7.5. FREDHOLM ALTERNATIVE
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Corollary 7.5.4 Let A be an m × n matrix. Then A maps Rn onto Rm if and only if the only solution to AT x = 0 is x = 0. ¡ ¢ ¡ ¢⊥ Proof: If the only solution to AT x = 0 is x = 0, then ker AT = {0} and so ker AT = Rm because every b ∈ Rm has the property that b · 0 = 0. Therefore, Ax = b has a solu¡ ¢⊥ tion for any b ∈ Rm because the b for which there is a solution are those in ker AT by Theorem 7.5.3. In other words, A maps Rn onto Rm . ¡ ¢⊥ Conversely if A is onto, then by Theorem 7.5.3 every b ∈ Rm is in ker AT and so if AT x = 0, then b · x = 0 for every b. In particular, this holds for b = x. Hence if AT x = 0, then x = 0. This proves the corollary. Here is an amusing example. Example 7.5.5 Let A be an m × n matrix in which m > n. Then A cannot map onto Rm . The reason for this is that AT is an n × m where m > n and so in the augmented matrix, ¡ T ¢ A |0 there must be some free variables. Thus there exists a nonzero vector x such that AT x = 0.
7.5.1
Row, Column, And Determinant Rank
I will now present a review of earlier topics and prove Theorem 7.3.4. Definition 7.5.6 A sub-matrix of a matrix A is the rectangular array of numbers obtained by deleting some rows and columns of A. Let A be an m × n matrix. The determinant rank of the matrix equals r where r is the largest number such that some r × r sub-matrix of A has a non zero determinant. The row rank is defined to be the dimension of the span of the rows. The column rank is defined to be the dimension of the span of the columns. Theorem 7.5.7 If A, an m × n matrix has determinant rank, r, then there exist r rows of the matrix such that every other row is a linear combination of these r rows. Proof: Suppose the determinant rank of A = (aij ) equals r. Thus some r × r submatrix has non zero determinant and there is no larger square submatrix which has non zero determinant. Suppose such a submatrix is determined by the r columns whose indices are j1 < · · · < j r and the r rows whose indices are i 1 < · · · < ir I want to show that every row is a linear combination of these rows. Consider the lth row and let p be an index between 1 and n. Form the following (r + 1) × (r + 1) matrix ai1 j1 · · · ai1 jr ai1 p .. .. .. . . . air j1 · · · air jr air p alj1 · · · aljr alp Of course you can assume l ∈ / {i1 , · · · , ir } because there is nothing to prove if the lth row is one of the chosen ones. The above matrix has determinant 0. This is because if p∈ / {j1 , · · · , jr } then the above would be a submatrix of A which is too large to have non