Finding eigenvalues For $A$

Question

Can you help me to find a way to get a quick answer for the following problem:

if $A$ is Real and non-uniform square square matrix and $A^t=A^2$(transposed). Find the eigenvalues For $ A$?

Thank you.

Answer 1

From above, $A^{t} = A^{2} \implies A = (A^{t})^{t} = (A^{2})^{t} = (A^{t})^{2} = (A^{2})^{2} = A^{4}$

If $A= Q \Lambda Q^{t} $ then $A^{4} = (Q\Lambda Q^{t})^{4} = Q \Lambda^{4} Q^{t}$ and $\Lambda^{4} \implies \lambda_{i}^{4} ,1 \leq i \leq n $

Answer 2

Set $v$ to be an eigenvalue of length one, with eigenvalue $\lambda$.

Now try this: multiply by $v^t$ on the left and $v$ on the right. Then $$ v^t A^t v = v^t A^2 v $$

This can be simplified using matrix multiplication associativity, and that $ (AB)^t = B^t A^t $. Can you get an equation that's purely in $\lambda$? Feel free to ask if you don't understand something :)