TITLE: Prove that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^T$.
QUESTION [5 upvotes]: Prove that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^T$.

I'm stucked here, i've approached the problem by looking at $\det(A-\lambda I)=0\iff\det(A^T-\lambda I)=0$. I tried some cases, and I can see it when the matrix is triangular since the main diagonal remains the same when $A$ is transposed, but this hasn't shown me a way to proceed. Any hints or ideas?

REPLY [24 votes]: Here is an easier proof that avoids determinants:
First note that $(AB)^T = B^T A^T$ and use this to prove that $A$ is invertible if and only if $A^T$ is invertible. Thus
\begin{align*}
\lambda \text{ is an eigenvalue of } A &⟺ (A - \lambda I) \text{ is not invertible}\\
&⟺ (A - \lambda I)^T \text{ is not invertible}\\
&⟺ A^T - \lambda I \text{ is not invertible}\\
&⟺\lambda \text{ is an eigenvalue of } A^T.
\end{align*}