TITLE: Eigenvalues and eigenvectors of $A_{1}$ and $ A_{2}=A_{1}^{T}$
QUESTION [0 upvotes]: We have a positive integer $n$ and two $n\times n$ matrices of real numbers, $A_{1}$ and $A_{2}$. For $j=1, 2$, we have the eigenvalues and eigenvectors $\lambda _{j}$ and $x_{j}$ of $A_{j}$. Show that, if $ A_{2}=A_{1}^{T}$, then $\lambda _{1}= \lambda _{2}$ or $x_{1}, x_{2}$ are vertical.
We have that $det(A_{1}-\lambda I_{n})=det(A_{2}-\lambda I_{n})$. Am I right? What can I do to solve this problem? I am really stuck.

REPLY [1 votes]: Hint
For any matriz $C_{n\times n}$ we know that $\det C=\det C^T$. Now, see that
$$(A_1-\lambda I)^T=A_1^T-\lambda I$$
Can you finish?
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