If c is an eigenvalue of $A$, is it an eigenvalue of $A^{\mathbf{T}}$? I am given a square matrix $A$, and I need to prove that if c is its eigenvalue, then it is also an eigenvalue of its transpose. How should I approach this? Clearly $Av$=$cv$, but I am not sure how to bring transpose into the equation.

A simple way would be to look at $\left | A^T-cI \right |$. $$\left | A^T-cI \right | = \left| (A-cI)^T \right| = \left| (A-cI) \right|$$
EDIT
$\left| B \right|$ denotes the determinant of $B$. First note that for any matrix $B$, $\left| B \right| = \left| B^T \right|$. 
(This is true since you get the same determinant if you find the determinant along the row or column.)
$c$ is an eigenvalue of $A$ iff $\left| A - cI\right| = 0$.
Note that $\left| A^T - cI \right| = \left| A^T - cI^T \right| = \left| (A - cI)^T \right| = \left| A - cI\right| = 0$.
Hence, $c$ is an eigenvalue of $A^T$.
