# Does $\langle Tv, w \rangle = \langle v, w \rangle$ imply that $T = I$?

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?

## 3 Answers

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

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

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