On the proof of $\langle T(v),v \rangle = 0$ for all $v \in V \iff T(v)=0$ for all $v \in V $

Question

For a complex inner product space $V$ and a linear map $T:V \to V$ $$\langle T(v),v \rangle = 0 \text{ for all } v \in V \iff T(v)=0 \text{ for all } v \in V $$

One proof makes use of the identity

$$ \begin{align} \langle T(u),w \rangle = & \frac{\langle T(u+w),u+w \rangle - \langle T(u-w),u-w \rangle}{4} \\ &+ \frac{\langle T(u+iw),u+iw \rangle - \langle T(u-iw),u-iw \rangle}{4}i \end{align}$$

and observing each term on the right is of the form $\langle T(v),v \rangle$.

Where does the above identity come from? Is there an easy way to see it is true? Are there any proofs that don't rely on this identity?

I verified the identity by expanding the RHS but didn't recieve any insight.

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Reference

Axler, S. (2015). Linear Algebra Done Right. (Third ed.). Theorem 7.14 (p. 210). New York: Springer

Answer 1

The identity is the complex polarization identity.