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I have the following question: "Let $V$ be a Hilbert space and let $T$ be a linear operator on $V$. If $S$ is any linear operator on $V$ that satisfies $\langle Tv,w \rangle = \langle Sv,w \rangle$ for all $v,w \in V$, then $S = T$."

My attempt at the problem went something like this: Notice that the equality $\langle (T-S)v,w \rangle = \langle Tv,w \rangle - \langle Sv,w \rangle = 0$ holds for all $v,w \in V$. If $S \neq T$, then $T-S$ is not the zero linear operator and there exists some $v_0 \in V$ such that $(T-S)v_0 = w_0 \neq 0$. But then $0 < \langle (T-S)v_0,w_0 \rangle$ by the properties of the inner product and this is a contradiction.

I have been informed that this argument is incorrect, but cannot seem to find the flaw, nor a correct proof. Any help would be appreciated.

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Your argument is not incorrect: you just have to write $\langle (T-S)v_0,w_0\rangle$ as $\lVert (T-S)v_0\rVert^2$. –  Davide Giraudo Oct 14 '12 at 18:38
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up vote 0 down vote accepted

Fix $v\in V$. Taking $w:=(T-S)v$, we get that $\langle (T-S)v,(T-S)v\rangle=0$ hence $(T-S)v=0$ by properties of positive definiteness of an inner product.

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So it looks like my argument was correct after all, just not particularly elegant. Thanks! –  James Miller Oct 14 '12 at 18:39
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