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Let $V$ be a complex innerproduct space and $T$ be a linear operator on $V$.

If $\langle T(v),v \rangle =0$ for every $v \in V$, show that $T$ is a zero operator.

I figured out that the only eigenvalue of $T$ is zero, which implies that the monic polynomial of $T$ is $x^n$ for some $n$. I have to prove that $n=1$, but I don't know how to proceed.

I also considered Jordan form and the basis $B=\{v_1, v_2, ...,v_n\}$ such that $[T]_B^B$ is the Jordan form of $T$, but it doesn't work.

Does anyone have ideas?

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    So the inner product space is assumed finite dimensional? Turns out it's true even in infinite dimensions. See http://math.stackexchange.com/questions/315942/prove-that-there-is-no-non-zero-linear-operator-on-c2-such-that-alpha?noredirect=1&lq=1, http://math.stackexchange.com/questions/524970/if-the-expectation-langle-v-mv-rangle-of-an-operator-is-0-for-all-v-is-t?noredirect=1&lq=1, – 2017-02-04
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    Thanks a lot. Then what if $T$ is self-adjoint and $T^k(v)=0$ for some $k>1$ and $v\in V$? – 2017-02-04
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    (next to the above comment)It might not need to be $T=0$, but $T(v)=0$ in this case? – 2017-02-04
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    That is true but seems a separate question. – 2017-02-04

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