Topics In Analysis

4.5 An Application To Matrices

The following is a theorem of major significance.

Proof: First suppose A is one to one. Consider the vectors,

where e_k is the column vector which is all zeros except for a 1 in the k^th position. This set of vectors is linearly independent because if

n ∑ ckAek = 0, k=1

then since A is linear,

( ) ∑n A ckek = 0 k=1

and since A is one to one, it follows

3 ∑n 3 This follows because A0 =A (0 +0)= A0 +A0 and so A0 =0.Therefore, since A is one to ckek = 0 one, this conclusion follows. k=1

which implies each c_k = 0. Therefore,

must be a basis for Fⁿ because if not there would exist a vector, yspan and then by Lemma 4.4.10, would be an independent set of vectors having n + 1 vectors in it, contrary to the exchange theorem. It follows that for y ∈ Fⁿ there exist constants, c_i such that

( ) ∑n ∑n y = ckAek = A ckek k=1 k=1

showing that, since y was arbitrary, A is onto.

Next suppose A is onto. This means the span of the columns of A equals Fⁿ. If these columns are not linearly independent, then by Lemma 4.4.2 on Page 150, one of the columns is a linear combination of the others and so the span of the columns of A equals the span of the n − 1 other columns. This violates the exchange theorem because

would be a linearly independent set of vectors contained in the span of only n − 1 vectors. Therefore, the columns of A must be independent and this equivalent to saying that Ax = 0 if and only if x = 0. This implies A is one to one because if Ax = Ay, then A = 0 and so x − y = 0.

Now suppose AB = I. Why is BA = I? Since AB = I it follows B is one to one since otherwise, there would exist, x≠0 such that Bx = 0 and then ABx = A0 = 0≠Ix. Therefore, from what was just shown, B is also onto. In addition to this, A must be one to one because if Ay = 0, then y = Bx for some x and then x = ABx = Ay = 0 showing y = 0. Now from what is given to be so, it follows

A = A and so using the associative law for matrix multiplication,

A (BA )− A = A(BA − I) = 0.

But this means

x = 0 for all x since otherwise, A would not be one to one. Hence BA = I as claimed. This proves the theorem.

This theorem shows that if an n×n matrix, B acts like an inverse when multiplied on one side of A it follows that B = A⁻¹and it will act like an inverse on both sides of A.

The conclusion of this theorem pertains to square matrices only. For example, let

( ) 1 0 ( 1 0 0 ) A = ( 0 1 ) ,B = 1 1 − 1 (4.5.18) 1 0

(4.5.18)

Then

( ) BA = 1 0 0 1

but

( ) 1 0 0 AB = ( 1 1 − 1 ) . 1 0 0