Proof to show that a matrix $A$ is Hermitian

Question

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.

Answer 1

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 $(\forall x\in\Bbb C^n):v(x)=0$.

Answer 2

$$<Ax|x> \implies (xA)^{T,*}.x= A^{T*}~~ x^{T,*}. x =A^{T*} <x|x>~~~~(1)$$ 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.