# Proof to show that a matrix $A$ is Hermitian

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.

• You are wrong. Vector v is not independent of x, so you can not assume "if x = v". Apr 20, 2020 at 13:33
• 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. Apr 20, 2020 at 13:41
• 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.
– Ian
Apr 20, 2020 at 13:44

$$ \implies (xA)^{T,*}.x= A^{T*}~~ x^{T,*}. x =A^{T*} ~~~~(1)$$ Next $$= x^{T*}. A x= A~~~~(2)$$ $$$$ being real means Eq. (1) and (2), are identical, which means $$A^{T*}=A^\dagger =A$$ implying that $$A$$ is Hermitian.
• Indeed $xA$ makes no sense, what you wanted to say is $\langle Ax,x \rangle = (Ax)^* x$ where $*$ is the conjugate transpose (that $T*$ notation is something I have never seen before), but that does not do what you wanted to do here, since you arrive at $x^* A^* x$.
No, it is not correct. It is fine until you assert that $$\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$$.