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Let $A$ be a Hermitian square complex matrix of size $n$ then $\langle Ax,x \rangle=0$, $\forall x \in \Bbb C^n$ if and only if $A=0$

I want to know if the condition that $A$ is Hermitian is required for the statement to still hold, please?

Thank you

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    For prettier parentheses, use `\langle` and `\rangle` (producing $\langle Ax, x\rangle$) instead of `<` and `>` (producing $$) – 2017-01-10

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Yes, the condition is required, as $$A=\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$$ shows.

For the matrix above, you can see that if $$z=\begin{bmatrix}z_1\\ z_2\end{bmatrix}$$

then $$\langle Az, z\rangle = \left\langle \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}\cdot \begin{bmatrix}z_1\\ z_2\end{bmatrix}, \begin{bmatrix}z_1\\ z_2\end{bmatrix}\right\rangle = \left\langle\begin{bmatrix}z_2\\ -z_1\end{bmatrix}, \begin{bmatrix}z_1\\ z_2\end{bmatrix}\right\rangle = z_2z_1 - z_1z_2 = 0$$

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    but A in this counter-example is not a complex matrix it's a real matrix – 2017-01-10
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    @Sweetstack Of course it is a complex matrix. $1$ is a complex number, isn't it? Or are you saying that $1\neq \mathbb C$? – 2017-01-10
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    @Sweetstack Why do you think $A$ is hermitian? – 2017-01-10
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    oh alright I am sorry your counter-example neglects the statement if A isn't hermition thanks sir – 2017-01-10
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    @Sweetstack I don't understand that sentence. But you're welcome :) – 2017-01-10
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    I mean that yeah the counter-example works I wasn't thinking straight with my other comments earlier – 2017-01-10
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    @Sweetstack Oh, OK. Happy to help! – 2017-01-10
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    Do we know that weaker condition would not suffice? – 2017-01-10