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As the title says the goal is to prove: If $X^*AX\in \mathbb{R}$ for all $X$, then $A$ is Hermitian.

This is my attempt:

$X^*AX=P\in \mathbb{R}$ then $(X^*AX)^*=(AX)^*X=X^*A^*X=P^*=P^t$. Furthermore, let $X$ be invertible, then $A=(X^*)^{-1}PX^{-1}$ and $A^*=(X^*)^{-1}P^tX^{-1}$. If I can show that $P=P^t$, then I am done, but I haven't been able to do that. I suspect one has to choose the right $X$ (since the above it is true for all invertible $X$.

I would appreciate if someone can give me a hint on how to go on or a different approach to prove the statement. Ideally, I don't want to invoke theorems. Thanks!

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    $\begingroup$ If $X$ is $n\times 1$, $A$ is $n\times n$ then $X^*AX$ is $1\times 1$, i.e. it is in $\Bbb R$. When you say $P = X^*AX$, it goes without saying that $P^t = P$ and $P^* = P$, since $P$ is just a number! $\endgroup$ Mar 16, 2021 at 2:50
  • $\begingroup$ How can $X$ be invertible if it's not square? If it is square then how can $X^{\ast}AX$ be a scalar? $\endgroup$
    – anon
    Mar 16, 2021 at 3:03
  • $\begingroup$ @runway44 that's a good point. Then the argument presented won't work. $\endgroup$
    – Schach21
    Mar 16, 2021 at 3:07

1 Answer 1

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First we show that if $\langle AX, X \rangle =0 ~~~\forall ~X$ then $A\equiv0$. It suffices to show that $\langle AX, Y \rangle =0~~~\forall ~X,Y$ , because \begin{align} &\langle AX, Y \rangle =0~~~\forall~ Y \\ &\Rightarrow AX=0 ~~~\forall~ X \\ &\Rightarrow A \equiv0 \end{align} Let $\alpha, \beta \in \mathbb{C}$ be arbitrary, by hypothesis $\forall ~X,Y$ \begin{align} &\langle A(\alpha X+\beta Y),\alpha X+\beta Y \rangle =0\\ &\Rightarrow |\alpha|^2 \langle AX,X \rangle + |\beta|^2 \langle AY,Y\rangle+\alpha\bar\beta\langle AX,Y\rangle+ \bar\alpha\beta\langle AY,X \rangle=0 \\ &\Rightarrow \alpha\bar\beta\langle AX,Y\rangle+ \bar\alpha\beta\langle AY,X \rangle=0 \end{align} Now putting $\alpha=\beta=1$ then $\alpha=i,~\beta=1$ in the above equation we get two equations in two unknowns, solving we get $\langle AX,Y \rangle=0~~~\forall~X,Y$ this implies $A\equiv 0$. Thus we get $\langle AX, X \rangle =0~~~\forall~X \Rightarrow A\equiv0$.

Note that $\overline{\langle X, AX \rangle}=\langle X, AX \rangle$. \begin{align} &\overline{\langle X, AX \rangle}=\langle AX, X\rangle ~~~\forall ~X\\ &\Rightarrow \langle X, AX \rangle =\langle X, A^*X \rangle ~~~\forall ~X\\ &\Rightarrow \langle X, (A-A^*)X \rangle =0 ~~~\forall ~X \\ &\Rightarrow (A-A^*)X=0 ~~~\forall ~X \\ &\Rightarrow A=A^* \end{align}

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  • $\begingroup$ why is $\overline{\langle X, AX \rangle}=\langle X, AX \rangle$ true? $\endgroup$
    – Schach21
    Mar 16, 2021 at 4:25
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    $\begingroup$ By assumption, it is a real number. $\endgroup$ Mar 16, 2021 at 4:30

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