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Let $A$ and $B$ are nxn matrices and $x \in C^{n}$. If $Ax = Bx$ for all $x$ then $A = B$. To prove this I have selected $x$ from Euclidean basis B = {$e_{1},e_{2},...,e_{n}$}.
Then $Ae_{i} = Be_{i}$ implies $i^{th}$ column of A = $i^{th}$ column of B for all $1 \leq i \leq n$.
Hence $A = B.$ Is this proof complete?

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Yep, that's pretty much it. – Qiaochu Yuan May 29 '11 at 13:59
Yes, this is perfectly correct, complete and probably the easiest way of doing it. You could also argue that $(A-B)x = 0$ for all $x$ implies $A - B = 0$ but this boils down to the same. – t.b. May 29 '11 at 14:00
A remark: What this problem shows is that the map from $n$-by-$n$ matrices to linear transformations on $\mathbb C^n$, given by $A\mapsto(x\mapsto Ax)$, is injective. – Jonas Meyer Dec 28 '11 at 5:05
up vote 1 down vote accepted

For the sake of an answering: Yes, your proof is correct.

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