# Prove that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^T$.

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?

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*}
Step 1: Prove that $\det(A^T) = \det A$ for all square matrices $A$.
Step 2: Prove that $(A-\lambda I)^T = A^T - \lambda I$.