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Let $A$ be and $n\times n$ matrix over $\mathbb{R}$ or $\mathbb{C}$. Suppose $B$ is an $n\times n$ matrix over the same field as $A$ and $B\vec{x}=A\vec{x}$ for all $\vec{x} \in \mathbb{R}^n$. Prove that $A=B$, that is $a_{ij}=b_{ij}$, for all $i,j$.

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This looks like homework. What have you tried? – Qiaochu Yuan Oct 27 '10 at 12:51
Well, $\mathbf{x}=\mathbf{I}\mathbf{x}=\sum x_i \mathbf{e}_i$... ($x_i$ is the i-t component of $\mathbf{x}$, and $\mathbf{e}_i$ is the i-th axis vector). Seriously, what exactly have you tried to solve this problem? You don't even seem to be making any effort. – J. M. Oct 27 '10 at 12:51
You should try to see what happen when you multiply your matrix A with a basis vector. And by "basis vector" I suggest you try with for example v = (1, 0, 0, ... , 0). – Djaian Oct 27 '10 at 13:20
up vote 1 down vote accepted

You don't even need to look at the coordinate entries of your matrices. If E and F are vector spaces, and A and B are linear maps from E to F such that for all x in E: Ax=Bx, then by linearity (A-B)x = 0 for all x. Now what can you say about the linear map A-B, and what does that imply about A and B?

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Depending on your definition of equality of linear maps it is necessary to look at the coefficients. The analogous statement for polynomial functions from F_p to F_p, for example, is false. – Qiaochu Yuan Oct 27 '10 at 13:30
because linear map A-B is identically zero then it implies that (a_{ij}=b_{ij}, forall i,j. – alvoutila Oct 27 '10 at 13:34

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