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A question on my linear algebra assignment says:

Let $A\in M_n(\Bbb C)$ and assume $\langle Ax,x\rangle\in\Bbb R\ \forall x\in\Bbb C^n$, where $\langle\cdot,\cdot\rangle$ denotes the complex dot product. Show that $A=A^*$, i.e $A$ is Hermitian, where $A^*$ is the adjoint of $A$.

Can someone tell me if what I have done in the following is correct? The proof in the solution is different from mine, so I am curious if what I have done is right.

Since $\langle Ax,x\rangle\in\Bbb R\ \forall x\in\Bbb C^n,\langle Ax,x\rangle$ is equal to its complex conjugate, which, using the conjugate symmetry of the inner product, means $\langle Ax,x\rangle =\langle x,Ax\rangle =\langle A^* x,x \rangle$.

So, $0=\langle Ax,x\rangle -\langle A^* x,x\rangle=\langle(A-A^*)x,x\rangle$ using the linearity in the first vector of the inner product. But if we call $(A-A^*)x=v$, then we have $\langle v,x\rangle=0\ \forall x\in\Bbb C^n$. In particular, if $x=v$, then $\langle v,v\rangle=0$, which happens precisely when $v$ is the zero vector. Hence $(A-A^*)x=0\ \forall x\in\Bbb C^n$, which means that $A-A^*$ must be the zero matrix, and thus $A=A^*$, so $A$ is Hermitian.

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  • $\begingroup$ You are wrong. Vector v is not independent of x, so you can not assume "if x = v". $\endgroup$
    – Zerox
    Apr 20, 2020 at 13:33
  • $\begingroup$ em...Something seems not true since $<(A-A*)x,x>=0$ can't ensure the following results directly. Maybe something could achieve this, but this is in you assignment. So you have to fill this gap. $\endgroup$
    – Weijun Yin
    Apr 20, 2020 at 13:41
  • $\begingroup$ For example, notice that if $A-A^*$ were a pure 90 degree rotation, $\langle (A-A^*)x,x \rangle$ would also be identically $0$ but $A-A^*$ as a matrix wouldn't be zero. $\endgroup$
    – Ian
    Apr 20, 2020 at 13:44

2 Answers 2

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$$<Ax|x> \implies (xA)^{T,*}.x= A^{T*}~~ x^{T,*}. x =A^{T*} <x|x>~~~~(1)$$ Next $$<x|Ax>= x^{T*}. A x= A<x|x>~~~~(2)$$ $<AX|x>$ being real means Eq. (1) and (2), are identical, which means $$A^{T*}=A^\dagger =A$$ implying that $A$ is Hermitian.

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  • $\begingroup$ ...What is going on in that first line? That seems to be saying that a number is equal to a matrix which is extremely wrong. $\endgroup$
    – Ian
    Apr 20, 2020 at 16:17
  • $\begingroup$ Indeed $xA$ makes no sense, what you wanted to say is $\langle Ax,x \rangle = (Ax)^* x$ where $*$ is the conjugate transpose (that $T*$ notation is something I have never seen before), but that does not do what you wanted to do here, since you arrive at $x^* A^* x$. $\endgroup$
    – Ian
    Apr 20, 2020 at 16:25
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No, it is not correct. It is fine until you assert that $\bigl\langle(A-A^*)x,x\bigr\rangle=0$ for each $x\in\Bbb C^n$. But then, if you say that $v=(A-A^*)x$, you must keep in mind that $v$ is a function; it depends on $x$. It would be more clear to call it $v(x)$. But then, what you roved was that $(\forall x\in\Bbb C^n):\bigl\langle v(x),x\bigr\rangle=0$ and, just from this, you cannot deduce that $(\forall x\in\Bbb C^n):v(x)=0$.

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