Can anyone help give an example of a complex matrix $A \in \Bbb C^{n\times n}$ such that $x^HAx > 0$ for any non-zero $x\in \Bbb C^n$ but $A$ is non-Hermitian? Here $x^H$ denotes the conjugate transpose of $x$. Thanks!
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If I recall correctly, when you focus on $\mathbb{C}^n$ this is actually impossible, whereas the analogous statement for $\mathbb{R}^n$ is possible. (At any rate, any example will have to be non-normal...) – Ian Dec 11 '16 at 03:56
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Check this for what Ian mentioned in their comment. – Sangchul Lee Dec 11 '16 at 04:26
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No such example exists. If $x^HAx$ is real for all $x\in\mathbb{C}^n$, then $A$ is Hermitian.
Proof: Let $S=\frac{1}{2}(A+A^*)$ and $T=\frac{1}{2i}(A-A^*)$. Observe that $$S^*=\overline{\left(\frac{1}{2}\right)}(A+A^*)^*=\frac{1}{2}(A^*+A)=S$$ and $$T^*=\overline{\left(\frac{1}{2i}\right)}(A-A^*)^*=-\frac{1}{2i}(A^*-A)=T,$$ that is, $S$ and $T$ are Hermitian, hence $x^HSx$ and $x^HTx$ are real for all $x\in\mathbb{C}^n$. Since $A=S+iT$ we have $$x^HAx=x^HSx+i(x^HTx),$$ so if $x^HAx$ is real then $x^HTx=0$ for all $x\in\mathbb{C}^n$. This implies $T=0$, so $A=S$ is Hermitian.
Heath Winning
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