Upper triangular matrix $A$ is diagonal, if every eigenvector of $A$ is eigenvector of $A^T$. 
Let $A\in \mathbb{R}^{4\times 4}$ upper triangular, such that every eigenvector of $A$ is also an eigenvector of $A^T$ (transpose). Prove that $A$ is a diagonal matrix.

Attempt We should derive $a_{ij}=0,~i\neq j.$ Let $x\neq 0$ be an eigenvetor of A, corresponding to eigenvalue $\lambda.$ Then, by hypothesis, $Ax=A^Tx=\lambda x$, but $(A-A^T)x=0$ does not lead me somewhere.
Thanks in advance for the help!
 A: Perhaps the easiest way to see this is to observe that the first standard basis vector is an eigenvector of the upper triangular matrix $A$. By your hypothesis it must also be an eigenvector of $A^\text{T}$. This forces the first column of $A^\text{T}$ to be $0$ after the first entry. This is the same as saying that the first row of $A$ is $0$ after the first entry.
Now we know that the second standard basis vector of $A$ is also an eigenvector so we may continue in a similar fashion.
I hope that helps. If you have questions about it then do ask.

What is the intuition for my proof?
I know that a standard basis $e_i$ is an eigenvector of $A$ precisely if the $i$th column of $A$ is $0$ except perhaps in the $i$th row. Similarly the $e_i$ is an eigenvector of $A^\text{T}$ precisely if the $i$th row of $A$ is zero except perhaps in the $i$th column. My proof depends only on these two facts and some reflection on the shape of an upper triangular matrix.
My first thought, before I realized the above, was to ask myself what happens when the matrix is in Jordan normal form as I completely understand the eigenvectors in that case. I realized that all the Jordan blocks must have size one in that case and then that the reason for that is the one I gave you.
