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Let $A$ and $B$ are $n\times n$ 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?

Chris Tang
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Vinod
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    Yep, that's pretty much it. – Qiaochu Yuan May 29 '11 at 13:59
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    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 05:05
  • @JonasMeyer Great remark! When I read the title of the question I thought "Of course - the map $A\mapsto(x\mapsto Ax)$ is injective", without realizing that this is exactly what the OP wants to prove :D – Filippo Feb 01 '21 at 09:24

2 Answers2

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

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Let $\langle x|y\rangle:=\sum_{i=1}^n\overline{x_i}y_i$ be the standard inner product on $\mathbf{C}^n$.

To make the title of the question even more obvious, one can use the fact that $\langle e_i|A\cdot e_j\rangle=A_{ij}$ for all $i,j\in\{1,\ldots,n\}$. Thus, if $A\cdot x=B\cdot x$ for all $x\in\mathbf{R}^n$, $A_{ij}=\langle e_i|A\cdot e_j\rangle=\langle e_i|B\cdot e_j\rangle=B_{ij}$ for all $i,j\in\{1,\ldots,n\}$.

Filippo
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