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

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.

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.
For any matriz $$C_{n\times n}$$ we know that $$\det C=\det C^T$$. Now, see that
$$(A_1-\lambda I)^T=A_1^T-\lambda I$$