The following is a theorem of major significance.
Theorem 4.5.1 Suppose A is an n×n matrix. Then A is one to one if and only if A is onto. Also, if B is an n × n matrix and AB = I, then it follows BA = I.
Proof: First suppose A is one to one. Consider the vectors,
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then since A is linear,
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and since A is one to one, it follows
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which implies each ck = 0. Therefore,
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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
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
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But this means
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
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Then
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but
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