Show that A and A^{T} have the same eigenvalues. what if anything can we say about the associated eigenvectors of A and A^{T}?
$\displaystyle \det (A - \lambda I) = \det (A - \lambda I)^T = \det (A^T - \lambda I) $.
The eigenvectors of $\displaystyle A^T$ are generally different to those of $\displaystyle A$. However, there is a relationship between the two:
If $\displaystyle v_i$ is an eigenvector of $\displaystyle A$ corresponding to the eigenvalue $\displaystyle \lambda_i$ and $\displaystyle w_j$ is an eigenvector of $\displaystyle A^T$ corresponding to the eigenvalue $\displaystyle \lambda_j$ then $\displaystyle v_i^T w_j = 0 ~ (\lambda_i \neq \lambda_j)$.
Proof:
$\displaystyle A v_i = \lambda_i v_i \Rightarrow v_i^T A = \lambda_i v_i^T \Rightarrow v_i^T A w_j = \lambda_i v_i^T w_j$ .... (1)
$\displaystyle A^T w_j = \lambda_j w_j \Rightarrow v_i^T A^T w_j = \lambda_j v_i^T w_j $ .... (2)
(1) - (2): $\displaystyle 0 = \lambda_i v_i^T w_j - \lambda_j v_i^T w_j $ and the result is easily seen.