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?

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 $Ax=\lambda x$.
