Lin Alg - Matrix multiplication (Proof by contrapositive)

[mattmns]

Hello, here is the question my book is asking:

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 A \neq B, then there is some X such that AX \neq BX

Proof:

Assume A \neq B
Then A^j \neq B^j for some j, where A^j, B^j are the j-th columns of A and B.
Then, let X = E^j be the unit vector with 1 in the j-th spot, the same j where A^j \neq B^j
So AX = AE^j = A^j and
BX = BE^j = B^j
and so AX \neq BX for X = E^j as A^j \neq B^j

Thus if A \neq B there is some X such that AX \neq BX
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!

 

[AKG]

test\pi $test\pi$

 

[0rthodontist]

I would call that sufficient.

 

[Galileo]

The proof is good. Proving directly might be more transparant though. You already noted the crucial AE^j=A^j, so if AX=BX for all X, then A^j=AE^j=BE^j=B^j for j=1,2,...,n. So A=B.

 

[mattmns]

Yep, I saw that just as I was posting my proof. Thanks.