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Suppose $V$ is an inner product space and $T \in \mathcal{L}(V)$. Suppose $\langle Tv, w \rangle = \langle v, w \rangle$ for all $v, w \in V$. Does this imply that $T = I$?

EDIT: Thanks for the replies. In that case, does $\langle Tv, w \rangle = \langle Sv, w \rangle$ imply that $T = S$? Because:

$ \hspace{0.33 in} \langle Tv, w \rangle = \langle Sv, w \rangle$ for all $v, w \in V$

$ \Longrightarrow \langle Tv-Sv, w \rangle = 0$ for all $v, w \in V$

$ \Longrightarrow Tv-Sv = 0$ for all $v \in V$

$ \Longrightarrow T-S = 0$

$ \Longrightarrow T=S$

And the same should hold for $\langle v, Tw \rangle = \langle v, Sw \rangle$, because of additivity in the right slot as well, right?

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Yes. Just take $w=Tv-v$ to see that $\|Tv-v\|^{2}=0$ so $Tv=v$.

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$\langle Tv,w\rangle-\langle v,w\rangle=\langle Tv-v,w\rangle=0$ for any $v,w\in V$. Hence $Tv-v=0$ for any $v\in V$ hence $T=Id$.

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Yes. $$\forall v,w\quad \langle Tv,w\rangle = \langle v,w\rangle \Leftrightarrow \\ \forall v,w\quad \langle (T-I)v,w\rangle = 0 \Leftrightarrow \\ \forall v\quad (T-I)v = 0 \Leftrightarrow \\ T-I =0$$

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