122
RANK OF A MATRIX
7.4.2
Subspaces
It turns out that the span of a set of vectors is something called a subspace. We will now give a different, easier to remember description of subspaces and will then show that every subspace is the span of a set of vectors. Definition 7.4.8 Let V be a nonempty collection of vectors in Fn . Then V is called a subspace if whenever α, β are scalars and u, v are vectors in V, the linear combination, αu + βv is also in V . Theorem 7.4.9 V is a subspace of Fn if and only if there exist vectors of V, {u1 , · · · , uk } such that V = span (u1 , · · · , uk ) . Proof: Pick a vector of V, u1 . If V = span {u1 } , then stop. You have found your list of vectors. If V 6= span (u1 ) , then there exists u2 a vector of V which is not a vector in span (u1 ) . Consider span (u1 , u2 ) . If V = span (u1 , u2 ) , stop. Otherwise, pick u3 ∈ / span (u1 , u2 ) . Continue this way. Note that since V is a subspace, these spans are each contained in V . The process must stop with uk for some k ≤ n since otherwise, the matrix ¡ ¢ u1 · · · uk having these vectors as columns would have n rows and k > n columns. Consequently, it can have no more than n pivot columns and so the first column which is not a pivot column would be a linear combination of the preceding columns contrary Pk Pk to the construction. For the other half, suppose V = span (u1 , · · · , uk ) and let i=1 ci ui and i=1 di ui be two vectors in V. Now let α and β be two scalars. Then α
k X i=1
ci ui + β
k X
di ui =
i=1
k X
(αci + βdi ) ui
i=1
which is one of the things in span (u1 , · · · , uk ) showing that span (u1 , · · · , uk ) has the properties of a subspace. This proves the theorem. The following corollary also follows easily. Corollary 7.4.10 If V is a subspace of Fn , then there exist vectors of V, {u1 , · · · , uk } such that V = span (u1 , · · · , uk ) and {u1 , · · · , uk } is linearly independent. Proof: Let V = span (u1 , · · · , uk ) . Then let the vectors {u1 , · · · , uk } be the columns of the following matrix. ¡ ¢ u1 · · · uk Retain only the pivot columns. That is, determine the pivot columns from the row reduced echelon form and these are a basis for span (u1 , · · · , uk ) . The message is that subspaces of Fn consist of spans of finite, linearly independent collections of vectors of Fn . The following fundamental lemma is very useful. Lemma 7.4.11 Suppose V is a subspace of Fn and {x1 , · · · , xr } is a linearly independent subset of V while V = span (y1 , · · · , ys ) . Then s ≥ r. In words, spanning sets have at least as many vectors as linearly independent sets. Proof: Since {y1 , · · · , ys } is a spanning set, there exist scalars cij such that xj =
s X i=1
cij yi