Answer
See solution
Work Step by Step
If $\lambda$ is an eigenvalue of $A^T$, $A^Tx=\lambda x$ and $(A^T-\lambda I)x=0$.
Taking a transpose of a square matrix does not change the values of the main diagonal, so $(A^T-\lambda I)^T=(A-\lambda I)$ and $(A^T-\lambda I)=(A-\lambda I)^T$.
If $\lambda$ is an eigenvalue of A but not $A^T$,
$(A-\lambda I)x=0$ has a nontrivial solution, which means $(A-\lambda I)$ is singular.
But $(A^T-\lambda I)x=0$ has only the trivial solution, which means $(A^T-\lambda I)$ is invertible.
But $(A^T-\lambda I)$ is merely the transpose of $(A-\lambda I)$. An invertible matrix cannot be a transpose of a singular matrix, so $\lambda$ must be the eigenvalue of both or neither matrices.
