Solution verification: Let $A$ be a matrix; $A^TA=AA^T$. If $x$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A^Tx=\lambda x$ Here is the proposition I wish to show holds true:

Let $A$ be an $n \times n$ matrix such that $A^TA=AA^T$. Show that if $x$ is an eigenvector of matrix $A$ with eigenvalue $\lambda$, then $x$ is an eigenvector of $A^T$ with eigenvalue $\lambda$. $\it {\text{Hint: How are the expressions}}$ $||Ax-\lambda x||$ $\it{\text{and}}$ $||A^Tx-\lambda x||$ $\it{\text{related?}}$


$\textbf{Solution}$: First, we assume that $x$ is an eigenvector of $A$ with eigenvalue $\lambda$ such that $Ax= \lambda x$. Using the hint, if $x$ is an eigenvector of $A^T$ with the same eigenvalue, we require $A^Tx=\lambda x$. Thereby, if this is the case, then $||Ax-A^Tx||=0$. We try to show this by the following:
$$
\begin{align}
||Ax-A^Tx||^2 & = (Ax-A^Tx)^T(Ax-A^Tx) \\
&=(x^TA^T-x^TA)(Ax-A^Tx) \\
&=x^TA^TAx-x^TAAx-x^TA^TA^Tx+x^TAA^Tx
\end{align}
$$
As $A^TA=AA^T$, we have that $x^TA^TAx=x^TAA^Tx=||Ax||^2$.
$$
\begin{align}
||Ax-A^Tx||^2 & = 2||Ax||^2-x^TAAx-x^TA^TA^Tx \\
&= 2||Ax||^2 -x^TA \lambda x -(Ax)^TA^Tx \\
&= 2||Ax||^2 -x^TA \lambda x -(\lambda x)^TA^Tx \\
&=2||Ax||^2 -x^TA \lambda x -\lambda x^TA^Tx \hspace{3mm}(\text{as}\hspace{1mm} \lambda^T=\lambda) \\
&=2||Ax||^2 -\lambda x^TA  x - x^TA^T \lambda x \\
&=2||Ax||^2 -(\lambda x)^TA  x - x^TA^T (\lambda x) \\
&=2||Ax||^2 -(Ax)^TA  x - x^TA^T (Ax) \\ 
&=2||Ax||^2 -||Ax||^2 - ||Ax||^2 \\
&=0
\end{align}
$$
This shows that vector $(Ax-A^Tx)$ is orthogonal to itself, i.e it must be the zero vector. Hence $Ax=A^Tx$ and since $Ax=\lambda x$, we have that $\lambda x = A^T x$ and we are done. Is this correct/ any other thoughts on my proof?
 A: Your proof is correct, however we must stress that $A$ is a real matrix. You can make the proof a bit simpler:

*

*First prove that if $A$ is normal ($AA^T = A^TA$), then $\lVert Ax\rVert = \lVert A^Tx\rVert$.

*Show that $A + \lambda I$ is normal for all $\lambda$

*$0 = \lVert (A - \lambda I)x\rVert = \lVert (A - \lambda I)^Tx\rVert = \lVert (A^T - \lambda I)x\rVert \implies A^Tx = \lambda x$.

These steps are individually really easy to prove.
A: Shorter argument from the hint:
\begin{aligned}
||A'x-\lambda x||^2&=(A'x-\lambda x)'(A'x-\lambda x)\\
&=x'AA'x-\lambda x'Ax-\lambda x'A'x+\lambda^2x'x\\
&=x'A'Ax-\lambda x'(Ax)-\lambda(Ax)'x+\lambda^2x'x\\
&=(\lambda x)'(\lambda x)-\lambda x'(\lambda x)-\lambda^2x'x+\lambda^2x'x\\
&=(\lambda^2-\lambda^2-\lambda^2+\lambda^2)x'x=0
\end{aligned}
The third line uses $AA'=A'A$ and the fourth line uses $Ax=\lambda x$.
A: Since this question is title and tagged "solution verification", I will permit myself to comment on the form of the presentation of the proof, rather than on the substance of the expression manipulation (you can check for yourself whether the latter is correct).

Solution: First, we assume that $x$ is an eigenvector of $A$ with
eigenvalue $\lambda$ such that $Ax=\lambda x$.

The initial "First assume" suggests you are either doing a case-by-case proof, or that you are doing one direction at a time of a double implication. Neither of these is the case here (and indeed no "second" or "next" ever follows). The assumption is simply the hypothesis of the statement to be proven, so assuming it is almost automatic (only if you want to start by stating and proving some auxiliary result would you not start by assuming the hypothesis).
Also "such that" is usually used to introduce a restricting condition (we can find some $v$ such that...), but that is not the case here; what follows is just a restatement of what the hypothesis means, so "which means that" is more appropriate.
All in all I would reformulate "We are given an eigenvector $x$ of $A$ with eigenvalue $\lambda$, which means that $x\neq0$ and that $Ax=\lambda x$" (you can safely bet the condition $x\neq0$ is going to be used somewhere.)

Using the hint, if $x$ is an eigenvector of $A^T$ with the same eigenvalue,
we require $A^Tx=\lambda x$.

The hint is a question, so what do you mean by using the question without first providing at least a tentative answer? One can surmise that the answer is equality of the two norms (what other relation could one imagine?) but then you still need to say and prove that (the latter will involve invoking the commutation hypothesis). But it does not seem that you are doing anything with the hint here. The rest of the sentence is turning the desired conclusion into a hypothesis (which is rarely a good idea), and then requiring that same conclusion to be true; what does this even mean? In conclusion, this sentence can be greatly improved by removing it.

Thereby, if this is the case, then $\|Ax−A^Tx\|=0$.

It seems you are saying that what needs to be proved is equivalent to $Ax=A^Tx$, and that this will be implied by $\|Ax−A^Tx\|=0$, which you will then establish. But what you say is that $Ax=A^Tx$ implies $\|Ax−A^Tx\|=0$, which is also true, but not the direction that will be useful for your argument.
If your computation succeeds in showing that $\|Ax−A^Tx\|^2=0$ (and it seems to be the case), then you do have the ingredients of a proof. But you have not used the hint, and your argument is not clearly presented.

For the record, I think that the hint was to establish that for $B=A-\lambda I$, which by hypothesis commutes with its transpose, and any vector $x$ one has $$\|Bx\|^2=(Bx\mid Bx)=(x\mid B^TBx)=(x\mid BB^Tx)=(B^T x\mid B^Tx)=\|B^Tx\|^2,$$ so that $Bx=0$ is zero (if and) only if $B^Tx=0$ which gives the result.
