2
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
+50
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.