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Suppose we've got two linear maps $\ X:\mathcal{H}\rightarrow\mathcal{H}\ $ and $\ Y:\mathcal{H}\rightarrow\mathcal{H}$, where $\mathcal{H}$ is some finite-dimensional Hilbert space. Let's say $\mathrm{dim}(\mathcal{H})=n$ and we've got some inner product $< \bullet , \bullet >$.

Somebody hands me basis for $\mathcal{H}$. I'll write this as $\begin{Bmatrix} \mathbf{v}_{j} \end{Bmatrix}^{n}_{j=1} \subset \mathcal{H} $.

He tells me that the operators $X$ and $Y$ have the property:

$<\mathbf{v}_{j},X(\mathbf{v}_{k})>=<\mathbf{v}_{j},Y(\mathbf{v}_{k})>$

.....for all $j,k$ (with respect to this basis he's handed me).

Is it true then that $X=Y$? I believe the answer is yes.

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    $\begingroup$ If $x=\sum a_iv_i$ then, multiplying by $\overline{a_i}$ and adding, $\langle v_j,X(x)\rangle=\langle v_j,X(\sum a_iv_i)\rangle=\langle v_j,Y(\sum a_iv_i)\rangle=\langle v_j,Y(x)\rangle$. Therefore, $\langle v_j,X(x)-Y(x)\rangle=0$. $\endgroup$ – Nathanson Mar 27 '15 at 2:02
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    $\begingroup$ If $X(x)-Y(x)=\sum b_j v_j$, then multiplying by $b_j$ and adding, $0=\langle X(x)-Y(x),X(x)-Y(x)\rangle=\|X(x)-Y(x)\|$. Hence $X(x)-Y(x)=0$. $\endgroup$ – Nathanson Mar 27 '15 at 2:04
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Fixed $k$, for all $j$, we have $\langle v_j,X(v_k)\rangle=\langle v_j,Y(v_k)\rangle \Leftrightarrow \langle v_j,X(v_k)\rangle - \langle v_j,Y(v_k)\rangle =0 \Leftrightarrow \langle v_j,(X-Y)(v_k)\rangle=0$. But this implies that $(X-Y)(v_k)$ is orthogonal at all $v_j$, hence $(X-Y)(v_k)=0$. That is, $X$ and $Y$ coincide in a base. Therefore, $X=Y$.

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