Hello, here is the question my book is asking:(adsbygoogle = window.adsbygoogle || []).push({});

Let A, B be two m x n matricies. Assume that AX = BX for all n-tuples X. Show that A = B.

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So I decided to try and prove the contrapositive, which is (unless I am mistaken): If [itex]A \neq B[/itex], then there is some X such that [itex]AX \neq BX[/itex]

Proof:

Assume [itex]A \neq B[/itex]

Then [itex]A^j \neq B^j[/itex] for some j, where [itex]A^j, B^j[/itex] are the j-th columns of A and B.

Then, let [itex] X = E^j[/itex] be the unit vector with 1 in the j-th spot, the same j where [itex]A^j \neq B^j[/itex]

So [itex]AX = AE^j = A^j[/itex] and

[itex] BX = BE^j = B^j[/itex]

and so [itex]AX \neq BX[/itex] for [itex]X = E^j[/itex] as [itex]A^j \neq B^j[/itex]

Thus if [itex]A \neq B[/itex] there is some X such that [itex]AX \neq BX[/itex]

So, as the contrapositive is logically equivalent, we have just showed that if AX = BX for all X, then A = B. Where A, B are two m x n matricies, and X is an n-tuple.

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First, is the contrapositive correct, and if so then is the proof correct. The whole thing looks perfectly sufficient to me. Thanks!!

Also, I just realized that it is very easy to just prove it directly, but I am still curious to see if this is a sufficient proof. Thanks!

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# Lin Alg - Matrix multiplication (Proof by contrapositive)

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