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!

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    $\begingroup$ What you are trying to prove follows if you can show that $M = A - A^T = 0$. $\endgroup$ – Winther Dec 11 '16 at 23:20
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    $\begingroup$ Your attempt assumes that $x$ has the same eigenvalue for $A^t$ that it has for $A$, which is not given as a hypothesis. $\endgroup$ – Gerry Myerson Dec 23 '16 at 20:48
  • $\begingroup$ @GerryMyerson, your observation is correct. Although $A,~A^t$ have the same eigenvalues, they do not need to correspond to the same eigenvectors. $\endgroup$ – Nikolaos Skout Dec 23 '16 at 20:55
  • $\begingroup$ $A$ and $A^{t}$ have the same eigenvalues. Additionally, if both have the same eigenvectors, then $A = A^{t}$. $\endgroup$ – Alex Silva Dec 23 '16 at 21:03
  • $\begingroup$ @Alex, that's what we're meant to prove. It may work to use the argument in math.stackexchange.com/questions/1762563/… $\endgroup$ – Gerry Myerson Dec 23 '16 at 21:09

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.

  • $\begingroup$ It works. Thank you! +50 $\endgroup$ – Nikolaos Skout Dec 24 '16 at 15:52

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