$\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$As a homework assignment we had to show that if $\lambda$ is eigenvalue for the matrix $A$, then it is an eigenvalue for $A^t$.
To prove this, I followed this idea: $$ \Brak{AA^t v, v} = \Brak{A v, A v} = \Brak{\lambda v, \lambda v} = \lambda^2 \Brak{v, v} = \Brak{\lambda^2 v, v}. $$ But my professor told me that my way of showing it is inconsistent, namely I need a proof for $\Brak{A^tA v, v} = \Brak{\lambda^2 v, v}$.
Could anyone point out any way to prove this?
Remember that $\det(A) = \det(A^t)$, moreover $(A + \lambda I)^t = (A^t + \lambda I)$, therefore
$$ \det(A + \lambda I) = \det(A+\lambda I)^t = \det(A^t+\lambda I) $$
So, if $\lambda$ is a solution of $\det(A + \lambda I) = 0$, it is also a solution of $\det(A^t+\lambda I)$. Rewording, if $\lambda$ is an eigenvalue of $A$ then it is also an eigenvalue of $A^t$