11012

1 212 13123 4 011311 101132 123412341234 510213 110124 +

222 121115012 111 555 211502 000 221125022

=

535525 323313 535525 535525 212212 535525

+

2761121021 631648 112443 1264151015 1044101518 2183151830

+ 4221004 211502 211502 105525010 000000 4221004 =

36118271230 116514513 85814610 271414451230 1256121620 301310302039

. NowweproceedontoillustratethemajorproductoftypeIV vectors Example 1.1.29: Let X=

121123 312311 113111 231201 342010 424100 501111

Theproductof2

Theproductof3rdrowofXwiththe3rdcolumnofY.

Theproductof3rdrowofXwith1stcolumnofY 1121 1024 0103 501111 1100 1011 0101

= 1100 1024 51121011111011 0103 0101 =[55105]+[0103]+[2212] =[781110].

Theproductof3rdrowofXwith2ndcolumnofY. 21 31 10 501111 21 12 11

= [5][21]+[01] 21 31 11112 10 11 =[105]+[10]+[44] =[159].

6681716112068 88915191021148 4651612614107 76101819822106 85152621929128 79122420725167 7811101592094

XY= 79

Onsimilarlineswecanfindthetransposeofmajorproductof TypeIVvectors.

Nowweproceedontojustshowthemajorproductmoment ofatypeIVvector.

Example 1.1.30: Suppose X=

121321 231212 142322 413211 232323 341142 212213 and Xt =

1214232 2341341 1123212 3232312 2121241 1221323

.

Productof1strowofXtwith2ndcolumnofX. 21 31 42 121423213 32 41 12 =

42 2113 121423212 3132 41 =[83]+[2621]+[24] =[3628].

Theproductof1strowofXtwith3rdcolumnofX. 321 212 322 1214232211 323 142 213 =

322 321211 1214232213 212323 142

=[745]+[202218]+[426] =[312829].

Theproductof2ndrowofXtwith1stcolumnofX. 1 2 1 2341341 4 1123212 2 3 2 =

1 231413441 2 112232122 3 = 826236 321428 .

Theproductof2ndrowofXtwith2ndcolumnofX. 21 31 42 2341341 13 1123212 32 41 12

Theproductof3rdrowofXtwith1stcolumnofX. 1 2 32323121 21212414 12213232 3 2 = 720431 422228 518629 .

Theproductof3rdrowofXtwith2ndcolumnofX. 21 31 323231242 212124113 122132332 41 12 = 125271924 73311512 83261536 = 4128 3920 3724 .

inthedefinitionoflinearbicodesandlinearn-codes respectively.

Inthissectionwerecallthenotionofbimatrixandillustrate themwithexamplesanddefinesomeofbasicoperationson them.

DEFINITION 1.2.1: A bimatrix AB is defined as the union of two rectangular array of numbers A1 and A2 arranged into rows and columns. It is written as follows AB = A1 A2 where A1 A2 with A1 =

aaa aaa aaa and A2 =

111 11121 111 21222 111 12

n n mmmn

222 11121 222 21222 222 12

aaa aaa aaa ‘ ’ is just the notational convenience (symbol) only.

n n mmmn

Theabovearrayiscalledambynbimatrix(writtenas B(mn)sinceeachofAi(i=1,2)hasmrowsandncolumns). Itistobenotedabimatrixhasnonumericalvalueassociated withit.Itisonlyaconvenientwayofrepresentingapairof arrayofnumbers.

Note: IfA1=A2thenAB=A1A2isnotabimatrix.A bimatrixABisdenotedby12 ijij aa.IfbothA1andA2arem nmatricesthenthebimatrixABiscalledthemnrectangular bimatrix.

Butwemakeanassumptionthezerobimatrixisaunionof twozeromatricesevenifA1andA2areoneandthesame;i.e., A1=A2=(0).

Example 1.2.1: Thefollowingarebimatrices

i.AB= 301021 121110 isarectangular23bimatrix. ii.B 30 A'11 20 isacolumnbimatrix. iii.A"B=(3,–2,0,1,1)(1,1,–1,1,2) isarowbimatrix.

InabimatrixAB=A1A2ifbothA1andA2aremn rectangularmatricesthenthebimatrixABiscalledthe rectangularmnbimatrix.

DEFINITION 1.2.2: Let AB = A1 A2 be a bimatrix. If both A1 and A2 are square matrices then AB is called the square bimatrix.

If one of the matrices in the bimatrix AB = A1 A2 is a square matrix and other is a rectangular matrix or if both A1 and A2 are rectangular matrices say m1 n1 and m2 n2 with m1 m2 or n1 n2 then we say AB is a mixed bimatrix.

Thefollowingareexamplesofasquarebimatrixandthemixed bimatrix.

Example 1.2.2: Given AB=

301411 211210 110001 isa33squarebimatrix. A'B=

11002001 20011010 00030103 10123200 isa44squarebimatrix.

Example 1.2.3: Let AB=

3012 112 0011 021 2100 004 1010 thenABisamixedsquarebimatrix. Let B

2011 20 A'0101 43 1021 , A'Bisamixedbimatrix. Nowweproceedontogivetheoperationsonbimatrices.

LetAB=A1A2andCB=C1C2betwobimatriceswe sayABandCBareequalwrittenasAB=CBifandonlyifA1and C1areidenticalandA2andC2areidenticali.e.,A1=C1andA2 =C2.

IfAB=A1A2andCB=C1C2,wesayABisnotequal toCB,wewriteABCBifandonlyifA1C1orA2C2.

Example 1.2.4: Let AB= 320012 211001 and CB= 111201 000102 clearlyABCB.Let AB= 001042 112300 CB= 001000 112101 clearlyABCB. IfAB=CBthenwehaveCB=AB.

Wenowproceedontodefinemultiplicationbyascalar. GivenabimatrixAB=A1B1andascalar,theproductof andABwrittenABisdefinedtobe AB= 111n m1mn

aa aa

bb bb

111n m1mn

eachelementofA1andB1aremultipliedby.Theproduct ABisthenanotherbimatrixhavingmrowsandncolumnsifAB hasmrowsandncolumns. Wewrite AB=ijij ab =ijij ab =AB. Example 1.2.5: Let AB= 201011 331210 and=3then 3AB= 603033 993630 If=–2for AB=[312–4][01–10], AB=[–6–2–48][0–220].

LetAB=A1B1andCB=A2B2beanytwomn bimatrices.ThesumDBofthebimatricesABandCBisdefined asDB=AB+CB=[A1B1]+[A2B2]=(A1+A2)[B2+ B2];whereA1+A2andB1+B2aretheusualadditionof matricesi.e.,if AB= 11 ijijab and CB= 22 ijijab then AB+CB=DB= 1212 ijijijij aabbij. Ifwewriteindetail

AB=

aabb aabb CB=

1111 111n111n 1111 m1mnm1mn

aabb aabb

2222 111n111n 2222 m1mnm1mn

AB+CB= 12121212 11111n1n11111n1n 12121212 m1m1mnmnm1m1mnmn

aaaabbbb aaaabbbb .

Theexpressionisabbreviatedto

DB=AB+CB =(A1B1)+(A2B2) =(A1+A2)(B1+B2).

Thustwobimatricesareaddedbyaddingthecorresponding elementsonlywhencompatibilityofusualmatrixaddition exists.

Note:IfAB=A1A2beabimatrixwecallA1andA2asthe componentsofABorcomponentmatricesofthebimatrixAB.

Example 1.2.6: (i)Let AB= 311401 102012 and

CB= 101331 221021, then, DB=AB+CB = 311101 102221 401331 012021 = 212730 121033 . (ii)Let AB=(32–101)(0110–1) and CB=(11111)(5–1203), AB+CB=(43012)(50302). Example 1.2.7: Let AB=

6131 2202 1113 and CB=

2414 4121 3031 .

AB+AB=

12262 4404 2226 =2AB CB+CB=

4828 8242 6062 =2CB.

Similarlywecanadd

18393 6606 3339 . Note: Additionofbimatricesaredefinedifandonlyifboththe bimatricesaremnbimatrices. Let AB= 301111 120021 and CB=

AB+AB+AB=3AB=

3111 2121 0030 .

TheadditionofABwithCBisnotdefinedforABisa23 bimatrixwhereasCBisa32bimatrix.

ClearlyAB+CB=CB+ABwhenbothABandCBaremn matrices. AlsoifAB,CB,DBbeanythreemnbimatricesthenAB+ (CB+DB)=(AB+CB)+DB.

Subtractionisdefinedintermsofoperationsalreadyconsidered forif

AB=A1A2 and BB=B1B2 then AB–BB=AB+(–BB) =(A1A2)+(–B1–B2) =(A1–B1)(A2–B2) =[A1+(–B1)][A2+(–B2)]. Example 1.2.8: i.Let AB=

3152 1211 0332 and BB=

8192 4229 1311 AB–BB=AB+(–BB). = 31528192 12114229 03321311 = 31815292 12421129 03133211

= 5244 5018 1043 .

ii.Let

AB=(1,2,3,–1,2,1)(3,–1,2,0,3,1) and BB=(–1,1,1,1,1,0)(2,0,–2,0,3,0) then AB+(–BB)=(2,1,2,–2,1,1)(1,–1,4,0,0,1).

Nowwehavedefinedadditionandsubtractionofbimatrices. Unlikeinmatriceswecannotsayifweaddtwobimatricesthe sumwillbeabimatrix.

Nowweproceedontodefinethenotionofn-matrices.

DEFINITION 1.2.3: A n matrix A is defined to be the union of n rectangular array of numbers A1, …, An arranged into rows and columns. It is written as A= A1 … An where Ai Aj with Ai =

aaa aaa aaa i = 1, 2, …, n.

11121 21222 12

iii p iii p iii mmmp

''isjustthenotionalconvenience(symbol)only(n3).

Note:Ifn=2wegetthebimatrix.

Example 1.2.9: Let A= 31012110 00110110 10015102 01017103

Aisa4-matrix.

Example 1.2.10: Let A=A1A2A3A4A5 =[100]

1 2 1 0 0

312 011 978 2135 0102

798110 12097 05718 46601

; Aisa5-matrix.InfactAisamixed5-matrix.

Example 1.2.11: Considerthe7-matrix A= 201111 110104

SUPERBIMATRICES AND THEIR PROPERTIES

Inthischapterweintroducethenotionofsuperbimatricesand explainsomeofitsproperties.Wealsogivethetypeofproducts definedonthem.Alsothenotionofsemisuperbimatricesand symmetricsemisuperbimatricesareintroduced.

DEFINITION 2.1: Let A1 and A2 be any two supermatrices, we call A = A1 A2 to be a superbimatrix; ‘ ’ is just the symbol.

Note: FurtherifA1=A2,asnonpartitionedmatricesthenthey musthavedistinctpartitions.IfA1=A2andtheyhavethesame setofpartitionsthenwedon’tcallA=A1A2tobea superbimatrix.

Wefirstillustratethisbythefollowingexamples.

Example 2.1: Let A1=

3102 1160 0101

and A2=

11010 20201 52105 11012 beanytwosupermatricesA=A1A2isasuperbimatrix. A= 3102 1160 0101

11010 20201 52105 11012 isasuperbimatrix.

Example 2.2: Let A1=

3012 0103 1152 0021 and A2=

3012 0103 1152 0021

betwosupermatrices.A=A1A2isasuperbimatrix.Wesee clearlyA1andA2areidenticalbutonlythepartitiononA1and A2isdifferent.HenceAisasuperbimatrix.

Example 2.3: Let A1=

301 211 520

and A2=

301 211 520 =A1.

ClearlyA=A1A2isnotasuperbimatrix.

Example 2.4: LetA=A1A2beasuperbimatrixwhere A1=[3012|112|15] and A2=

21 05 11 01 32 10

.

Example 2.5: LetA=A1A2whereA1=[3101|50231] andA2=[301|220531|011].Aisasuperbimatrixin whichweseebothA1andA2arerowsupermatrices.

Example 2.6: LetA=A1A2where A1=

3 0 2 1 10 1 5 4

1 2 3 4 5 6 7 aretwosupercolumnmatrices.ThenAisasuperbimatrix.

andA2=

Example 2.7: LetA=A1A2whereA1=[31|205]and A2=

1 1 0 5 1

betwosupermatrices.ClearlyA=A1A2isnotasuper columnbimatrixorasuperrowbimatrix.

Nowwehaveseenseveralexamplesofsuperbimatricesandwe seeeachofthemisofaspecifictype,sonowweproceedonto definethem.

DEFINITION 2.2: Let A = A1 A2 where A1 = [a11 | a12 |… | a1n] and A2 = [a21 a22 a23 | a24 … a2m] are both distinct super row matrices. Then we define A = A1 A2 to be a super row bimatrix.

Thesuperbimatrixgiveninexample2.5isasuperrowbimatrix. Nowweproceedontodefinethenotionofsupercolumn bimatrix.

DEFINITION 2.3: Let A = A1 A2 where both A1 and A2 distinct column supermatrices,

a a a a a a and

A1 =

1 1 1 2 1 3 1 4 1 5 1 m

A2 =

a a a a

2 1 2 2 2 3 2 n

.

A is a superbimatrix which we define as the super column bimatrix or a column superbimatrix.

Thesuperbimatrixgiveninexample2.6isasupercolumn bimatrix.

Nowweproceedontodefinesquaresuperbimatrix.

DEFINITION 2.4: Let A = A1 A2 be a superbimatrix. If both A1 and A2 are distinct m m square supermatrices then we call A = A1 A2 to be a square superbimatrix.

Thesuperbimatrixgiveninexample2.2isasquare superbimatrixortobemorespecificAisa44issquare superbimatrix.

Note: IfinthesquaresupermatrixA=A1A2ifwehaveA1to beammsquaresupermatrixandA2tobeannsquare supermatrix(mn)thenwecallA=A1A2tobeamixed squaresuperbimatrix.

Wenowillustratethisbyasimpleexample.

Example 2.8: LetA=A1A2where A1=

and A2=

3102 1011 2100 5112 ;

bothA1andA2aresquaresupermatricesofdifferentorder.Thus A=A1A2isamixedsquaresuperbimatrix.

Nowweproceedontodefinerectangularsuperbimatrixand mixedrectangularsuperbimatrix.

DEFINITION 2.5: Let A = A1 A2 if both A1 and A2 are distinct m n rectangular superbimatrices then we define A = A1 A2 to be a rectangular superbimatrix. If A1 is a m1 n1 rectangular supermatrix and A2 is a m2 n2 rectangular supermatrix with m1 m2 (or n1 n2) then we call A = A1 A2 to be a mixed rectangular superbimatrix.

Theexample2.1isasuperbimatrixwhichisamixed rectangularsuperbimatrix.

Nowweproceedontogiveanexampleofarectangular superbimatrix.

2.9: LetA=A1A2where

beanytworectangularsupermatricesofsameorderi.e.,bothA1 andA2are46supermatrices.WedefineA=A1A2tobea rectangularsuperbimatrixof46order. Nowlastlyweproceedontodefinethenotionofmixed superbimatrix.

DEFINITION 2.6: Let A = A1 A2 where A1 and A2 are supermatrices if A1 is a square supermatrix and A2 is a rectangular supermatrix then we define A = A1 A2 to be a mixed superbimatrix.

Thesuperbimatrixgiveninalltheexamplesisnotamixed superbimatrix.Sonowweproceedontogiveanexampleofthe same.

Example 2.10: LetA=A1A2where A1=

31025 10112 21051 51321 32003 and A2=

401112 210215 313010 125210

aretwosupermatriceswhereA1isa55squaresupermatrix andA2isa46rectangularsupermatrix.ThenA=A1A2is amixedsuperbimatrix.

Nowhavingseenexamplesanddefinitionsofseveraltypesof superbimatricesnowweproceedontodefineoperationsonthem andtheconditionsunderwhichoperationsaredefinedonthem.

Wefirstillustratebysomeexamplesbeforeweabstractly definethoseconcepts.

Example 2.11: LetA=A1A2andB=B1B2betwo superbimatrices.Suppose

A=A1A2=[312|0151] 301 211 and B=B1B2=[0-10|10–15] 001 205 .

ThenwecandefinebiadditionofthesuperbimatricesAandB.

A+B=(A1A2)+(B1B2) =(A1+B1)(A2+B2) =[312|0151]+[0–10|10–15] 301 211 + 001 205 . =[302|1146] 302 016 .

WeseebothAandBaremixedsuperbimatricesandA+Bis alsoamixedsuperbimatrixofthesametype.

Nowwegiveyetanotherexample.

Example 2.12: LetAandBbeanytwosuperbimatriceswhere

A=A1A2=[311|2] 01 52 and B=B1B2=[3|112] 01 52 .

ClearlyA+BcannotbedefinedforA1andB1thoughare thesupermatricestheyenjoydifferentpartitions.Similarlywe seeB2andA2aresupermatricesyetonthemaredefined differentpartitionssoadditionofthemcannotbedefined.Thus weseeunlikematricesofsameordercanbeadded;incaseof supermatricesforadditiontobecompatibleweneedthe matricesshouldbeofsameorderandalsotheyshouldhavethe samepartitiondefinedonthem.

Nowweproceedontodefineadditionofsuperbimatrices.

DEFINITION 2.7: Let A = A1 A2 and B = B1 B2 be any two superbimatrices. For their addition A + B to be defined we demand the following conditions to be satisfied.

1.A1 and B1 should be supermatrices of same order and the partition on A1 and B1 must be the same or identical then alone A1 + B1 is defined.

2.A2 and B2 should be supermatrices of the same order and the partitions on A2 and B2 must be identical then alone the sum of A2 and B2 can be defined.

Thus when A = A1 A2 and B = B1 B2 the sum of A and B is defined to be A + B = (A1 A2) + (B1 B2) = (A1 + B1) (A2 + B2).

Thusonlywhenalltheaboveconditionaresatisfiedwehave thesumoradditionoftwosuperbimatricestobedefinedand existing.

Note: IfA=A1A2isasuperbimatrixthenwehave

A+A=(A1A2)+(A1A2) =(A1+A1)(A2+A2) isalwaysdefinedand A+A=(A1+A1)(A2+A2) =2A12A2.

Thus ntimes AA=nA=nA1nA2.

Example 2.13: Let A= 32 10 05

11302 30521 11325 =A1A2 beamixedrectangularsuperbimatrix. A+A=(A1A2)+(A1A2) =(A1+A1)(A2+A2) = 32 10 05 + 32 10 05 11302 30521 11325 + 11302 30521 11325 = 6422604 20601042 010226410 =2A12A2. Thus8A=8A18A2 = 24168824016 8024040168 04088241640 .

Nowadditionofmixedsquaresuperbimatrices,mixed columnsuperbimatricesetc.,canbedefinedinasimilarway, providedtheyenjoythesameorderandidenticalpartition.

Example 2.14: Let A=A1A2 =

3512 0051 2012 0101

31005 10112 30110 and B=B1B2 =

5312 1111 0051 1203

01112 11103 20101 betwomixedsuperbimatrices. A+B=(A1A2)+(B1B2) =(A1+B1)(A2+B2) =

3512 0051 2012 0101

+

5312 1111 0051 1203 31005 10112 30110 + 01112 11103 20101

NowwehavetodefinethetransposeofasuperbimatrixA= A1A2.

DEFINITION 2.8: Let A = A1 A2 be any superbimatrix. The transpose of the superbimatrix A denoted by AT is defined to be AT = (A1 A2)T = 12 ATT A . Clearly the transpose of a superbimatrix is again a superbimatrix. If A is a mixed rectangular superbimatrix then its transpose, AT will also be a mixed rectangular superbimatrix.

Ithasbecomepertinenttomentionherethathowevera columnsuperbimatrixtransposewouldbearowsuperbimatrix andarowsuperbimatrixtransposewouldbeacolumn superbimatrix.

Nowweproceedontoillustratethemwithexamples.

Example 2.15: LetA=A1A2=[3011|–15231][101 |520|11102]bearowsuperbimatrix.ThetransposeofA denotedby AT =

3 0 1 1 1 5 2 3 1

1 0 1 5 2 0 1 1 1 0 2

= TT 12 AA.

ClearlyATisacolumnsuperbimatrix.

Example 2.16: Let B=B1B2=

1 2 1 0 1 1 3 5 1 2 6 beacolumnsuperbimatrix.ThetransposeofBwhichisBT=[3 111|22|11|0][1|2–1011|35126]=TT 12 BB. ClearlyBTisarowsuperbimatrix.

3 1 1 1 2 2 1 1 0

Example 2.17: LetC=C1C2beanysuperbimatrixwhere C1=

31021 62105 10116 and C2=

310 116 215 101 232 105

Cisamixedrectangularsuperbimatrix. CT=(C1C2)T = TT 12CC =

361 120 011 201 156

312121 111030 065125 .

CTisalsoamixedrectangularsuperbimatrix. Example 2.18: LetD=D1D2whereDisamixedsquare superbimatrixwith D1=

311 052 101 and D2=

3012 1201 0110 1301

. Now DT=(D1D2)T = TT 12DD = 301 150 121

3101 0213 1010 2101

WeseeDTisalsoasquaremixedsuperbimatrix.

.

Nowweareinterestedtoknowwhattodefineorcallthe followingtypeofbimatrices.

Example 2.19: LetA=A1A2where A1=

31256 02010 11531 and A2=

3110 1011 1100 0221 whereA1isjusta35matrixandA2isasquaresupermatrix. SinceA1isnotasupermatrixwecannotdefineAtobea superbimatrixsinceA2isasupermatrixwecannotdefineAto beabimatrix.

Sowedefineanewnotioncalledsemisuperbimatrixin suchcases.

DEFINITION 2.9: Let A = A1 A2 where A1 is just a simple matrix and A2 is a supermatrix then we define A = A1 A2 to be a semi superbimatrix.

Example 2.20: Let A=A1A2= 3112 0510

3120 5111 2026 1015 whereA1isjusta24matrixandA2isasquaresupermatrix, Aisdefinedasthesemisuperbimatrix.

Example 2.21: LetA=A1A2whereA1=[323|1005]and A2=[110111].WeseeAisasemisuperbimatrixcalledasa rowsemisuperbimatrix.

Example 2.22: LetA=A1A2where A1=

1 0 1 2 2 1 6 and A2=

3 1 4 5

.

Aisasemisuperbimatrixwhichwedefineasacolumnsemi superbimatrix.

Example 2.23: LetA=A1A2where A1= 321353 100121 and A2=

12353701 76351211 10101523 .

Aisamixedrectangularsemisuperbimatrix.

Example 2.24: LetA=A1A2where A1=

3120 1010 3121 5011 and A2=

312 001 101

Aisamixedsquaresemisuperbimatrix.

Example 2.25: LetA=A1A2where A1=

3010 1111 2101 5110 and A2=

812 011 101 110 100 010 218

AisamixedsemisuperbimatrixforA1isjustasquarematrix andA2isasuperrectangular73matrix.Thuswecandefine7 typesofsemisuperbimatricesviz.rowsemisuperbimatrix, columnsemisuperbimatrix,nnsquaresemisuperbimatrix,m

nrectangularsemisuperbimatrix,mixedsquaresemi superbimatrix,mixedrectangularsemisuperbimatrixandmixed semisuperbimatrix.

Example 2.26: Let A= 311 101 012 345 =A1A2.

Aisa23rectangularsemisuperbimatrix.

Example 2.27: Let A=

1031 1011 1052 1203

0123 1234 5678 4321 =A1A2.

Aisa44squaresemisuperbimatrix.

Weseeasincaseofsuperbimatricesthetransposeofasemi superbimatrixisalsoasemisuperbimatrix.Thetransposeofa rowsemisuperbimatrixisacolumnsemisuperbimatrixandthe transposeofacolumnsemisuperbimatrixisarowsemi superbimatrix.

Example 2.28: Let A=A1A2 =[3100035][1123|554|412]

bearowsemisuperbimatrix. AT=(A1A2)T

TT 12 AA=

3 1 0 0 0 3 5

isacolumnsemisuperbimatrix.

1 1 2 3 5 5 4 4 1 2

Example 2.29: LetA=A1A2where A1=

3 1 1 1 0 2

andA2=

1 2 3 4 5 6 7

AT=(A1A2)T = TT 12AA =[311102][123|4567].

AisclearlyacolumnsemisuperbimatrixbutATisarowsemi superbimatrix.

Example 2.30: LetA=A1A2beamixedsquaresemi superbimatrixwhere A1=

301 111 010

and A2=

2105 1128 0213 5830

. AT=(A1A2)T = TT 12AA = 110 011 3102105 1128 0213 5830

. ClearlyATisalsoamixedsquaresemisuperbimatrix. Example 2.31: LetA=A1A2beamixedrectangularsemi superbimatrixwhere A1=

312 011 110 220 210 103 and A2= 1358130 2169211 . ThetransposeofA AT=(A1A2)T = TT 12AA where T 1 A= 301221 111210 210003

ATisalsoamixedrectangularsemisuperbimatrix.

Example 2.32: LetA=A1A2bea73rectangularsemi superbimatrixwhere A1=

101 011 100 001 010 111 578 and A2=

212 101 110 115 235 123 100

. AT = TT 12 AA= 1010015 0100117 1101018

2111211 1001320 2115530

isalsoa37rectangularsemisuperbimatrix. Asincaseofsuperbimatricesthesumoftwosemi superbimatricescanbeaddedi.e.,A=A1A2andB=B1B2 canbeaddedifandonlyifbothA1andB1andsameorder matricesandA2andB2aresameordersupermatriceswithsame oridenticalpartitiononit.

Example 2.33: LetA=A1A2andB=B1B2beanytwo semisuperbimatrices.Here A1=

012 345 678 and A2=

31021 11502 13101 and B1=

010 111 210 and B2=

11010 01235 67891 .

A+B=(A1A2)+(B1B2) =(A1+B1)(A2+B2)

= 012 345 678 + 010 111 210 31021 11502 13101 + 11010 01235 67891 = 022 456 888

42031 12737 510992 . A+Bisalsoasemisuperbimatrixofthesametype. Supposewehave A=A1A2=

012 011 110

311110 420101 101012 tobeamixedsuperbimatrixthenwehavealwaysA+Ais definedandis2A.InfactA+A+…+A,n-timesinnA.We see A+A= 012 011 110 + 012 011 110 311110 420101 101012 + 311110 420101 101012

99

NownA=nA1nA2foranyn>1.Also

A–A=(A1A2)–(A1A2) =(A1–A1)(A2–A2) = 000 000 000

000000 000000 000000 .

ThuswegetthedifferenceofA–Atobeazerosuperbimatrix. Nowif A=A1A2 = 123 010 101

32105231 12011311 beamixedsemisuperbimatrix.Then A+A=(A1A2)+(A1A2) =(A1+A1)(A2+A2) = 123 010 101 + 123 010 101 32105231 12011311 +

32105231 12011311 = 246 020 202

642010462 24022622 =2A12A2=2A.

ThusweseesumofasemisuperbimatrixAwithitselfis2A. OnsimilarlineswecansayifAisasemisuperbimatrixthenA +A+…+A,n-timesisnA=nA1nA2.

Havingdefinedtransposeandsumofsemisuperbimatrix wheneveritisdefinedweproceedontodefinesomeproductof thesesuperbimatricesandsemisuperbimatrices.

Asincaseofsupermatricesweincaseofsuperbimatrices andsemisuperbimatricesfirstdefinethenotionoftheproduct ofasuperbimatrixwithitstranspose.Wealsoforthisneedthe simpledefinitionofsymmetricsuperbimatrices,symmetricsemi superbimatrices,quasisymmetricsuperbimatricesandquasi symmetricsemisuperbimatrices.

DEFINITION 2.10: Let A = A1 A2 be a superbimatrix. We call A to be a symmetric superbimatrix if both A1 and A2 are symmetric supermatrices.

Example 2.34: LetA=A1A2beasuperbimatrixwhere A1=

and

A2=

6120 1314 2151 0417

,

sincebothA1andA2aresymmetricsupermatricesweseeA= A1A2isasymmetricsuperbimatrix.

DEFINITION 2.11: Let A = A1 A2 be a superbimatrix, A is said to be quasi symmetric superbimatrix if and only if one of A1 or A2 is a symmetric supermatrix.

Thefollowingresultisobviouseverysymmetricsuperbimatrix istriviallyaquasisymmetricsuperbimatrix.Howeveraquasi symmetricsuperbimatrixisneverasymmetricsuperbimatrix.

Example 2.35: LetA=A1A2where A1=

32105 21321 13512 02143 51232 and A2=

4123 1234 2341 3412

beasuperbimatrix.ClearlyA1isasymmetricsupermatrix whereasA2isonlyasquaresupermatrix.HenceAisonlya quasisymmetricsuperbimatrix.

DEFINITION 2.12: Let A = A1 A2 be a semi superbimatrix. If both A1 and A2 are symmetric matrices and A1 or A2 is a super

102

symmetric matrix then we all A to be a symmetric semi superbimatrix ‘or’ used in the mutually exclusive sense.

Example 2.36: LetA=A1A2where A1=

54123 41234 12412 23131 34212 and A2=

0101 1431 0321 1115

.

AisasymmetricsemisuperbimatrixasA1isasymmetric supermatrixandA2isasymmetricmatrix.

DEFINITION 2.13: Let A = A1 A2 be a semi superbimatrix. If only one of A1 or A2 is a symmetric matrix (or a symmetric supermatrix) then we call A to be a quasi symmetric semi superbimatrix.

Example 2.37: LetA=A1A2beasemisuperbimatrixwhere A1=

21305 15162 31712 06131 52214 and

A2=

312567 123456 234561 345612 456123 561234 isaquasisymmetricsemisuperbimatrix.

Example 2.38: LetA=A1A2where A1=

0101 1234 1010 4321 and A2=

12345 25432 34243 43431 52316

.

Aisasemisuperbimatrixwhichisalsoaquasisymmetricsemi superbimatrix.

Weshowlaterthebyproductwhichwedefineon superbimatriceswegetaclassofsymmetricbimatrices.

Wefirstillustratetheproductoftwosuperbimatrices.

Example 2.39: LetA=A1A2andB=B1B2beanytwo superbimatrices.Here

AB= 203014 111101 120110

01 30 10 11 20 01 3103301 4511011 3410101 1224256

103 311 512 110 011 100 010 = 20301104 01 1111011101 30 12011200 3103301110 103 4511011011 311 3410101100 512 1224256010 = 025004 312101 613100

Weseetheproductoftwosuperbimatricesisonlya bimatrix. Tothisendwedefinetwonewnotions.

DEFINITION 2.14: Let A = A1 A2 be a mixed rectangular m n superbimatrix with m < n. If in both A1 and A2 partition is only between the columns i.e., only vertical partition then we call A to be a row superbivector.

Example 2.41: LetA=A1A2beamixedrectangular superbimatrixwhere A1=

301213321 111310110 211410010 and A2=

3151011110 1172120011 0083232578 2194341234

;

WecallAtobearowsuperbivector.Itisclearthepartitionof thematricesareonlybytheverticallinesi.e.,betweenthe columnsandnopartitioniscarriedoutbetweentherows.

DEFINITION 2.15: Let A = A1 A2 be a mixed rectangular m n superbimatrix with m > n. If we have on both A1 and A2 only horizontal partitions then we call A to be a column superbivector. When we say horizontal partitions it takes place only between the rows.

Example 2.42: LetA=A1A2arectangularsuperbimatrix, where

A1=

1011 2120 0101 3125 1234 5678 0123 1105 2571 and A2=

110 111 012 345 678 901 011 101 110 123 456 isacolumnsuperbivector. Nowweproceedontodefinethenotionofproductofthese typeofsuperbivectors.

DEFINITION 2.16: Let A = A1 A2 a column superbivector and B = B1 B2 be a row superbivector. Now how to define the product BA. BA is defined if the following conditions are satisfied.

If A = A1 A2 where

111

A1 = 1

1 1 1 m

A A and A2 = 2

A A and in B = B1 B2 we have B1 = 1

2 1 2 m

11 1 m B B and B2 = 2

22 1 m B B then first B.A = (B1 B2) (A1 A2) = B1 A1 B2 A2 , B1A1 and B2A2 is the super vector product defined only if the number of columns in each of 1Bi , 1 i m1 is equal to the number of rows in each of 1 j A ; 1 j m1 and the number of columns in each of 2Bi , 1 i m2 is equal to the number of rows in each of 2 j A , 1 j m. Now BA = B1 A1 B2 A2 = 1

A A 2

11 1 m B B 1

1 1 1 m

A A = 11

22 1 m B B 2

2 1 2 m

2222 11 mm B ABA = C D; we see C and D are not super vectors they are only just matrices. Thus C D is only a bimatrix and not a superbimatrix.

1111 11 mm B ABA 22

Wehaveillustratedthistypeofproductinexample2.39and 2.40.Theproductdefinedusingrowsuperbivectorsandcolumn superbivectorsresultsonlyinbivectorsandthisproductwillbe knownastheminorproductofsuperbivectors.

Nowweproceedontodefinethenotionofmajorproductof thesesuperbivectors.Beforewedefinethemabstractlywegive someexamplesofthem.

AB=(A1A2)(B1B2) =A1B1A2B2 =

123 011125131 301001010 110112011 501 21 30 11231115131 14111210101 25 30

= 125131 123123123 001010 011011011 112011 301130125130131 110011001011010 501150112050111 212312111215131 301113021300101 231115131 111111 111210101 141414 231115131 252525 111210101 303030

=

4513144 113001 47173104 124141 611275166 5734310363 6933315393 342325232 675955535 911712710767 6933315393

=CDwherebothCandDaresuperbimatricesandnot superbivectors. Weillustratethesamewithonemoreexample. Example 2.44: LetA=A1A2andB=B1B2betwo superbivectorswhereA=A1A2isacolumnsuperbivector andB=B1B2isarowsuperbivectorwith

A2=

12 34 56 10 01 11 10 11

isacolumnsuperbivector.NowB=B1B2where B1=

10110512 21011011 31101012 and B2= 1011111111 1100110101

istherowsuperbivector.NowthemajorbyproductofABis definedasAB=(A1A2)(B1B2)=A1B1A2B2 =

201 111 370 110 201 351 111 010 101 001

10110512 21011011 31101012

=

513211036 62222535 1773107151013 31121523 513211036 16648615913 62222535 21011011 41211524 31101012 3211331313 7433773737 116551111511511 1011111111 1100110101 2111221212 1011111111 2111221212 =C1C2.Thustheresultantofacolumnsuperbivectorwitha rowsuperbivectorisasuperbimatrix.

Nowweproceedontodefinethemajorbyproductof superbivectors.

DEFINITION 2.17: Let A = A1 A2 be a column superbivector and B = B1 B2 be a row superbivector. The major byproduct of these two superbivectors is defined to be AB where AB = (A1 A2) (B1 B2) = A1 B1 A2 B2 is compatible if and only if the number of columns in A1 must be equal to the number of rows in B1 and the number of columns in A2 must be equal to the number of rows in B2 respectively. Then the resultant bimatrix is

118

always a superbimatrix. Thus the major byproduct yields a superbimatrix which is neither a row superbivector nor a column superbivector.

Nowtheimmediateapplicationofthesemajorbyproductof superbivectorsistheproductofasuperbivectorwithits transpose.Theexample2.43and2.44areillustrationsofmajor byproductofsuperbivectors.

Nowweproceedontogiveanexampleofasuperbivector withitstranspose.

Example 2.45: LetA=A1A2beacolumnsuperbivector.Let ATbeitstranspose.ThebyproductAATgivesasuperbimatrix whichisneitheracolumnsuperbivectornorarow superbivector.GivenA=A1A2where A1=

310 116 011 210 123 101 010 101 and A2=

2104 1160 0011 1011 0523 1101 2021

. Now AT=(A1A2)T = TT 12AA

T 1 A= 31021101 11112010 06103101 and T A2=

2101012 1100510 0611202 4011311 AAT=(A1A2)(A1A2)T =(A1A2)(TT 12 AA) =A1 T 1 AA2 T 2 A. =

310 116 011 210 123 101 010 101

31021101 11112010 06103101 2104 1160 0011 1011 0523 1101 2021

2101012 1100510 0611202 4011311

=

31031031023101101 11611111611162010 01106101100113101 31021101 21011121012102010 06103101 123 101 010 101

123123 31021101 101101 11112010 010010 06103101 101101 2101012 2104112104005121040 1160061160112011602 4011311 00112100110101 10111110110051 05230605231120 11014011011131

00112 10110 05232 11011 2101012 1100510 202120212021 0611202 4011311

121

=

102175313 2387119515 17215111 71154212 5195414424 35124202 11112010 35124202 213461778 3386717214 4622513 6723525 1717553887 7212833 81435739

=S1S2=SweseebothS1andS2aresymmetric supermatricesthusAAT=Sisasymmetricsuperbimatrix.Thus thisproductofacolumnsuperbivectorwithitstranspose

101 234 103 110 511 200 311 100

1234 1011 3610 1101 2135 0012 3121 1101

.

1234 1011 3610 1101 2135 0012 3121 1101

.

11312031 20611011 31103120 41015211

12115231 03011010 14301010 101 234 103 110 511 200 311 100

= 12115231 101101101 03011010 234234234 14301010 12115231 10303103010311010 14301010 . 110 51112 20003 31114 100

110110 115231511511 011010200200 301010311311 100100

12341131123420312341 10112061101110110111 36103110361031236100 11014101110152111011 1131 2135 2061 0012 3110 3121 4101

2031 21352135 1011 00120012 3120 31213121 5211 11312031 20611011 110111011101 31103120 41015211 =

26416241 629145174132 4141018261 15126241 614862710175 242210462 41364176113 12115231 3081873311157 834210362 184469151179 72938253 33101583913188 1131213542 156175184155 72938253

125

=S1S2 =ATA=S.

ThisSisasymmetricsuperbimatrix.

THEOREM 2.1: Let A = A1 A2 be a column superbivector. Then AAT is a symmetric superbimatrix. Proofisleftasanexerciseforthereader. Hint:Let A= 1

1 1 1 2 1 n

A A A

2 1 2 2 2 n

A A A 2

=A1A2 bethegivencolumnsuperbivector.Now AT=(A1A2)T =A T 1A T 2 = 1

222 12n AAA AAT=(A1A2)(A1A2)T =(A1A2)(A T 1A T2) =A1A T 1A2A T 2 = 1

111 12n AAA 2

1 1 1 2 1 n

A A A 1

2 1 2 2 2 n

1T1T1T1T 123n AAAA 2

2T2T2T2T 123n AAAA

A A A 2

=

1 1 1111

1tT1tT1tT 11121n 1tT1tT1tT 22222n 1tT1tT1tT n1n2nn

AAAAAA AAAAAA AAAAAA

22T22T22T 11121n 22T22T22T 21222n 22T22T22T n1n2nn

AAAAAA AAAAAA AAAAAA 2 2 2222

.

Wesee 11TT 1i AA=11T i1 AA;i=1,2,…,n1.Also 22TT jk AA= 22T kj AA,1k,jn2.ThusweeasilyseetheproductAATgives asymmetricsuperbimatrix. Nowweproceedontodefinetheminorproductofsemi superbimatrix,tothisendfirstwedefinesemisuperbivector.

DEFINITION 2.18: Let A = A1 A2 be any semi supermatrix. We say A is a semi superbivector if the supermatrix A1 (or A2) is just partitioned only vertically or horizontally ‘or’ in the mutually exclusive sense. The other matrix may be a square matrix or a rectangular matrix or a column vector or row vector.

Example 2.47: LetA=A1A2where A1= 321051 120563 and A2=

31101311 11012130 00113251 .

127

Aisasemisuperbimatrix,becauseA2happenstobearow supervectorwecallAtobearowsemisuperbimatrix.

Note: EvenifA1isnotarectangularmatrixstillwecallAtobe onlyarowsemisuperbimatrix.

Example 2.48: LetB=B1B2where B1=

2011 1201 1102 1012 and B2=

3156 1021 7654 3210 1121 0725

;

thenwecallBtobeacolumnsemisuperbimatrixthoughB1is justa44squarematrix.

Nowwecalltheserowsemisuperbimatrixandcolumnsemi superbimatrixassemisuperbivectors,evenifthenonsuper componentisasquarematrixoracolumnvectororarow vectororarectangularmatrix.

Nowweillustratetheminorproductofsemisuperbivector.

Example 2.49: LetA=A1A2andB=B1B2betwosemi superbimatrices;where A1=

3012 1101 5013 and

A2=

31253143 10110111 01021002 . AistherowsemisuperbivectorandB=B1B2with B1=

31 06 11 20 and B2=

21 30 11 12 21 10 01 15 bethecolumnsemisuperbivector.Now AB=(A1A2)(B1B2) =A1B1A2B2 betheminorproductofsemisuperbivectors. A1B1A2B2=

3012 1101 5013

31 06 11 20

beasemisuperbivectorandB=B1B2with B1=

0 1 2 3 0 1 5 and B2=

0512 1202 1010 0101 1111 0101 1010 1111 0110 beacolumnsemisuperbivector.Now AB=(A1A2)(B1B2) =A1A2B1B2 =[1230156]

0 1 2 3 0 1 5

=[43]

42 530512 101202 01

+

353 1010 263 0101 114 1111 105 +

50710101 10111010 01011111 20100110

= 22441268687362 33151659591100 43 051255551120 120265651111 =[43]

129422 7391025 61187 6656

.

ClearlyABisonlyausualbimatrixandisnotasemi superbimatrix.Thusweseetheminorproductyieldsonlya bimatrix,thesemisuperqualityislostbythisproduct. Nowweproceedontodefinetheminorproductofsemi superbivectors.

DEFINITION 2.19: Let A = A1 A2 and B = B1 B2 be two semi superbivectors. The minor product of AB is defined if and only if in the product AB = (A1 A2) (B1 B2) = A1B1 A2B2, the usual matrix product A1B1 of the A1 and B1 is defined i.e., if A1 is a m n matrix then B1 must be a n t matrix and A2B2 is defined only if A2 is a row super vector say A2 = 2

222 12||| n AAA and B2 is a column super vector such that if B2 = 2

B B B then each of the product of matrices 22 ii AB is defined for i =1, 2, ..., n2 and all of them are of same order, i.e., A2B2 = 2

B B B

2 1 2 2 2 n

2 1 2 2 2 n

222 12||| n AAA 2

= 22

222222 1122 nn ABABAB since each 22Aii B is only a ordinary simple matrix and not a supermatrix we see A2B2 = 2 22 1

n ii i ABis just a single m2 n2 matrix. This us see AB = A1B1 A2B2 is only a bimatrix.

Nowwejustdefinemajorproductofsemisuperbivectors. Wefirstillustrateitbyexamplesbeforewegofortheabstract definition.

Example 2.51: LetA=A1A2andB=B1B2betwosemi superbivectorswhereA=A1A2with A1=

31057210 01012316 11100101 and A2=

3101 1110 1101 2120 1010 1231 1110 0110 1001

beacolumnsemisuperbivectorandB=B1B2with B1=

110 201 310 401 510 601 710 801 and

121430 310103101303101111 111011110111110230 201011 121 110211021102 030 212021202120 111 101010101010 201

430 111 230 011 123111231211231430 111001110301110111 011010110110110230 100121001011001011 1111 31010101 11101010 0011 1111 1102 0101 2120 1010 1010 0011 12311111 11100101 01101010 10010011 = 79118 89311 1803

594131123445 2627712222 5535631234 494131314343 2326602121 6114121434354 2627712222 1413411111 3224411122

.

Clearlythemajorproductyieldasemisuperbimatrixwhich isnotaroworcolumnsemisuperbivector. Nowwegiveyetanotherexamplebeforeweproceedto givetheabstractdefinition.

2.52: LetA=A1A2andB=B1B2beanytwo semisuperbivectorsgivenby

Abethecolumnsemisuperbivector. B=B1B2

11001111024 231231231 02110110101 120120120 20201011211 11001111 110110110 02110110 315315315 20201011 517517517

024 101 211 110110110 011011011 11001111 101101101024 02110110 01001001010 20201011 100100100 001001001 111111111

1 211 =

466 136 93 101

485335615512 15221331226 13111221125 1351118492111118 197151126132151728 13111221125 22311120312 31202120235 02110110101 11001111024 20201011211 33312230336 isasemisuperbimatrix.

Thismajorproductconvertsproductofsemisuperbivectorsin tosemisuperbimatrixwhereasminorproductmakesthe productofsemisuperbivectorsintojustabimatrix.

DEFINITION 2.20: Let A = A1 A2 and B = B1 B2 be two semi superbivectors. The major product of the two semi superbivectors A and B is AB, is defined as AB = (A1 A2) (B1 B2) = A1B1 A2B2 if

1.A1B1 must be compatible with respect to usual matrix product that is if A1is a m n matrix then B1 must be a n t matrix.

2.A2B2 is defined only if A2 is a super column vector and B2 is a super row vector such that the number of columns in A2 must be equal to the number of rows in B2

NowwefindtheproductofAwithATorATwithAwhichever iscompatiblewhereAisasemisuperbivector.

Example 2.53: LetA=A1A2beasemisuperbivectorgiven by

weseeATAisasymmetricsemisuperbimatrix.Thisproduct helpsonetoconstructanynumberofsymmetricsemi superbimatrices.

Example 2.54: LetA=A1A2becolumnsemisuperbivector. ThenwecanfindAAT.Given A=A1A2 =

3012 1101 5201 0110

0123 1101 5312 0101 3301 1200 0310 1010 1111 0101 0110 1001 tobeacolumnsemisuperbivector. Now AT=(A1A2)T = TT 12AA =

3150 0121 1001 2110

015031011001 113132301110 201000111010 312110001101

. Now AAT=(A1A2)(A1A2)T =(A1A2)(TT 12 AA) =A1 T 1 AA2 TA2 =

30123150 11010121 52011001 01102110

0123 1101 5312 0101 3301 1260 0310 1010 1111 0101 0110 1001 015031011001 113132301110 201006111010 312110001101 =

145171 5381 178302 1122

01230150012331 11011131110132 53122010531206 01013121010110 015031 33011131330132 12602010126006 312110 0310 1010 1111 0101 0110

0310 0150101031 1131111132 2010010106 3121011010 10011001 0123011001 1101301110 5312111010 0101001101 011001 3301301110 1260111010 001101 0310 1010011001 1111301110 0101111010 0110001101 1001

=

145171 5381 178302 1122 144114614526433 4310273313212 1110395261710611547 425242302211 67264199937434 1431729411279281 531039121014340 216037122011 6311279424222 425242302211 314138412120 327141012102

.

AATisasymmetricsemisuperbimatrix.Thusthismajor productwhenAismultipliedbyitstransposewhereAisonly semibivectoryieldsasymmetricsemisuperbimatrixwhichis notabivector.

Likeincaseofsuperbimatriceswecaneasilyprovethe followingtheorem.

THEOREM 2.2: (1) Let A = A1 A2 be a column semi superbivector then AAT is a symmetric semi superbimatrix. (2) If A = A1 A2 is a row semi superbivector then ATA is a symmetric semi superbimatrix.

Theexamples2.53and2.54substantiatetheabovetheorem. Nowweproceedontodefinetheproductofsuperbimatrices

146

whicharenotsuperbivectorsandsemisuperbimatriceswhich arenotsemisuperbivectors.

1B2beanytwo

=

031112 110211 211011 001101 124012 001101 111301 001103 110010

1011243 0110120 1001012 3101010 1110111 0101105 301101150 110110101 011001210 211101100 121011010 010111001 100010102 522102012

101112 101010 110101 012512 101110 100101 312112 011060 101211 =

000 111 1011243 222 000 000 1011243 111 111 1011243000 111

+

313131 1001101011020 1110110101112 010101 2401241012420 0110010100112 1111 01101 0101 100 1010

11 20 01 1012 10

+

= 112112112 3101010 211211211 1110111 011011011 0101105 100100100 3101010 012012012 1110111 101101101 0101105 3101010301301301 1110111103103103 0101105010010010 303030 101112 111111 101010 010101 212121 121011211122 011010101010 101010 101112 525252 101010

110101 110111011101 012512 011001100110 101110 100110011001 100101 11011101101 10110121011 01111010111 00101000010

1011011 5110112 1101110 1000101 110101 012512 210221022102 101110 100101 +

312112 150150150 011060 101101101 101211 210210210 100100100 31211 010010010 01106 001001001 10121 102102102

2 0 1 312112 011060 012012012 101211

=

0000000 1011243 2022486 0000000 0000000 1011243 1011243 0000000 1011243

+ 1331372 0110120 1111132 1001012 4224288 1001012 1111132 1001012 0110120

+ 44133211 7413236 1211216 3102110 13123111 3202115 9404135 34043115 1110111

303336 202122 101010 303234 303131 101010 101112 7075710

+ 222714 113622 210202 222714 311312 213723 101110 432916

+ 3671312 413323 635284 312112 011060 101211 514534 213482

=

57446913 8534599 434471214 4103122 55365919 52143610 1152641010 44053217 2231474 8812113512 7281067 946496 83710510 6254103 415944 716756 13412181618

.

Clearlytheresultantundertheminorproductoftwo superbimatricesisasuperbimatrix.

Thusweseetheminorproductoftwosuperbimatrices resultsinasuperbimatrix.Wehaveobservedfromtheexample 2.55thatonlyfortheproductABtobecompatibleweneedthe numberofcolumnsinAmustequalnumberofrowsofBbut alsothewaythecolumnsofAarepartitionedmustbeidentical withthewaytherowsofBarepartitioned.

Thenalonewehavetheproducttobedefined.Wegiveyet anotherexampleofaminorproductoftwosuperbimatrices beforeweproceedontodefinethem.

Example 2.56: LetA=A1A2andB=B1B2beanytwo superbimatrices.WewillfindtheminorproductAB.

GivenA=A1A2where

0123101 2110354 1011102 0100141 1102111 1015101 1210111 2021001 3100010 and

A1=

A2=

311110 100101 211110 032101 110121 520102

.

B=B1B2with

101215 021612 111101 253111 131101 421312 006213 and

B1=

B2=

11111 22011 34010 41101 06205 10103

.

AB=(A1A2)(B1B2) =A1B1A2B2 =

0123101 2110354 1011102 0100141 1102111 1015101 1210111 2021001 3100010

101215 021612 111101 253111 131101 421312 006213 311110 100101 211110 032101 110121 520102

11111 22011 34010 41101 06205 10103

=

012012 101215 211211 021612 101101 111101 010010 101215 110110 021612 101101 111101 121101121 202021202 310111310

+ 3131 0325303111 1113111101 0101 2125321111 5113151101 0101 253111 1010 131101 0000

215 612 101

+ 0101 5442154312 0200602213 4141 1142111312 0100601213 1111 421312 0101 006213 1010

156

3131 11111 1010 22011 2121 0303 11111 1111 22011 5252

+ 34010 11101110 41101 01010101 06205 11101110 10103 34010 21012101 41101 01210121 06205 01020102 10103 =

243814 33411313 212316 021612 122827 212316 25415310 4246212 32412417

+

71810434 393303 384212 131101 5137323 112816656 131101 253111 000000

+ 006213 20102923922 0012426 1681014511 427525 006213 427525 006213 421312 55344 11111 44233 66033 33122 99577

+

711316 51204 711316 119224 5136014 61307

=

9221914511 262236371238 59189414 17131221614 10171616615 13292411715 7101221516 67139416 74515519

12166510 62315 1115549 1715257 8167216 15108714

=S1S2=ABisonceagainasuperbimatrix.

DEFINITION 2.21: Let A = A1 A2 and B = B1 B2 be two superbimatrices. The minor byproduct of the two superbimatrices AB = (A1 A2) (B1 B2) = A1 B1 A2 B2 is defined if and only if the following conditions are satisfied.

1.The number of columns in Ai is equal to the number of rows in Bi; i = 1, 2.

2.The partition of Ai along the columns is equal or identical with the partition of Bi along the rows i = 1, 2.

WeseetheminorbyproductofABwhenitexistsforthe superbimatricesAandBisagainasuperbimatrix.Theexamples 2.55and2.56showexplicitlyhowthisminorbyproductofany twosuperbimatricesaredefinedresultinginasuperbimatrix. Nowweproceedontofirstillustratebyexampleshowthe productofthetransposeofasuperbimatrixwithasuperbimatrix isdefined.

Example 2.57: LetA=A1A2beasuperbimatrixwhere

A1=

101010 210212 021021 112120 110210 201201 512101 and A2=

1131014 5102010 7210101 8121101 2010221 3570160

. Now AT=(A1A2)T =(AT1AT2) =

1201125 0121101 1012012 0201221 1122100 0210011

157823 112105 301217 120100 001121 110026 401110

. Now

AAT=(A1A2)(A1A2)T =(A1A2)(AT1AT2) =A1AT1A2AT 2 =

101010 210212 021021 112120 110210 201201 512101

1201125 0121101 1012012 0201221 1122100 0210011 1131014 5102010 7210101 8121101 2010221 3570160

157823 112105 301217 120100 001121 110026 401110 =

111 1201125 222 000 111 1201125 111 222 120112 5555

+

0101012110011 1010101201102 212121 1201122110121 1010101201102 010101 0121101 121212 1012012

+ 0201221 010010010 1122100 212212212 0210011 021021021 0201221 120120120 1122100 210210210 0210011 201201201 0201221 1122100 101101101 0210011 111111 157823 515151 112105 727272 811578182813 201122010205 157823 353535 112105

+

1122100 1944564 2454211 2445421 1524542 0612453 0411232

269928 62637411020 93753581431 94158651629 210141646 8203129634

+ 10237321 240200 302338 7236415 303459 210815950

1714466 110026 401110 401110 6211512 66001236

+

164

.

= 3335237 3146781015 36108425 578116611 2846768 3102661015 71551181532 29916201135 93137431226 163756621839 204362722144 111218211427 3526394427120

Weseetheresultantisasymmetricsuperbimatrix.Wesee bytheminorbyproductofsuperbimatriceswiththeirrespective transposeswecangetmoreandmoresymmetric superbimatrices.

=

3113 1001 2102 5211 0300

31250 10123 10010 31210 81310 00125 10010 21212 51003 60110

801256 100110 310201 121101 050230 =

333 31250 111 222 31250 555 3125 0000

+ 11101112113 00100001000 10101012103 21102101210 10123 303030 10010

+

1.Number of rows of B1 is equal to number of columns in A1 (A1 and B1 are usual matrices).

020121 101001 210110 702121 31201150 60110012 11011001 20010101 32101020 40210101 . B=B1B2=

901235 12120 20211 01101 10010 01111 10100 10110 21101

. AB=(A1A2)(B1B2)

63963 21221 11121 20110 22432 22312

+

00000 42202 21101 21101 00000 21101

= 94191526 42617 31232 21528 64111019

11914134 121510143 64553 65451 9811114 811893 isasemisuperbimatrixwhichisnotsymmetric.

2.60: LetA=A1A2andB=B1B2beanytwo semisuperbimatrices.Wefindtheirminorproduct.

172

and B=B1B2 =

1432 2101 3010 4212

0230133123 2113020010 3021311101 1302202120 arethegivensemisuperbimatrices.

AB=(A1A2)(B1B2) =A1B1A2B2 =

0312 1020 0111 1010 2101

1432 2101 3010 4212 3145 0101 1010 1001 0110 1110 1011 0111 1101 1100

0230133123 2113020010 3021311101 1302202120

=

17737 7452 9323 4442 81177 313131 0230133123 010101 2113020010 101010 101010 010101 0230133123 111111 2113020010 101010 010101 1102301113311123 1121130112011010 + 454545 3021311101 010101 1302202120 101010 010101 101010 3021311101 101010 1302202120 111111 111111 0130213011101101 0013022000200120

19221817251523121713 3415222130 3251444224 1532335243 5134331111 5364464234 4553646344 6436533231 3645355253 2343153133

.

Weseetheresultantisalsoasemisuperbimatrix.Thusthe minorproductoftwo(compatibleunderproduct)semi superbimatricesisasemisuperbimatrix.

Example 2.61: LetA=A1A

1B2beanytwo

B1=

10301 10111 01010 11121 01312 01101 20011 10501 01110

andB2=

3012 1120 2011 5100 1010

.

AB=(A1A2)(B1B2) =A1B1A2B2 =

012121011 101210101 210011001 156561121 100010110 231101001 102110101

3012 1120 2011 5100 1010

=

10301 10111 01010 11121 01312 01101 20011 10501 01110 02156 10111 01010 21021 41401 10121

01210012301 10110101111 21001210010 156156 10301 100100 10111 231231 01010 102102

+ 1212 11121 2121 01312 0101 5656 011101121 100110312 1111

+ 01101 10111011 20011 01010101 10501 10011001 01110 1121011121101 0110200110011 1001101001501 0101010101110 A

A1=

0123 1011 0101 1200 5121 0110 and A2=

101511 210610 012101 121030 110222 501101 612010

.

AT=(A1A2)T=(T 1 AT 2 A) T 1 A=

010150 101211 210021 311010 and T A2=

1201156 0112101 1021012 5610210 1103201 1010210

(A1A2)(A1A2)T=(A1A2)(T 1 AT 2 A) =A1 T 1 AA2 TA2

=

0123 1011 0101 1200 5121 0110

010150 101211 210021 311010 101511 210610 012101 121030 110222 501101 612010

1201156 0112101 1021012 5610210 1103201 1010210 =

1454283 531181 412221 212572 8827313 311232 101010 1201156 212121 0112101 010101 121212 1201156 111111 0112101 505050 1201156 616161 0112101

+

2731741263 3137631461 7652234 43210615 121426822 6631222 3145225

+

1010210 0000000 1010210 0000000 2020420 1010210 0000000 =

1454283 531181 412221 212572 8827313 311232

29338515129 334277171614 8774545 574159613 1517591499 12164692832 91451393242

.

WeseeAATisasymmetricsemisuperbimatrix.Wegiveyet anotherexample.

Example 2.63: LetA=A1A2beasemisuperbimatrixwhere A2=

012341 234100 341010 410103 101034 010101 101010 111001 and

A1=

3024 1101 2210 6002 1101

AT=(A1A2)T=(A T 1A T2) =

.

02341011 3126113410101 0120124101011 2010031010100 4102140103010 10034101

AT=(A1A2)T(A1A2) =(A T 1A T2)(A1A2) =A T 1A1A T2A2 =

.

31261 01201 20100 41021

3024 1101 2210 6002 1101 02341011 13410101 24101011 31010100 40103010 10034101

012341 234100 341010 410103 101034 010101 101010 111001

=

3024 1101 2210 6002 1101

31261 01201 20100 41021 012341 234100 341010 410103 101034 010101 101010 111001

02341011 13410101 24101011 31010100 40103010 10034101 =

2978267 73483 849124 26812408 73483 0101 2323 02341011 3434 13410101 4141 1010 0101 02341011 1010 13410101 1111

+

2323 4141 24101011 1010 31010100 0101 1010 0101 24101011 1010 31010100 1010

+ 4141 0000 40103010 1010 10034101 0303 3434 0101 40103010 1010 10034101 0101 =

2978267 73483 849124 26812408 73483

13410101 31318112325 41825163437 11116174145 02341011 13410101 02341011 15751112

+

THEOREM 2.3: Let A = A1 A2 be a semi superbimatrix. AT be the transpose of A. Then AAT and ATA are in general two distinct symmetric superbimatrices.

Theproofisleftasanexercisetothereader;howeverwegivea smallhintfortheinterestedreadertoworkoutfortheproofof thetheorem.

Hint: LetA=A1A2beasemisuperbimatrixwhereA1isam nsimplematrixandA2isastsupermatrix. Now AT=(A1A2)T =(A T 1A T2)

issuchasA T 1isanmmatrixandA T2isatssupermatrix AAT=B1B2isasymmetricsemisuperbimatrixsuchthatB1 isammsymmetricsimplematrixandB2isasssymmetric supermatrix.ThusAAT=B1B2isasymmetricsemi superbimatrix. OntheotherhandweseeATA=C1C2whereC1isan nsymmetricsimplematrixandC2isattsymmetric supermatrix.ThusATAisalsoasymmetricsuperbimatrix. ClearlyAATATAingeneral. IfbothA1andA2inA=A1A2aresquarematriceswe wanttofindoutwhatisAATandATAincaseofsemi superbimatrices.Tothisendwefirstgivenanexamplebefore weproceedtofindthegeneralrule.

Example 2.64: LetA=A1A2beamixedsquaresemi superbimatrixwhere A1=

30110 11011 00101 21010 10101 and

A2=

012345 123450 120101 010011 101101 010110

WefindAT=(A1A2)T=(A T 1A T2) =

31021 01010 10101 11010 01101

AAT=(A1A2)(A1A2)T =(A1A2)(AT1AT2) =A1AT1A2AT 2 =

30110 11011 00101 21010 10101

.

011010 122101 230010 341011 450101 501110

31021 01010 10101 11010 01101

012345 123450 120101 010011 101101 010110

011010 122101 230010 341011 450101 501110 =

114174 44142 11202 74062 42223 0101011010 1212122101 1212 0101011010 1012102101 0101

+ 230010 234234 341011 345345 450101 010010 230010 001001 341011 110110 450101 011011

+

Weseetheminorproductyieldsasymmetricsemi superbimatrix.FurtherweseeAATATA.Thepartitionofthe supermatrixcomponentinATAisdifferentfromAAT.Itisleft asanexerciseforthereadertofindATA.

SUPER TRIMATRICES AND THEIR GENERALIZATIONS

Inthischapterweintroducethenotionofsupertrimatricesand giveheresomeoftheirpropertiesandoperationsonthem.

DEFINITION 3.1: T = T1 T2 T3 is defined to be a super trimatrix if each of the Ti is a supermatrix for i = 1, 2, 3. We demand either each Ti must be a distinct matrix or each Ti must have a distinct partition defined on it, Ti Tj if i j, 1 i, j 3.

Example 3.1: LetT=T1T2T3where T1=(103|11234|013), T2=(15|513|32013) and T3=(111111|0010|11111); clearlyTisasupertrimatrix.

Example 3.2: LetD=D1D2D3where

D1=

1 2 3 4 7 8 9

Disasupertrimatrix.

,D2=

3 1 0 1 1 0 1 3 2

andD3=

1 2 3 4 5 6 7 8 9

;

Example 3.3: LetT=T1T2T3where T1=

312 101 513 , T2=

1011 1012 0101 2102 and T3=

05976 16815 27624 38532 49401

Tisasupertrimatrix.

Example 3.4: LetU=U1U2U3where

;

U1=

4942151 7621360 8500571 1713282 and U3=

3453211 1011131 2110341 , U2=

5131 3211 2001 1062 3103 1515

;

Uisasupertrimatrix. Example 3.5: LetV=V1V2V3where V1=

012 120 210 , V2=

31 10 21 02 15 16

andV3=

1 2 1 3 4 7

Visasupertrimatrix.

Wehaveseen5examplesofsupertrimatrices.

DEFINITION 3.2: Let T = T1 T2 T3 be a super trimatrix we call T to be a row super trimatrix if each of the Ti is a row supermatrix; for i = 1, 2, 3.

Thesupertrimatrixgiveninexample3.1isarowsuper trimatrix.

DEFINITION 3.3: Let D = D1 D2 D3 where D is a super trimatrix. If each of the Di is a column supermatrix, i = 1, 2, 3 then we call D to be column super trimatrix.

Thesupertrimatrixgiveninexample3.2isacolumnsuper trimatrix.

DEFINITION 3.4: Let V = V1 V2 V3 be a super trimatrix. If each Vi is a square supermatrix having a different order then V is a mixed square super trimatrix.

Thesupertrimatrixgiveninexample3.3isamixedsquare supertrimatrix.

Note: IfinasupertrimatrixT=T1T2T3ifeachofthe supermatrixisanmmsquaresupermatrixthenwecallTtobe anmmsquaresupertrimatrix.

Example 3.6: LetS=S1S2S3beasupertrimatrixwhere

2

10101 01100 S10111 01110 10100 and 3

98765 43210 S01234 56789 97531

ClearlySisasquaresupertrimatrix.Thisisnotamixed squaresupermatrix.

ItisinfactclearthatSisa55squaresupertrimatrix.

DEFINITION 3.5: Let M1 M2 M3 be a super trimatrix. If each Mi is a rectangular supermatrix having a different order for i = 1, 2, 3 then we call M to be a mixed rectangular super trimatrix.

Theexamplegivenin3.4isamixedrectangularsupertrimatrix. Example 3.7: LetT=T1T2T3bearectangularsuper trimatrixwhere 1

3101121 4011020 T 6201131 7130001

2

1369635 0478527 T 2587418 1200139 and 3

0401100 1510000 T 2601011 3110110

.

Tisa47rectangularsupertrimatrixwhichisnotamixed rectangularsupertrimatrix.

Example 3.8: LetN=N1N2N3beasupertrimatrixwhere 1

1231 1101 0101 1010 N1100 1001 0110 0011 1111

,2

12 21 13 31 11 22 33 N 44 55 66 10 01 11 00 and

01234 12340 23401 11402 01025 N 98776 66554 44332 22115 55443

.

3

WeseeeachNiispartitionedonlyhorizontallyandnever verticallyi.e.,eachNiispartitionedonlyalongtherowsfori= 1,2,3. Example 3.9: LetC=C1C2C3beasupertrimatrix;where 1

1321231223 C2111232211 3231232122 2

218121141 127187158 C393118989 562591408 075331711 and 3 2134093812 C 3115127193 . EachCiispartitionedonlyverticallyandneverhorizontallyfori =1,2,3.

DEFINITION 3.6: Let T = T1 T2 T3 be a super trimatrix. Of each Ti is only partitioned horizontally i.e., only along the rows or in between the rows, then we call T to be a column super trivector.

Thesupertrimatrixgiveninexample3.8isacolumnsuper trivector.

DEFINITION 3.7: Let T = T1 T2 T3 be a super trimatrix. If each of Ti is only partitioned vertically i.e., only along the columns or in between the columns, then we call T to be a row super trivector.

ThesupertrimatrixCgiveninexample3.9isarowsuper trivector.

Nowhavingdefinedseveraltypesofsupertrimatriceswe defineoperationsonthem.Itisimportanttomentionherethat evenitisverydifficulttodefineadditionofsupertrimatricesfor ifweneedtodefineadditionweneednotonlyhavetheorderto bethesamebutalsothepartitiondefinedonthemmustbe identicalotherwisewecannotevendefinesimpleadditionof supertrimatrices.

Example 3.10: Let T=T1T2T3

=

7 1 3 5 2 8 7

3236 0224 3141 2161 2311 3512

31369 13212 22031 62223 42375

T+Salsohappenstobethesametypeofsupertrimatrix. Note: IfT=T1T2T3isasupertrimatrixthen T+T=(T1T2T3)+(T1T2T3) =(T1+T1)(T2+T2)(T3+T3) =2T12T22T3 =2T isasupertrimatrix.ThusifwetakeT+T+…+T(ntimes)= nT=nT1nT2nT3. Example 3.11: Let T=T1T2T3 =

21 01 12 21

1131 0124 1230 2011

101101 211200 313000 404100 525110 beasupertrimatrix. T+T=(T1T2T3)+(T1T2T3) =(T1+T1)(T2+T2)(T3+T3)

31210123 01311111 11020101 00102110 012 103 3121011111 1011112000 0101101213 752 021 31 1111247 2001352 31 5011010 6010110 31012101 01113121 11010201 00211010

Thusweareabletodefineaproductbutinallcaseswemaynot beinapositiontodefineaproductofsupertrimatrices,whichis acolumnsupertrimatrix.

1B2B3bea rowsupertrivectorwhere 1

1012112 B1100034 0111105 , 2 21011312 B 21103101 and

2134 0110 1010 010112110 0101 100201000 1111 221010121 1101 102011210 2001 1002 0210 3101013102131012 2121102120021234 3100113101131005 1012112 111110111 1100034 002002002 0111105 101 01

1012112 101101 1100034 0010010 0111105 111111 20210120131202 33211033310331 141414 212121 21011312 313131 21103101 010101 2101131 1515 2110310 2 15 1

1181149913123 321211121 231122231 202212210 433334441 232324320 202235430 214134530 431412121

.

Weseetheproductofacolumnsupertrivectorwitharowsuper trivectorwhendefinedresultsinasupertrimatrixwhichisnota supertrivector. Weproceedontodefinesomemoreconceptsbeforewedefine theproductofsupertrimatrixwithitstransposeandsoon.

DEFINITION 3.8: Let A = A1 A2 A3 be a super trimatrix. Then the transpose of A denoted by AT = (A1 A2 A3)T = 123 TTT AAA is again a super trimatrix.

Example 3.14: LetT=T1T2T3

weseeT1andT2aresupermatriceswhereasT3isa simple(ordinary)matrix.SoTisnotatrimatrixorasuper trimatrix.

Example 3.15: LetT=T1T2T3 41123150121 02350101010 35472332120 11021321112 09505110234 92616055758 35131510 29041741 18215332 01422143

.

WeseeT1andT3aresimplematriceswhereasT2isa supermatrix.ThusTisnotasupertrimatrixoratrimatrix.So wedefineanewnotioncalledsemisupertrimatrixwhichwill accommodatebothexamples3.14and3.15.

DEFINITION 3.9: Let T = T1 T2 T3, where some of the Ti’s are supermatrices and some of the Tj’s are ordinary matrices 1 i, j 3. Then we call T to be a semi super trimatrix i.e., a semi super trimatrix T has at least one of the matrices Ti to be a supermatrix, 1 i j 3.

Thematricesgiveninexamples3.14and3.15aresemisuper trimatrices.Nextconsiderthefollowingexamples.

Example 3.16: T=T1T2T3

trimatrixoramixedsquaresupertrimatrix.

DEFINITION 3.10: Let S = S1 S2 S3 be a square super trimatrix or a mixed square super trimatrix. If each of Si is a symmetric supermatrix then we call S = S1 S2 S3 to be a symmetric trimatrix (1 i 3).

Theexamples3.16and3.17aresymmetricsupertrimatrices.

Nowhavingdefinedthenotionofsymmetricsupertrimatrices wenowproceedontodefinequasisymmetricsupertrimatrices.

DEFINITION 3.11: V = V1 V2 V3 be a square super trimatrix or a mixed square super trimatrix or a mixed super trimatrix. We call V to be a quasi symmetric super trimatrix if at least one of the supermatrices V1, V2 or V3 is a symmetric supermatrix.

Nowweillustratethisbysomeexamples.

Example 3.19: LetV=V1V2V3where 1

39645 92012 V, 60765 41692 52528 2

0123456 1423450 2265432 V3357012 4440901 5531013 6022137 and 3

310964329 121876453 V225140112 512345678 701014052

.

ClearlyVisasupertrimatrixbutVisonlyaquasisymmetric supertrimatrixasV1andV2aresymmetricsupermatricesbutV3 isonlyasupermatrix.

Nowhavingseenexamplesofquasisupertrimatriceswe proceedontodefinethenotionofquasisemisupertrimatrices.

DEFINITION 3.12: Let V = V1 V2 V3 be a semi super trimatrix. If V1, V2 and V3 are square matrices or super square matrices then we call V to be a mixed square semi super

trimatrix. If V1, V2 and V3 are n n square matrices or n n supermatrices then we call V to be a square semi super trimatrix or an n n semi super trimatrix.

Example 3.20: Let T=T1T2T3 = 313012 201235

beasemisupertrimatrix.ClearlyTisa22orsquaresemi supertrimatrix. Example 3.21: LetV=V1V2V3beasemisupertrimatrix where 1

3215 6789 V 7542 0132

. 2

310125 112011 029206 V 213291 531056 123421 and 3

312 V011 101 .

ClearlyVisamixedsquaresemisupertrimatrix.

Nextwedefinemixedsupertrimatrix.

DEFINITION 3.13: Let T = T1 T2 T3 where some of Ti is a square supermatrix or matrix and the rest are rectangular supermatrix or matrix then we call T to be a mixed super trimatrix.

Example 3.22: LetV=V1V2V3where 1

3145 7023 V 8482 9710

, 2

312578912 012302310 V 120520131 111031015 and 3

345678 421056 561623 V; 630120 713145 820521

Visasemisupertrimatrixwhichisamixedsemisuper trimatrix.

Example 3.23: LetS=S1S2S3where

1

3121 1102 S 3975 1864 and 3

12345678 S91234567 89123456 , 2

3125 1100 9922 S2135 1234 5678 9123

.

Sisasemisupertrimatrixwhichisamixedsemisuper trimatrix.

Example 3.24: LetV=V1V2V3beasemisupertrimatrix where 1 135792468 V 246813579, 2

3124 0102 5762 V 9018 3210 1115

220

and 3

12345 63121 70111 V 85601 97892 13456

.

WeseethethreematricesV1,V2andV3arerectangular matricesoforder29,64and65respectively.Viscalled ordefinedasamixedrectangularsemisupertrimatrix.

Example 3.25: LetT=T1T2T3beasemisupertrimatrix where 1

3126781 T0123456 7891230 , 2

1234567 T3211432 5780123 and 3

0123456 T7891120 3010125 .

Weseeeachofthe3matricesT1,T2andT3are37matrices or37supermatrices.ThusTisa37rectangularsemisuper trimatrix.

Example 3.26: LetT=T1T2T3beasemisupertrimatrix where

1 31021 T 11253, 2

789 123 456 T 312 564 897 and 3

3123451 2010112 T 1246813 0357905

Tisamixedrectangularsemisupertrimatrix.

Nowweseejustacolumnsemisupertrimatrixandrowsemi supertrimatrix.

Example 3.27: LetA=A1A2A3where A1=[123456], A2=[0101|01|013] and A3=[987|6432|10].

Aisasemisupertrimatrixwhichwecallassimplerowsemi supertrimatrixorsimplerowsemisupertrivector.

Example 3.28: LetV=V1V2V3where

Visasemisupertrimatrixwhichisasimplecolumnsemisuper trivectororasimplecolumnsemisupertrimatrix.Wedefine dualpartitionincaseofsupertrimatrices.Itisimportantto mentionherethemaindifferencebetweenasimplerowor columnsupertrimatrixandaroworcolumnsupertrivectoris thatwhenwesaysimplecolumnorrowsupertrimatrixeachof itscomponentsinT=T1T2T3arejustasimple1ni,i= 1,2,3rowmatricesorsimplemi1columnmatrices. Wefirstseehowtheproductisdefined.

DEFINITION 3.14: Let T = T1 T2 T3 be a simple row semi super trimatrix, where 1,, i

ii in Taa; 1 i 3. If Ti is partitioned between rows r and r + 1, s and s + 1 and t and t + 1 and so on. We know 123123 T TTTT TTTTTTT where 1 2 i

i i T i i n

a a T a

; i = 1, 2, 3. Now each T Ti will be partitioned between the columns r and r + 1, s and s + 1 and t and t + 1 and so on. We call this type of partition carried on say from a 1 ni row to any

ni 1 column vector to be a dual partition. Thus if a row supermatrix 12 Aaaan has some partition and if a column supermatrix 1 2 n

b b B b has the dual partition then we have a well defined product AB. Thiswewillfirstillustratebyanexample. Example 3.29: LetT=T1T2T3whereA1=[230|1547 |20],A2=[1|2345|01]andA3=[1110|12015|01] beasimplerowsupertrimatrix.LetB=B1B2B3where 123

3 11 002 120 201 BBandB=211 050 111 322 15 0 beasimplecolumnsupertrimatrix.Then AB=(A1A2A3)(B1B2B3) =A1B1A2B2A3B3

Weseethoughwehavethecompatibilitywithrespecttoeach cell,yettheresultantissomenewstructurewhichisnever definedfortheyarenotsupertrimatricesorvectororany knownmathematicalstructure.

ThusasincaseofmatricesifABisdefineditmayhappen thatBAisundefinedlikewiseweseehereABisdefinedbutBA isundefined.

Nowweillustratebyasimpleexampletheminorproductof twosupertrimatrices.

12 2121 305102310 1034 43 1100125 1234 0261301 1102 123 2021 045 5102 12310111 11001010 11520120 10100 2013 00010 1501 10001 0407 01110

+ 21 310011 01 10101 30 =63151730 12340125 0000620416 204410 335814

724913353172103 111211126010101 111612107711093003 = 217919 2420 39251230

17710125 32482 201113234 .

Weseetheresultantisonlyatrimatrix.ThustheproductTSof therowsupertrivectorTwithacompatiblecolumnsuper trivectorSisonlyatrimatrixandnotasupertrimatrix.

Example 3.31: LetT=T1T2T3bearowsupertrivector andS=S1S2S3beacolumnsupertrivector.

TS=[T1T2T3][S1S2S3] =T1S1T2S2T3S3 where 351312420 T 106101011 301111010 160000101 010121110 133103121 241010100 101101010 012210101 and

411112101310 010010112210 100111303100 201011001001

. Now TT = T 123 TTT = TTT 123 TTT 4012 30 1100 12143 1001 40300 1010 11110 1111 02010 2011 15101 1130 01120 0100 10310 0231 00202 3210 10112 1100 21 0001 TT = T 123123 TTTTTT

DEFINITION 3.15: Let T = T1 T2 T3 be a super trimatrix, if every Ti is symmetric supermatrix i = 1, 2, 3 then we say T is symmetric super trimatrix. Clearly if T is a symmetric super trimatrix then T should be either a mixed square symmetric super trimatrix or square symmetric super trimatrix.

Weillustratethisbythefollowingexamples.

Example 3.34: LetT=T1T2T3beasupertrimatrixwhere 1

0123 1456 T 2502 3627

, 2

0101 1202 T 0045 1253 and 3

0125 1078 T 2761 5815

.

WeseeT=T1T2T3isasquaresymmetricsupertrimatrix. FurthereachTi;1i3are44symmetricsupertrimatrix.

Example 3.35: LetV=V1V2V3where

1

31230 17012 V, 20501 31023 02131 2

101230 050110 107202 V 212910 3101116 002062 and 3

9212307 2051061 1531206 V2117115 3021351 0601582 7165126

beasymmetricsupertrimatrix.ClearlyVisamixedsquare symmetricsupertrimatrix.

Example 3.36: LetT=T1T2T3where 1

31305 17210 T32037 01316 50769

,

2

12305 20161 T31210 06189 51096

.

012306 121012 219603 T 306121 010258 623187 and 3

WeseeeachofthematricesT1,T2andT3aresymmetric matrices.

FurthertheyarealsosupermatricesbutT=T1T2T3is notasymmetricsupertrimatrixasT2isonlyasymmetricmatrix butT2isnotsymmetricsupermatrixthoughT2isasupermatrix. ThusTisonlyasupertrimatrixwhichisnotsymmetric.

Fromthisweseeeachofthesupermatrixwhichis symmetricmustbepartitioned,suchthatitisasymmetric supermatrix.ThoughT2issymmetricitisnotsymmetric supermatrixasthepartitionhappenstoyieldanonsymmetric supermatrix.

Example 3.37: LetT=T1T2T3where 1

340156 T116215 031241

2

012345 160121 207602 T 316513 420172 512328 and 3

0123018 1012301 2170120 T=3209312 0313601 1023601 8102157

.

WeseeTisasupertrimatrixbutitisnotasymmetrictrimatrix. OnlyoneofthematricesT2aloneisasupersymmetricmatrix. Itisnotevenasquaresupertrimatrix.ThusTisonlyamixed supertrimatrix.

Example 3.38: LetT=T1T2T3where 1

11023 17906 T, 09121 20202 36111

2

410231 108963 087123 T 291201 362053 133130 and 3

310131 121234 015123 T 121012 332175 143253

TisamixedsquaresupertrimatrixbutTisnotasymmetric mixedsquaresupertrimatrix.ForT3isasupermatrixfurtherT3 isasymmetricmatrixbutafterpartitionT3isnotasymmetric supertrimatrix.T1isnotasymmetricmatrixonlyasupermatrix. T2isasymmetricsupermatrix.ThusTisonlyamixedsquare supertrimatrix.

Example 3.39: LetV=V1V2V3where 1

0123 1561 V 2604 3147 , 2

301 V012 121 and

248

WeseeTisamixedsquaresupertrimatrixwhichisalsoa mixedsquaresymmetricsupertrimatrix.

ThusweareinterestedinstudyingthosesupertrimatricesT =T1T2T3inwhichatleastoneofthemisasymmetric supermatrix.Tothisendwegivethefollowingdefinition.

DEFINITION 3.16: Let T = T1 T2 T3 be a super trimatrix we say T is a quasi symmetric super trimatrix if at least one of the Ti is a symmetric supermatrix 1 i 3.

1.It may so happen all the 3 matrices are symmetric matrices; yet all of them are not super symmetric, only one is a symmetric supermatrix.

2.It may so happen only one of the matrices Ti alone is a super symmetric matrix where as others are rectangular supermatrices.

3.It may so happen T is a square super trimatrix or a mixed square super trimatrix where only one of the Ti is a symmetric super trimatrix.

ThusinallthesecasesalsowecallTtobeaquasi symmetricsupertrimatrix.NowwehaveseenifTisrowsuper trivectorthenT.TTtheproductofTwithitstransposeyieldsa trimatrixwhichisnotasupertrimatrixbutitisasymmetric trimatrix.Thuswehaveamethodbywhichwecangenerate symmetrictrimatrices,ofcourseitmaybesquaresymmetric trimatrixoramixedsquaresymmetrictrimatrix.

Nowdowehaveanymethodofgeneratingsymmetricsuper trimatrices?Theanswerisyesandnowweproceedonto generatethembyaspecialproduct.

Example 3.40: LetT=T1T2T3beasupertrimatrixwhere 1

012111 101101 T310011 402100 520010

, 2

1230125 0111001 1000110 T 3101010 4212121 5010101 and 3

11011 22100 31001 T06102 11101 20102 01010 Now T TTTT 123123 TTTTTTT where

T 1

01345 10102 21020 T 11010 10101 11100

, T 2

101345 210120 310011 T010120 101011 201120 510011 and T 3

1230120 1216101 T0101110 1000001 1012120

. TTT = T 123123 TTTTTT = TTT 123123 TTTTTT = TTT 112233 TTTTTT

01345 012111 10102 101101 21020 310011 11010 402100 10101 520010 11100 101345 1230125 210120 0111001 310011 1000110 010120 3101010 101011 4202121 201120 5010101 510011 11011 221001230120 310011216101 061020101110 111011000001 201021012120 01010 =

0 1 301345 4 5

555105011 111 510011 000 000 151101011 111 1112310120 222 1230120 333 000 111 1230120 222 000 1216101 101010 0101110 2112121216101 1001010101110 616161 1111121116101 0100110011110 101010

256

111 2123201225 333 000 1123101215 222 5123501255

+ 010101 0111001 101010 1000110 101010 101010 0111001 010101 1000110 010101 0111001 101010 1000110 3453103451013450 1204211202121201 0115010110100111 120310120101120 011421011212011 120501120010120

0 1 1 3101010 0114210112120111 5010101

1011012538433 012111310011 125323804200 113212402100 012111310011 113212310011 25100050 1040020 000000 000000 520010 000000 12301251000110 246024100111001 369036150111001 00000000111001 12301251000110 246024101000110 510150510250111001 50119119119 11525252 9222222 11525252 9222222 11525252 9222222

1101113720351305 11011752011387113 000002210037305 110110000001010 1101131001513509 51139584 1362141 929523 515312 842112 413223

52131211111414 1311964913 1291235818 11636253 11452457 1498551012 141318371228 123

199519 9449215 59405PPPP 12021 9155111

.

WeseePisalsoasymmetricsupertrimatrix.HoweverweseeP andSi.e.,TTTandTTThavenorelation. ThusforagivensupertrimatrixTwecanobtaintwo symmetricsupertrimatrices.Nowweproceedontodefinethe notionofsemisupertrimatrixandthetypesofsemisuper trimatrices.

DEFINITION 3.17: Let T = T1 T2 T3, where at least one of the Ti is a supermatrix and at least one of the Tj (i j) is just a matrix and not a supermatrix (1 i, j 3) then we call T to be a semi super trimatrix.

Example 3.41: LetT=T1T2T3whereT1=[01234];T2 =[21|057|1113]andT3=[1103|893|12571].Tis asemisupertrimatrixforT1isjustarowvectorwhereasT2and T3aresuperrowvectors.

Example 3.42: LetS=S1S2S3where 12

3 2 1 3 0 S,S4 1 5 1 6 0 and 3

1 0 1 S1 1 1 1

.

SissemisupertrimatrixasS1andS3arecolumnsupervectors whereasS2isjustacolumnvector.

Example 3.43: LetQ=Q1Q2Q3where Q1=[111110], Q2=[2304157] and Q3=[31|705|1143|0];

QisasemisupertrimatrixforQ1andQ2arejustrowvectors whereasQ3isarowsupervector.

Example 3.44: LetV=V1V2V3where 123

1 1 11 0 00 V,VandV 1 12 2 21 5

;

VisasemisupertrimatrixasV1andV3arejustcolumnvectors whereasonlyV2isasupercolumnvector.

Example 3.45: LetV=V1V2V3where 123

3102 102 231156 V,VandV201 153481 110 1234

.

VisasemisupertrimatrixasV1andV3arejustsquarematrices whereasV2isasquaresupermatrix.

Example 3.46: LetW=W1W2W3where 12

3102 123 1156 W012,W3481 567 1234 and 3

102 W201 110 .

WisasemisupertrimatrixasW1andW3arejustsquare matriceswhereasW2isasquaresupermatrix.

Example 3.47: LetW=W1W2W3where 12

123 21 W012W 05 567 and 3

12345 67890 W09876 54321 11011

.

WisasemisupertrimatrixasW2isjustamatrixwhereasW1 andW3aresupermatrices.

Example 3.48: LetS=S1S2S3where 12

312 31011015 S12560,S311 36712111 567 and 3 2134567 S 1234890 .

Sissemisupertrimatrix,asS1isjustarectangularmatrixbutS2 andS3arerectangularsupermatrices.

Example 3.49: LetT=T1T2T3where

12

11738 312 40620 T567,T69515 890 78416 and 3

1234 5678 T 9012 3456

.

TisasemisupertrimatrixasT1isasquarematrix,T2andT3are supermatrices.

Justasincaseofsemisuperbimatrixwecanincaseofsemi supertrimatricesalsodefine5typesofsemisupertrimatrices.If inthesemisupertrimatrixT1T2T3allthe3matricesare justcolumnvectorswecallTtobeasemisupercolumnvector.

Examples3.42and3.44aresemisupercolumntrimatrices. IfintherowsemisupertrimatrixV=V1V2V3,Vi’sare matrices1i3;someofthemsuperrowvector;thenwecall Vtobeasemisuperrowtrivector.Theexamples3.41and3.43 aresemisuperrowtrivector.LetV=V1V2V3beasemi supertrimatrixwhereeachoftheViisannnsquarematrices 1i3,someofwhicharesupersquarematricesandothers justsquarematrices.WecallV=V1V2V3tobeasemi squaresupertrimatrix.

Wegivethefollowingexample.

Example 3.50: LetU=U1U2U3where 1

1234 5678 U, 9012 3456

270

beasemisupertrimatrix.Uisa44squaresemisuper trimatrix.

NextweconsiderasemisupertrimatrixT=T1T2T3, whereTi’saresquarematricesofdifferentorder;someofthe squarematricesaresupersquarematrices.WecallTtobea mixedsemisupersquarematrix.Theexamples3.45and3.46 aremixedsemisupersquaretrimatrices.

Nowweproceedontodefinethenotionofamixedsemi superrectangulartrimatrixandasemisuperrectangular trimatrix.

LetV=V1V2V3whereVi’saremn(mn) rectangularmatricessomejustordinaryandother supermatrices.ThuswecallVtobeanmnrectangularsemi supertrimatrix.

beasemisupertrimatrix.WeseeMisa37rectangularsemi supertrimatrix.

AsemisupertrimatrixV=V1V2V3issaidtobea mixedrectangularsemisupertrimatrixifVi’sarerectangular matricesorrectangularsupermatricesofdifferentorders.The semisupermatricesgivenintheexample3.48isamixed rectangularsemisupertrimatrix.

AsemisupertrimatrixWissaidtobeamixedsemisuper trimatrixifinW=W1W2W3someofthematricesWi’s aresquarematricesorsquaresupermatricesandsomeofthe Wj’sarerectangularmatricesorrectangularsupermatrices.

Example 3.52: LetT=T1T2T

3145 0123 4567 T8910 1112 1314 1516

;

3

Tisamixedsemisupertrimatrix. Example 3.53: LetS=S1S2S3where 1

3159213 0261014 S 1371115 2481216

, 2

31352 13011 S20101 31110 02526 and 3

123 102 201 S010 567 892 103

Sisamixedsemisupertrimatrix.

273

.

Nowhavingseenthe5typesofsemisupertrimatricesnow wedefinethenotionofsemisupertrimatrixwhichissymmetric andaquasisymmetricsemisupertrimatrix.

Example 3.54: LetT=T1T2T3whereTisasemisuper trimatrix.WesayTisasymmetricsemisupertrimatrixifeach oftheTiisasymmetricmatrixorasymmetricsupermatrix.

Example 3.55: LetV=V1V2V3where 1

20123 05321 V13756 22589 31690

, 2

123456 209875 392764 V 487312 576178 654281 and 3

1230 2501 V 3072 0126

isasymmetricsemisupertrimatrixasV1andV2aresymmetric supermatricesandV3isjustasymmetricmatrix.

Example 3.56: LetT=T1T2T3where

1

01111 12010 T10511 11170 10108

, 2

308 T056 863 and 3

012345 120101 207562 T 315071 406790 512108

.

HereT1andT2aresymmetricmatriceswhereasT3isa symmetricsupermatrix.ThusTisasymmetricsemi supermatrix. Nowweproceedontodefinethenotionofquasisymmetric semisupermatrixandillustrateitbysomesimpleexamples.

DEFINITION 3.18: Let T = T1 T2 T3 be a semi super trimatrix. If one of the Ti’s is symmetric supermatrix and one of the Tj’s is a symmetric matrix then we call T to be a quasi symmetric semi super trimatrix; i.e., the other Tk can be a square matrix which is not symmetric or a square supermatrix which is not symmetric or a rectangular supermatrix or a rectangular ordinary matrix (1 i, j, k 3).

Example 3.57: LetT=T1T2T3beasemisupertrimatrix where

275

1

30845 01234 T82151 43502 54128

, 2

312 043 501 T480 112 301 685 and 3

3401 4856 T 0512 1627

. Tisaquasisymmetricsupertrimatrix. Example 3.58: LetV=V1V2V3where 1

3012 0563 V 1671 2318

,

276

2

3101 1023 1621 7543 V1121 0351 6834 1138 0101 and 3

123456 298765 381024 V 470901 560034 654147

, weseeV1isasymmetricsquarematrix.V3isasymmetric squaresupermatrixwhereasV2isjustarectangularmatrix. ThusVisaquasisymmetricsemisupertrimatrix.

Example 3.59: LetS=S1S2S3whereS1isasquarematrix givenby 1

987643 210123 456789 S 101523 453101 070362

,

277

2

03015 32102 S01793 10910 52307 and 3

3451 4201 S= 50311 1110 isnotaquasisymmetricsemitrimatrixasS2isasymmetricasa matrixbutnotasymmetricsupermatrixbecauseofthepartition. S3isasymmetricmatrixandS1justanonsymmetricsquare matrix. Example 3.60: LetT=T1T2T3where 1

08976 81567 T95234 76357 67478

, 2

012305 165432 257109 T 341861 030658 529183

278

2340 1153 0118 S1110 2345 6789 1080

isnotacolumnsupervectorforitisdividedbothverticallyand horizontally.

Thusaspecialcolumnsupervectorisalwaysdividedor partitionedonlyhorizontallyandaspecialsuperrowvectoris alwayspartitionedonlyvertically.Thus 021098643 P111321357 202590432

isnotaspecialsuperrowvector.

Nowwegiveconditionsunderwhichtheproductofa specialcolumnsupervectoriscompatiblewithaspecialrow supervectorandsoon.

Wedefineboththeminorproductaswellasthemajor productofthesesupermatriceswhenspeciallythese supermatricesaresupertrimatrices.

DEFINITION 3.19: Let T = T1 T2 T3 be a semi super trimatrix if at least one of the Ti is a column super vector and one of the Tj’s is a simple column vector i.e., Tj is a m n matrix with m > n then we call T to be a special column semi super trimatrix or vector (1 i, j 3).

Wefirstillustratethisbyasimpleexamplebeforewedefine moreconcepts.

280

Example 3.61: LetT=T1T2T3where 1 31579110 T 206810129, 2

31251 01234 56789 98765 T 43210 11011 01110 30021 and 3

215 678 901 T 234 678 913

.

Tisaspecialcolumnsemisupertrivector.Clearlyweseeallthe 3matricesneednotbeanmnmatrixwithn>m.

Example 3.62: LetS=S1S2S3where 1

3125 6011 S 0723 0115

,

2

312 011 611 101 S 110 325 678 134 and 3

31235 10261 01078 87007 S11084 12345 67890 13011 81910

;

SisaspecialsemisupercolumntrivectorormatrixweseeS1is asquaresupermatrixbutS2isaspecialcolumnsupertrivector andS3isanmnmatrixwithm>n.

DEFINITION 3.20: Let P = P 1 P 2 P3 where at least one of the Pi’s is a special row super vector; at least one of the Pj’s is a special row vector i.e., Pj is an m n matrix in which n > m. 1 i, j 3. Then we call P to be a special semi super row trivector. Weillustratethisbythefollowingexamples.

282

Example 3.63: LetP=P1P2P3where 1

3120 1981 P 2763 5328

, 2

0123456789 P1012070676 3110611001 and 3 317934310 P 538681517 .

Pisaspecialsemisuperrowtrivector.

Example 3.64: LetS=S1S2S3where 2 34680 S 15791, 1

3125 0101 7625 S3014 1103 1011 1111 and

.

3

398763423401 112345678987 S 265432101231 478401235101

Sisaspecialsemisuperrowtrimatrixortrivector,asS2andS3 arerectangularmnmatriceswithm>n. Nowweproceedontoillustratebyexamplestheproducts ofsemisupertrimatrices.

Example 3.65: LetT=T1T2T3beaspecialsemisuper rowtrivectorandP=P1P2P3aspecialsemisupercolumn trivector.HereT=T1T2T3 3102057314 5411211002 0103120110 13510011 24011000 isspecialrowtrimatrix. P=P1P2P3 101 10 010 01 111 01121 012 10110 100 11011 011 01 101 20 000 isaspecialsemisupercolumntrimatrix.

10 57314 11 1211002 01 20110 20 101 1351 010012 2401 111 + 100 0011011 1000101 000 13420105184 569122150 101314212 686012 242012 101 100 134309 799 56983 354 10175 .

WeseeTPisjustatrimatrixwhichisnotasemisupertrimatrix.

Example 3.66: LetT=T1T2T3beaspecialsemisuper rowtrimatrixandV=V1V2V3beaspecialsemisuper columntrimatrix. T=T1T2T3

286

= 2345 1234 0112 151010101 200111001 310100011 411011000 110011101 311101010 010110101 beaspecialsemisuperrowtrimatrix. V=V1V2V3 101101 010100 111001 011 101100 100 010011 211 000110 310 211000 010100 100001

beaspecialsemisupercolumntrimatrix.Tofind

TV=(T=T1T2T3)(V=V1V2V3) =T1V1T2V2T3V3

287

011 2345 100 1234 211 0112 310 10 01 11 151010101 10 200111001 01 310100011 00 411011000 21 01 10 1101 0100 1001 1100111011100 3111010100011 0101101010110 1000 0100 0001 15 26116 2010 2084 3101 931 41

Nowweproceedontodefinethenotionofminorproductof specialsemisuperrowtrivectorandaspecialsemisuper columntrivector.

DEFINITION 3.21: Let T = T1 T2 T3 be any special semi super row trimatrix and V = V1 V2 V3 be another special semi super column trimatrix. The minor product TV = (T = T1 T2 T3) (V = V1 V2 V3) = T1V1 T2V2 T3V3 is defined if each of the product TiVi, 1 i 3 is defined.

Note: Inthecaseofminorproductoftwospecialsemisuper trimatricestheresultantisaonlytrimatrixandnotevenasuper trimatrixorasemisupertrimatrix.

Example 3.67: LetT=T1T2T3beaspecialsemisuper rowtrivectortofindtheminorproductofTTT . GivenT=T1T2T3 31025 10101 01010 30201 120131250 012010101 301101010 11301011010 20100110111 31010101110 00011011101 isaspecialsemisuperrowtrivector.

3103 31025 1010 10101 0102 01010 2010 30201 5101 103 210 021 120131250101 012010101310 301101010101 210 501 010 1230 1010 3100 0011 11301011010 1001 20100110111 0110 31010101110 1101 00011011101 1011 0111 1110 0101 398314 8306 3020 146014

292

12 103 01 210 30

101 0130211250 210 2011010101 501 1103101010 010 1230 11301 1010 20100 3100 31010 0011 00011 1001 01101 0110 110110111 1101 101111110 1011 01110

+ 0 1 0101 0 1

398314 523 8306 210 3020 309 146014 10313026 352220 122602 125412112 55601211 461111121 10122112

11100000 12210101 12210000 01110101 =

398314 8306 3020 146014

15763 45710 71093 782 69153 10213 3336

.

WeseetheminorproductofTTTisasymmetrictrimatrixandis notasupertrimatrix.InfactTTTisamixedsymmetric trimatrix.Thustheminorproductofspecialsemisuperrow trivectorwithitstransposeyieldsasymmetrictrimatrix.

Example 3.68: LetT=T1T2T3beaspecialsemisuper rowtrivector.TofindtheproductofTTT.Given 12301111 0123 00110101 T1230 20101010 2310 11010010 10211110010 01021000111 22100101011 , TT=(T1T2T3)T = T 1 TT 2 TT 3T

294

=

102 012 1021 201 2001 120 0123110 110 1230101 101 2311010 100 3001100 001 1011 010 1100 111 011

TTT=(T1T2T3)(T 1 TT 2 TT 3 T) =T1 T 1 TT2 T 2 TT3 T 3T 012 0123 123 1230 231 2310 300 1021 2601 123011113110 061101010101 201010101010 110100101100 1011 1100

.

TV=(T=T1T2T3)(V=V1V2V3) =T1V1T2V2T3V3 here 1

70215 T11001 20101 , 2

1031011012 T=0100110112 7011100211 and 3

01231034501 10110110111 T21011001000 30101100101 01010011010 bethespecialsemisuperrowtrivector.V=V1V2V3; 123

010 0110 101 1010 011 0111 102111 0010 011011 1001 V,V,V100001 0101 010101 0110 101110 1001 010 0101 101 0010 010 bethesemisupercolumntrivector.

102 70215011 TV11001100 20101010 101 0111 1010 0111 0010 1031011012 1001 0100110112 0101 7011100211 0110 1001 0101 0010 010 101 011 01231034501111 10110101111011 21011001000001 30101100101101 01010011010110 010 101 010

141191 21400111 4057 031010010 1010 100111001 0111 011100101 0110 1012 1001 0112 0101 0211 0010 01231 010 10110 111 101 21011 011 011 30101 01010

+ 034501101 110111110 000101000010 100101101 011010010 1411901110333 21400001010 40507770111 +

isaspecialsemisupercolumntrivector.Here 02311011001101 00010101110010 V 11110200111110 10001010000101 0213100131 1001010001 1011001010 0002001020 1111110111

bethespecialsemisuperrowtrivector. SV=(S1S2S3)(V1V2V3) =S1V1S2V2S3V3 3101 02311 4010 00010 5100 11110 0110 10001 1001 1012 0101 1100011001101 2101101110010 1021200111110 1111010000101 0110 2001

10111 01011 11010 0213100131 12110 1001010001 10101 1011001010 01010 0002001020 10112 1111110111 01010 31101 02010 16944 191354 0101565 11120 12312

011001 10121012 101110 01010101 200111 11001100 010000 21010110210101 10211011102110 11112001111111 01100100011000 2001

011001 101110 2001 200111 010000 101 1012 010 0101 110 1100 101 2101101 1021010 1111110 0110101 101 010 2001 110 101

02131001 10010100 101111011011110010 00020010 11111101 02131001 10010100 0101101011 10110010 1101011010 00020010 11111101 12110 10101 0101

02131001 1001211010100 1011010110010 00000101020010 11111101 021 1011210112 100 0101001010 101 3110131101 000 0201002010 111

31001 10100 10010 20010 11101

2337212272 2114121132 1216111152 3228122163 2335211252 1003011021 3448322383 1003011021 375124214115 2004021022

.

Weseethemajorproductyieldsasemisupertrimatrix. Thususingmajorproductofcompatiblespecialsemisuper trivectorwecangetmoreandmoresemisupertrimatrices whicharenottrivectors.Wegiveyetanotherexampleofhow themajorproductisdetermined.

S=S1S2S3 1010 2101 3012 0010 7101 0101311301450 1102111110110 2100021210100 1301110101 2410001110 beaspecialsuperrowtrivector.

TS=(T1T2T3)(S1S2S3) =T1S1T2S2T3S3 1010 310122101 110103012 230140010 7101

2150101215311 1011102101111 5112100511021 011011 0101311 100100 1102111 010010 2100021 110110 012 0101 210 1102 314 001

012 311 210 111 314 2100021 001 215301450 101110110 511210100 011 301450 100 110110 010 210100 110 012 301450 210 110110 314 210100 001

112543334743 1301110101 92325552725 3903330303 3711111211 3711111211 . 41012221312 51313331413 143254445954 132761116761 51121112321

Weseetheresultantisasemisupertrimatrix.

Example 3.72: Let T=T1T2T3 3101 1012 1110 1010 01010 0100 10101 1011 11100 0110 00011 1001 1010 1000 0100

01312 10100 01001 11000 21110 00110 10100 11011 01010 beaspecialsemisupercolumntrivector.

TT=(T=T1T2T3)T = T 1 TT 2 TT 3T 0110 1010 0110 1001 0101 31110101110 10101010001 01110110100 12000101000 010120110 101110011 310011100 100011011 201000010

TTT=(T1T2T3)(T1T2T3)T

313

.

=(T1T2T3

=T1 T 1 TT2

1

2 TT3

3T =

0110 01010 1010 10101 0110 11100 1001 00011 0101 3101 1012 1110 1010 31110101110 0100 10101010001 1011 01110110100 0110 12000101000 1001 1010 1000 0100 01312 10100 01001010120110 11000101110011 21110310011100 00110100011011 10100201000010 11011 01010

=

1011 311011010 101100101 011101010 120100000 0100

2011 0321 1230 1102 31110 31013101 10101 10121012 01110 11101110 12000 31110 1010101101001 0100011010010 12000 1011 0110 1001 1010 1000 0100

101110 3101 010001 1012 110100 1110 101000 101110 1010010001 0100110100 101000 1011 0110101110 1001010001 1010110100 1000101000 0100 010120 0131201312 101110 1010010100 310011 0100101001 100011 1100011000 201000 010120 101110 2111021110 310011 0011000110 1000 2010 11 00 010120 1010010100101110 1101111011310011 0101001010100011 201000

316

110 01312 011 10100 100 01001 011 11000 010 110 011 21110 100 00110 011 010 110 10100011 11011100 01010011 010

2011 0321 1230 1102 115431414331 56220413210 42321221211 32220211210 10101010001 44220312210 11211120101 43110202110 32220211211 31111101110 10101010101

317

1533154342 320131210 302110021 111230121 531372342 410022111 320131210 412241142 201121022

.

Weseetheresultantisasymmetricsemisupertrimatrix. Thusthemajorproductofaspecialsemisupercolumntrimatrix TwithitstransposeTTyieldsasymmetricsemisupertrimatrix.

Example 3.73: LetP=P1P2P3beaspecialsemisuper rowtrimatrix.TofindthevalueofPTP.Given

1011 0101 1010 0110 1100 31110101101 10100010100 01001111000 13011010101 1111110111 0100101001 1011010100

P=

318

T

10101 01011 P 10110 11000 3101 1013 1100 1001 0011 1010 0111 1010 1101 0000 1001

101 110 101 101 110 101 010 101 100 110

.

PTP=(P1P2P3)T(P1P2P3) =(T 1 PT 2 PT 3 P)(P1P2P3) = T 1 PP1 T 2 PP2P3 T 3P 1011 10101 0101 01011 1010 10110 0110 11000 1100

3101 1013 1100 1001 31110101101 0011 10100010100 1010 01001111000 0111 13011010101 1010 1101 0000 1001 101 110 101 101 1111110111 110 0100101001 101 1011010100 010 101 100 110 3121 1311 = 2131 1112

31110 3101103101100 1013011013001 13011 31110 11001100 10100 10011001 01001 00110011 13011 1010 0111 1010 1101 0000 1001

1010 310111110 101010100 011101001 130000011 1001 101101 3101010100 1013111000 010101 101101 1100 010100 1001 111000 0011 010101 1010 0111101101 1010010100 1101111000 0000010101 1001

1011111011110110111 1100101100111001001 1011011011010110100 1111110111 101101101 0100101001 110110110 1011010100 101101101 0101110101101010111 1010101010110101001 1001011001010010100 110110110 =

3121 1311 2131 1112 116441323504 611144242404 41210111201 44121111202 14012121101 32111212101 24112131201 32111212101 54221121302 00000000000 44121111202

2122120211 1211211112 2122120211 2122120211 1211211112 2122120211 0100101001 2122120211 1111110111 1211211112

ClearlyPTPisagainasymmetricsemisupertrimatrix.Thusthe majorproductofatransposeofaspecialsemisupermatrixP withitselfyieldsasymmetricsemisupermatrix.Nowwe proceedontodefinethenotionofmajorproductofsemi supermatrices.

Example 3.74: LetP=P1P2P3andT=T1T2T3be twosemisupertrimatricesforwhichmajorproductPTis defined.WefindthisproductforPandT.

Given

101101 010101 111000 010101 101110 010001 001010 isasemisupertrimatrix. Now T=

121 010 100 101

0153120 0100112 1011011 1101001 0010101 1111112 0000120 1011111 1110101 1011012 1101102 0112111 2011001 PT=(P1P2P3)(T1T2T3) =P1T1P2T2P3T3

=

3101 1202 0163 1000

121 010 100 101

(1ij Z)(2 ij Z) =

474 343 913 121

Wecalculate (1ij Z)=P2T2=

(1ij Z)(2 ij Z).

1110301 0010111 3100101 1001001 1101000 0101011 1110102 0153120 0100112 1011011 1101001 0010101 1111112 0000120

1 11 Z=[1|110|301]

0 0 1 1 0 1 0 =(1)(0)+(110) 0 1 1 +(301) 0 1 0 =(0)+(1)+(0)=1. 1 12 Z=[1|110|301]

1531 1001 0110 1010 0101 1111 0001 =[1][1531]+[110] 1001 0110 1010 +[301] 0101 1111 0001 =[1531]+[1111]+[0304] =[2946].

1 13 Z=[1|110|301]

20 12 11 01 01 12 20 =[1][20]+[110] 12 11 01 +[301] 01 12 20 =[20]+[23]+[23] =[66]. 1 21 Z=

0010111 3100101 1001001 1101000

0 0 1 1 0 1 0 =

0010111 00 3100101 011 1001001 10 1101000

Nowhavingdefinedthiswecanalsoasincaseofspecial semisupertrivectorsdefinetheproductofasemisuper trimatrixwithitstranspose.

Example 3.75: LetT=T1T2T3beasemisupertrimatrix where T=

21000 01010 10001 01100 30102

1011234 0100101 1010110 3101011 0111000 1010111 1001110 0100101 1131010 2001101 0101111 1101011 1011001 0010010 3103110

. TT =

20103 11010 00011 01000 00102

10130110 01011001 10101100 10011010 21100111 30110110 41010101

1201103 1011001 3000110 1111103 0110001 1011011 0111100 2100020103 0101011010 1000100011 0110001000 3010200102 1011234 10130110 0100101 01011001 1010110 10101100 3101011 10011010 0111000 21100111 1010111 30110110 1001110 41010101 0100101 11310101201103 20011011011001 01011113000110 11010111111103 10110010110001 00100101011011 31031100111100

333

1 12 Z=[1|0112|34]

01301 10110 01011 00110 11001 01101 10101

=[1][01301]+[0112]

10110 01011 00110 11001

+ [34] 01101 10101 =[01301]+[23123]+[43707] =[6711211]. 1 13 Z=[1|0112|34]

10 01 00 10 11 10 01

=[1][10]+[0112]

01 00 10 11

+[34] 10 01

335

Thus

51216 12010 20205 11021 605114 32671121176 63121213 71441431 1124132552 21123111 112451532 71351341 63121213 13334548 37344010 3354216 4445318 5423416 4011121 810686121

TTT =

Weseetheproductgivesusasymmetricsemisupertrimatrix. TheinterestedreadercanfindTTTwhichwillgiveyetanother symmetricsemisupertrimatrix.

Wehavegiventheexplicitworkingoftheproduct,themain motivationforitisthatwhilecalculatingforageneralcasewe

havetogivelotsofnotationsandwefeltitwouldonlyconfuse thereader.Itiscertainlyeasytofindthemajorproductsusing theseillustrationsforafterallwhatweareinterestedisthatthe readershouldbeinapositionstoworkitout.Thetheorybehind itisnotverydifficultbutthenotationalrepresentationsislittle cumbersome.

SUPER n-MATRICES AND THEIR PROPERTIES

Inthischapterweforthefirsttimeintroducethenotionofsuper n-matrices(n-anintegerandn>3)andgiveafewofits properties.Whenn=1wegetthesupermatrix,n=2,itisthe superbimatrixstudiedinchaptertwo.Thestudyofsupertri matrixhasbeenstudiedinchapterthree,thecasewhenn=3; whenwesaysupern-matrixwemeannapositiveintegerandn >3.Herewealsodefinethenotionofsemisupern-matrices andshowhowtheproductdefinedusingthemattimesyields onlyanordinaryorelementaryn-matrix.Furthersomeofthe productsinduceasymmetricsupern-matrixoraquasi symmetricsupern-matrix.

DEFINITION 4.1: Let V = V1 V2 … Vn (n > 3) denote n distinct super matrices, i.e., each Vi is a super matrix 1 i n. ‘ ’ is just a symbol. We define V to be a super n-matrix.

Example 4.1: LetV=V1V2V3V4where V1=[10|234|5],

V2=

0 1 1 1 3 2 5

, V3= 30 12 and V4=

301 111 202 535

. Visasuper4-matrix.Heren=4. Example 4.2: LetT=T1T2T3T4T5where T1= 3021531 1120780, T2=

3134 0110 1001 3152 1234 9831

,

T3=

31245 01051 63789 11012 20203

, T4=

85312 10112 84031 71011 83815 and T5=

3410 1301 5110 0231

.

Tisasuper5-matrix.Inthiscasewehaven=5.

Example 4.3: LetT=T1T2T3T4T5T6where T1=[0|101|234], T2=[1|2345|578|90], T3=[013|45|78910], T4=[6|12|301461], T5=[310|22501] and T6=[1|23|456|7181].

Tisasuper6-matrix.WeseeeachofthesupermatricesTi’sare rowsupervectors.

DEFINITION 4.2: Let T = T1 T2 … Tn (n > 3) be a super n-matrix. If each of the Ti is a super row vector i = 1, 2, …, n then we call T to be a row super n-vector or super row n-vector.

Theexample4.3isasuperrow6-vector.

Example 4.4: LetS=S1S2S3S4where S1=

1 2 3 4 5 6

,S2=

0 1 2 4 7 0 2 5 1

,S3=

1 2 5 3 7 6 4

andS4=

1 2 1 3 2 3 1 0 1

.

Sisasuper4-matrixorasuper4-columnvector.

DEFINITION 4.3: Let T = T1 T2 T3 … Tn be a super nmatrix (n >3). If each of the Ti is a super column vector (1 i n) then we call T to be a super n-column vector or column super n-vector or column super n-matrix.

Example4.4isasuper4-columnvector.

Example 4.5: LetV=V1V2V3V4V5where V1= 21 01,V2= 11 02,

V3=

123 456 789 V4=

1234 0125 7810 9642 and V5=

12011 01201 14013 25121 36100

.

Visasuper5-matrix.WeseeeachofthesupermatricesVi;1 i5aresquaresupermatrices.

DEFINITION 4.4: Let K = K1 K2 K3 … Kn (n > 3) be a super n-matrix if each of the Ki are t × t square matrices for i = 1, 2, …, n then we call K to be a t × t square super n-matrix. If on the other hand each of Ki is a mi × mi square matrix i = 1, 2, …, n then we call K to be a mixed square super n-matrix.

Theexample4.5isamixedsupersquaren-matrix(n=5).

Example 4.6: LetP=P1P2P3P4where P1=

1234 5678 9012 3456

,

P2=

1234 0101 1111 1100 and P4=

0123 1011 0101 1001 , P3=

1024 1101 1011 0111

.

Pisasuper4-matrix.InfactPisa4×4squaresuper4-matrix.

Example 4.7: LetS=S1S2S3S4S5where S1=

310109 011171 111088 , S2=

3412 1110 0001 1101 0109 1125

,

S3=

1793 2881 3972 4060 5152 6245 and S5=

210475 311966 111847 , S4=

11745114 21836210 31927318 41619017

.

Sisasuper5-matrix.WeseeeachoftheSi;1i5are rectangularsupermatricesofdifferentorder.

DEFINITION 4.5: Let S = S1 S2 S3 … Sn (n >3) be a super n-matrix. If each of the Si’s are rectangular super matrices of different order then we call S to be a mixed super rectangular n-matrix or a mixed rectangular super n-matrix.

Thesupern-matrixgiveninexample4.7isamixedrectangular n-matrix.Ifinthesupern-matrixS=S1S2…Sneachof theSi’sarem×t(mt)rectangularsupermatricesthenwecall Stobeanm×trectangularn-matrix.

Thesupern-matrixgiveninexample4.7isamixed rectangularn-matrix.

DEFINITION 4.6: Let T = T1 T2 T3 Tn (n > 3) be a super n-matrix. If some of the Ti’s are square supermatrices and

some of the Tj’s are rectangular super matrices (i = j); i j, i n, then we call T to be a mixed super n-matrix.

Example 4.8: LetK=K1K2K3K4K5beasuper5matrixwhere K1=

312 110 011 , K2=

30111 01201 11012 , K3=

3101 0210 1143 5612

, K4=

1650103 2749114 3838062 4927201 5016315 and K5= 11357917 02468048 . Kisamixedsupern-matrixheren=5.

DEFINITION 4.7: Let T = T1 T2 … Tn (n >3) be a super n-matrix we say T is a special super row n-vector if each of the Ti’s is a mi × ni matrix (ni >mi); 1 < i < n; with partitions done

only vertically i.e., only between the columns. No partition is made between the rows. If P = P1 P2 … Pn (n > 3) be a super n-matrix we say P is a special super column n vector if each Pi is a ti × si matrix with ti > si (1 i n) and each column matrix is partitioned only in between the rows i.e., horizontally and never partitioned in between the columns.

Example 4.9: LetS=S1S2S3S4S5where S1= 13103 24113, S2=

3147101 1058131 2169274 , S3=

112352014 011021172 243010110 301111101

, S4=

1012311231 1110110123 2115120110 , and S5=

11732715116 21823826202 11654937373 00416048621

. Sisasuper5-matrixwhichisaspecialsuperrow5-matrix.

Example 4.10: LetT=T1T2T3T4where

T1=

3101 1123 5678 91011 1211 0101 1110 0110

, T2=

210 311 987 654 321 011 100 011

,T3=

12 31 15 58 76 68 91 14 43 36 and T4=

131 111 012 311 111 114 423 301 115 012

.

358

Tisasuper4-matrixwhichisaspecialsupercolumn4-matrix.

Example 4.11: LetS=S1S2S3S4where S1=

2141 0315 1234 5678 9018 1114 4412

,S2=

311 024 131 146 789 104 412 116

, S3=

12345 67890 32148 11141 70813 31256 11011 and S4=

1231 1101 1011 2111 4231 1101 1503 1723

.

ClearlySisasuper4matrixbutSisnotaspecialsupercolumn 4matrixforS1andS3arepartitionedverticallyalso.S1andS3 areonlysupermatricesandarenotspecialsupercolumn vectors.

Example 4.12: LetV=V1V2V3V4beasuper4matrix;where V1=

2101718 0114617 1135806 0246194

, V2= 13579 24680 V3=

1930618 2821529 3712451 and V4= 1310111 2151011.

VisnotaspecialsuperrowvectorforthesupermatrixV1and V4arepartitionedhorizontallyalso.

Nowwecandefineminorproductofspecialsupernmatrices.Wefirstillustrateitandthendefinetheconcept.

Example 4.13: LetT=T1T2T3T4beaspecialsuper column4-matrixandV=V1V2V3V4beaspecialsuper row4matrixtheminorproductTVisdefinedasfollows.

T=T1T2T3T4

0123 12123 1111 01011 1234 11111 1110 23112 0111 34011 1114 10111 0010 12011 01101 12301 10110 01001 11010 01001 11101 bethegivenspecialsupercolumn4vector.S=S1S2S3 S4 = 021011 131113 01111011 10101101 10110110 1201110110 0111101001 0101101100 1110010000

12011 01101 110000011 12301 101000101 10110 000100110 01001 100010001 11010 100001111 01001 11101 =

1202121011 0113011113 1111 2302231011 3413341113 1010 01111 123123101123 10101 011011110011 10110 111111011111 01111011 112112112 10101101 011011011 10110110 01111 101 10101 111111110111 10110 011

12011101 01111010 01230123 01011011 11100100 12011101 1111011111111010 1234010123411011 11100100 1110120 0111011 1114010 0010111

111011101 011111010 111411011 001000100 10 01 0123 00 00 10 111101 123400 00 111010 011101 111400 001000

283237 131113 152124 31353511 41874715 021011 51643543 20211211 21322222 31432332 20211211 21322222 3643333201 2523322211 51166655412 1413312211 1322212101 5853352211 0101101100 512011325 201101322 412301654 210110122 201001212 311010113 201001212 311101333

.

366

Theresultantisasuper4-matrixwhichisnotaspecialsuper roworcolumnn-matrix.

Nowwedefinesymmetricsupern-matrixandasemi symmetricsupern-matrix.

DEFINITION 4.8: Let T = T1 T2 … Tn (n > 3) be a super n-matrix. If each of the Ti is a symmetric super n-matrix then we call T to be a symmetric super n-matrix: 1 i n,

Example 4.14: LetT=T1T2T3T4T5beasuper5 matrixwhere T1= 310 101,T2=

311 101 118 , T3=

1204 2152 0516 4264

, T4=

123 257 371 and T5=

123456 201101 312712 417035 501312 612520

Tisasymmetricsuper5-matrix.

Thusasymmetricsupern-matrixiseitherasquaresupernmatrixoramixedsquaresupern-matrix.Itmaysohappenthat inasupern-matrixT=T1T2T3…Tn(n>3)someof theTi’saresymmetricsupermatricessomeofthemjust supermatricesinsuchcasewedefineTtobeaquasisymmetric supern-matrix

Example 4.15: LetT=T1T2T3T4T5T6beasuper 6-matrixwhere T1=

3120 1101 2057 0170

, T2=

78101 81513 15101 01025 13150

,T3= 34 57 T4=

1210 2112 1130 0203

, T5=

111010 120100 105312 013017 101101 002710 and

T6=

10135 01201 12725 30213 51530

.

ClearlyT=T1T2T3T4T5T6isonlyaquasi symmetricsuper6-matrix.

DEFINITION 4.9: Let T = T1 T2 T3 … Tn (n >3) be an n-matrix in which some of the Ti’s are super matrices and some of the Tj’s are just matrices 1 i, j n. Then we call T = T1 T2 … Tn to be a semi super n-matrix.

Example 4.16: LetT=T1T2T3T4where T1=

31256 13011 20210 51321 61456

, T2=

012345 678901 301057 , T3=

21 03 12 and

T4=

31215 10132 31357 02406

Tisasemisuper4-matrix

.

Example 4.17: LetS=S1S2S3S4S5where S1=

31518 11680 02792 13016 , S2=

301 850 012 , S3=

360 543 218 101 867

, S4=

364815 606103 412410 901240 226112 and

370

S5=

310311 146102 112213 015682

S=S1S2S3S4S5isasemisuper5-matrix.

Example 4.18: LetS=S1S2S3S4where S1=

301 450 026 , S2=

3612 0106 4112 3016 2154 1028

, S3=

312 111 012 670 051 123 015 and

371

S4=

3264 2101 6211 4012

.

S=S1S2S3S4isasemisuper4-matrix.

DEFINITION 4.10: Let T = T1 T2 … Tn be a semi super nmatrix (n > 3). If each Ti is a m × p matrix with m < p and 1 i n with some of the Ti’s super matrices and some of the Tj’s just matrices then we call T to be a special semi super row nvector. If m > p then we call T = T1 T2 … Tn to be a special semi super column n-vector.

Weillustratethembythefollowingexamples

Example 4.19: LetV=V1V2V3V4V5beaspecial semisuperrow5-vectorwhere V1=

31013211 01102021 25111151 , V2= 246810 157921, V3=

100161 213448 322057 431326

, V4=

1051 1162 0173

372

Example 4.21: LetT=T1T2T3T4where

T1=

312 101 123 014 567 890 127

, T2=

1081 3168 1076 6102 1101 3146 7890 1134

, T3=

11111 00100 00010 00001 10000 01000 10100 00101 10101 and

beasemisupern-matrix(n=4)butTisnotaspecialsemi supercolumnn-matrixasT2isdividedverticallyalso.IfTisto bespecialsemisupercolumnn-vectoreachmatrixmustbe rectangularm×tmatrixwithm>twithonlyhorizontal partitionofthem×tmatrix.Ifthepartitionisalsoverticaleven forasingleTithenTisnotaspecialsemisupercolumnnvector.

Nowweillustratemajorandminorproductsofsupernmatricesbeforewegivetheabstractdefinition.

T3=

1234 0111 1101 0110 1201 0110

121 111 010 101 211 101 111 202 and T5=

T4=

210 111 011 012 600 101 111 001 010

bethespecialsemicolumn5-matrix.GivenV=V1V2V3 V4V5where V1= 131011 011100

=

131011 1212 011100 3030 131011 1212 011100 1414 1131111 301301 3000101 111111 0102101 2311132311111 1013001010101 1520101522101 123410412342011 011101001111011 110110111011100 011051101100010 1042011 120101012011011 011010101101100 5110010 1211101121101 1110110111010 0100111010110 101101 1101101211211 0110010101101 0111110111111 202202

379

.

246117373 6222121 6253032 1112111 6364043 1112111 143223141231 13222213621 01100101610 12122112611 24233224632 12122112611 13222213621 24224224022 283551232 261111212 66612120666 211220212 272331222 100000101 061111010

WeseetheminorproductofTwithVyieldsasupern-matrix(n =5).Itisnotaspecialsemin-matrixoraspecialsemicolumn n-vectororaspecialsemirown-vector.Itisasupern-matrix(n =5).

Nowwestudyunderwhatconditionsistheproductdefined andiscompatible.WeseeifTandVarespecialsemisupernvectorsthenbothTandVmusthavenelementsi.e.,n-matrices fortheproducttobedefined.SecondlyweneedinbothTiViis

381

compatiblewithrespecttominorproducti.e.,T=T1T2… TnandV=V1V2…Vn,1inwehaveTitobe specialcolumnsupervectorandVitobespecialrowsuper vectorsuchthatthenumberofcolumnsinTi=numberofrows ofVi;1in.Thirdlyweseetheresultantoftheminor productyieldsasupern-matrixandnotasemisupern-matrix providedbothTandVarespecialsupern-vectors.Nowwewill findthemajorproductofaspecialcolumnsupern-matrixwith itstranspose.

Example 4.23: LetT=T1T2T3T4T5beasemisuper column5-vector.TofindtheproductTTT.GivenT=T1T2 T3T4T5 =

30115 340110521 310 120011014 141 010221010 011 110111010 111 110010100 010 100100011 101 104511011 110 110200100 00010 1213111 1111011 1011101 3402111 0111001 1110101 1010111 2010110 1200100

TT=(T1T2T3T4T5)T =T1TT2TT3TT4TT5 T = 3101011 1411101 0111010 31011111 42111001 00000040 10210152 3112110100 0011100100 1500010010 1211101101 5140001100 111301121 210411002 111011110 311210000 101101111 110100110 111111100

.

3101111 34013401 4211100 12001200 0000004 01020102 1021015 110131011011111 110042111001100 100100010010004 104510210451015 3101111 4211100 11021102 0000004 1021015 1 3401 1 1200 0 0102 2 11011 11001 10010 10452 1 1 1102 0 2

31121 00111 3011530115 15000 1052110521 12111 51400 31121 110141101400111 210102101015000 110101101012111 51400 101003110100121 000110000011111 110111511011000 001001200100111 000105100010400 10100 00100 30115 10010 10521 01101 01100 10100 1101400100 2101010010 1101001101 01100 1010010100 0001100100 1101110010 0010001101 0001001100

121111121301112121 111210111411011102 101111101011110110 340340 1113011 011011 2104110 111111 1110111 101101

340 21 011 02 111 10 101 111301121 201201201 210411002 120120120 111011110 311 101 31113111 110 111 311 10111101101 11011110110 111 21112111 100131001 010110101 011110111 011010110 01000100

11 01 10 11

210000 101111 3111 100110 111100 210000 1011101111 1101100110 111100 2111 1001210000 0101101111 0111100110 0110111100 0100 10714134 71856425 1522111 4623122 1411101 3212021 4512112 2611687489 115233113 625312105 83332264 73122112 41221263 81106164211 935423116

3615247446911 153174363452 247194315701 7446421401 4334311301 4612120110 6351102201 9474312401 1500010010 1211101101 6421134235 432723233 222312231 117325473611 321422112 432723233 222312231 333613352 5311123125 1255942321 532421210 523422211 944732321 422321100 212212211 322312321 211201221 101101111

Theresultantisasymmetricsuper5-matrix. Nowweproceedontodefinetheproductofaspecialsemi columnsupern-matrixwithitstranspose.

Example 4.24: LetS=S1S2S3S4bethegiven specialsemisupercolumn4-vector.Tofindtheminorproduct ofSwithST Given S=

211 30101 012 41010 103 50011 901 01100 101 10001 010 3250 30112 1101 11010 0010 01110 1000 11100 0001 00111 0100 01010 1100 01011 0011 01101 1101 10001 1001

aspecialsemisupercolumn4-vector.

ST=(S1S2S3S4)T =S1TS2TS3TS4 T

391

= 211 012 201910 103 110001 901 123110 101 010

.

3010134501 4101001010 5001110010 0110001100 1000110101 3250 1101 0010 10003101001011 00012100011010 01005010000100 11000100100111 0011 1101 1001 30112 11010 01110310100001 11100011101110 00111101110010 01010111011100 01011200010111 01101 10001

=

201910 211211211 110001 012012012 123110 103201031911030 901119010009011 101121013111010 20 01011 12

1910 0100000101 3110 11121614 12182114 16212706 11020 44602

3101000 3011230112 0111011 1101011010 1011100 0111001110 1110111 1110011100 2000101 3101 0011100111 0111 01010010 1011 010111110 2000

000 011 10100 01011111 101 3101000 0111011 0110101101 1011100 1000110001 1110111 2000101 01 30112 10 11010 10 01110 00 11100 11 01 0011110 0101010 0101100 11 01 10 01101 10 10001 00 11

6351931 11121614 356221 12182104 56101240 16212706 1921282100 11020 3241020 44602 110001 38553025553 5301112132 5010000100 3101001011 0100100111 2100011010 5201012021 5110100211 5301112132 3201101122 1542441335 432212211 223222220 422311121 412131221 122112210 322122321 312221231 510110112

.

Weseetheresultantisasymmetricsemisuper4-matrix.One canbythisproductobtainseveralsymmetricsemisuper4matrices.

Thuswecandefinemajorproductorminorproductsincase ofsupern-matrices,n4asincaseofsupertrimatricesand superbimatrices.Thesametypeofoperationsarerepeatedasin caseofsupertrimatricesandsuperbimatrices.Thistypeofsuper n-matriceswillbehelpfulinthefuzzysupermodelapplications whenwehaveamultiexpertopinionwithmultiattributes. Thesematriceswillbebestsuitedfordatastorage.

FURTHER READING

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INDEX

A

Additionofsuperbimatrices,85

B

Bimatrices,7,64-5

C

Columnbimatrix,66

Columnsemisuperbimatrix,91-2

Columnsubvectors,17

Columnsuperbivector,110

Columnsupern-vector(matrix),352

Columnsupertrivector,202

Columnsupertrimatrix,198

D

Diagonalsubmatrices,14

Dualpartition,223-4

H

Heightofasupermatrix,9

M

Majorbyproductoftwosuperbivectors,118-9

MajorproductoftypeIsupervectors,29

MajorproductoftypeIVvectors,58-9

MinorproductofmomentoftypeIVrowvector,45-6

Minorproductofsemisuperbivector,133-5

Minorproductofspecialsemisupertrimatrix,290

Minorproductoftwosemisuperbimatrices,168-9

Minorproductoftwosuperbimatrices,159-160

Minorproductoftwosupervectors,27-8

MinorproductoftypeIIsupervectors,35

MinorproductoftypeIIIcolumnsupervectors,39-40

MinorproductoftypeIIIsupervectors,36-7

MinorproductoftypeIVsupervector,41-2

MinorproductoftypeIVvectors,51-2

Mixedbimatrix,66-7

Mixedrectangularsemisuperbimatrix,91-6

Mixedrectangularsupern-matrix,355-6

Mixedrectangularsupertrimatrix,199

Mixedrectangularsuperbimatrix,82

Mixedsemisuperbimatrix,91-6

Mixedsquarebimatrix,67

Mixedsquaresemisupertrimatrix,217-8

Mixedsquaresemisuperbimatrix,91-6

Mixedsquaresupern-matrix,353

Mixedsquaresupertrimatrix,198

Mixedsquaresuperbimatrix,82

Mixedsupern-matrix,355-6

Mixedsupertrimatrix,215,219

Mixedsuperbimatrix,83

N

n-matrices,74

O

Orderofasupermatrix,9,11

P

Partialtriangularmatrix,16

Partialuppertriangularmatrix,16

Partitioning,10

Q

Quasisymmetricsemisupertrimatrix,274

Quasisymmetricsemisuperbimatrix,103-4

Quasisymmetricsupertrimatrix,249

R

Rectangularsupertrimatrix,199

Rectangularbimatrix,66

Rectangularsemisuperbimatrix,92-3

Rectangularsupern-matrix,355-6

Rectangularsuperbimatrix,82

Rowsupertrivector,202

Rowbimatrix,66

Rowsemisuperbimatrix,91-2

Rowsuperbivector,110

Rowsupertrimatrix,198

S

Semisuperbivector,127

Semisupern-matrix,367

Semisupertrimatrix,214,266

Semisuperbimatrix,91

Simplediagonalmatrix,14-5

Simplematrix,8

Specialcolumnsemisupertrimatrix(vector),222-3,280

Specialrowsemisupertrimatrix(vector),222-4,282

Specialsemisupercolumnn-vector(matrix),372

Specialsemisuperrown-vector(matrix),372

Specialsupercolumnn-vector(matrix),356-7

Specialsuperrown-vector(matrix),356-7

Squarebimatrix,66

Squaresemisuperbimatrix,91-2

Squaresimplematrix,12-3

Squaresupern-vector(matrix),353

Squaresupertrimatrix,198

Submatrices,9-11

Supercolumnbimatrix,80

Superdiagonalmatrix,15

Supern-vector(matrix),349

Superrowbimatrix,80

Squaresuperbimatrix,81

Superrown-vector(matrix),352

Supertrimatrix,195-7

Superbimatrices,77

Supermatrices,7-11

Supervector(Supervector),17

Symmetricpartitioning,12-3

Symmetricsimplematrix,13

Symmetricsupern-matrix,367

Symmetricsupertrimatrix,216

Symmetricsuperbimatrix,101-2

Symmetricsupermatrix,101

Symmetricallypartitionedmatrix,12-3

Symmetricallypartitionedsymmetricsimplematrix,24

T

Transposeofasupertrimatrix,213

Transposeofasuperbimatrix,85

Transposeofasupermatrix,23

TypeIcolumnsupervector,17

TypeIIcolumnsupervector,18

TypeIIrowsupervector,18

TypeIIsupervector,18

TypeIIIsupervector,20

W

Widthofasupermatrix,9,11

ABOUT THE AUTHORS

Dr.W.B.Vasantha Kandasamy is an Associate Professor in the Department of Mathematics, Indian Institute of Technology Madras, Chennai. In the past decade she has guided 12 Ph.D. scholars in the different fields of non-associative algebras, algebraic coding theory, transportation theory, fuzzy groups, and applications of fuzzy theory of the problems faced in chemical industries and cement industries.

She has to her credit 646 research papers. She has guided over 68 M.Sc. and M.Tech. projects. She has worked in collaboration projects with the Indian Space Research Organization and with the Tamil Nadu State AIDS Control Society. This is her 41st book.

On India's 60th Independence Day, Dr.Vasantha was conferred the Kalpana Chawla Award for Courage and Daring Enterprise by the State Government of Tamil Nadu in recognition of her sustained fight for social justice in the Indian Institute of Technology (IIT) Madras and for her contribution to mathematics. (The award, instituted in the memory of Indian-American astronaut Kalpana Chawla who died aboard Space Shuttle Columbia). The award carried a cash prize of five lakh rupees (the highest prize-money for any Indian award) and a gold medal. She can be contacted at vasanthakandasamy@gmail.com You can visit her on the web at: http://mat.iitm.ac.in/~wbv

Dr. Florentin Smarandache is a Professor of Mathematics and Chair of Math & Sciences Department at the University of New Mexico in USA. He published over 75 books and 150 articles and notes in mathematics, physics, philosophy, psychology, rebus, literature.

In mathematics his research is in number theory, nonEuclidean geometry, synthetic geometry, algebraic structures, statistics, neutrosophic logic and set (generalizations of fuzzy logic and set respectively), neutrosophic probability (generalization of classical and imprecise probability). Also, small contributions to nuclear and particle physics, information fusion, neutrosophy (a generalization of dialectics), law of sensations and stimuli, etc. He can be contacted at smarand@unm.edu