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Let $A$ be a Hermitian square complex matrix of size $n$ then $\langle Ax,x\rangle=0$ for all $x \in \Bbb C^n$. Then $A=0$

I need a hint to prove it, if possible

Thank you

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    no that was a different question – 2017-01-10
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    Why was it different? It implies this one, right? A hint for a proof was given there, by the way. I will copy it to this question now. – 2017-01-10
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    In that question I asked if the condition that A is Hermitian is really necessary but that's it.. after trying to prove it myself I did one direction and couldn't do the other so I asked for a hint in this question – 2017-01-10
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    oh so sorry I didn't know that people could still post answers after I accept one.. so I didn't check for other answers in that post – 2017-01-10
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    I really didn't ask for a proof there so why would I have rechecked after accepting the answer to my question – 2017-01-10

2 Answers 2

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If $\lambda$ is an eigenvalue of $A,$ there exists $x\ne 0$ vector of $\mathbb{C}^n$ such that $Ax=\lambda x$. So, $$0=\langle Ax,x\rangle=\langle \lambda x,x\rangle=\lambda\langle x,x\rangle=\lambda \underbrace{\left\|{x}\right\|^2}_{\ne 0}\Rightarrow \lambda=0.$$ As $A$ is hermitian, there exists an unitary matrix $U\in\mathbb{C}^{n\times n}$ such $A=U^*\underbrace{D}_{0}U$. So, $A=0.$

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    Very good! @Sweetstack, this is exactly the former answer to the duplicate question (deleted now there). – 2017-01-10
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    oh ok thank you so much @DietrichBurde – 2017-01-10
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Hint

1) All Hermitian matrix can be diagonalized;

2) if $\lambda$ is a eigenvalue then $=\lambda |x|^2=0 \rightarrow \lambda =0$.

Can you finish?

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    @ Dietrich Burde: Your answer was published almost at the same time than mine. (+1) – 2017-01-10