[Math] Show that the eigenvectors of an orthogonal matrix and its transpose are the same

Question

I know that for an orthogonal matrix $A$, $A^TA = I$, but I'm having trouble using this to show the above. Any help would be much appreciated. Thanks in advance!

EDIT:
I forgot to say what I had tried already. Here's my attempt (but the logic might be wrong…) –

$Av = \lambda v$

$\frac{1}{\lambda}Av = v$

$A^Tv = A^T(\frac{1}{\lambda}Av) = \frac{1}{\lambda}A^TAv = \frac{1}{\lambda}v$

Answer 1

You have $A^T = A^{-1}$. In general, the eigenvectors of $A$ and $A^{-1}$ are the same (if $A$ is only assumed to be invertible). And it is really easy to see this. Just write down what it means for $x$ to be an eigenvector and apply $A^{-1}$ on both sides.