Eigenvalues and eigenvectors of $A_{1}$ and $ A_{2}=A_{1}^{T}$

Question

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.

Answer 1

Since the determinants are the same, you obtain the same characteristic polynomial for both matrices and therefore yield identical eigenvalues to both.

Answer 2

Let $A_1x_1 = \lambda_1x_1$ and $A_2x_2 = \lambda_2x_2$. Then $$ \lambda_1x_2^Tx_1 = x_2^T(\lambda_1x_1) = x_2^T(A_1x_1) = x_2^TA_1x_1 = (x_2^TA_1)x_1 = (x_2^TA_2^T)x_1 = \lambda_2x_2^Tx_1 $$ which in turn gives us $$ (\lambda_1-\lambda_2)(x_2^Tx_1) = 0 $$ which is what we wanted to show.

Answer 3

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?