When are the Eigenvectors of a A and $A^T$ the same? I have a matrix $A$ which has distinct, real eigenvalues. Will $A$ and $A^T$ have the same eigenvectors. I thought this result was false, but I am reading a paper which uses this as a fact when $A$ is a $2\times 2$ real matrix.
 A: With the hypothesis that the eigenvalues of $A$ are distinct, $A$ and $A^T$ have the same eigenvectors iff $A$ is normal ($A^T A = A A^T$), and with the further hypothesis that the eigenvalues of $A$ are real, $A$ and $A^T$ have the same eigenvalues iff $A$ is symmetric ($A^T = A$). So any non-symmetric matrix with real distinct eigenvalues is a counterexample; in the $2 \times 2$ case we can take, for example,
$$A = \left[ \begin{array}{cc} 0 & -2 \\ 1 & 3 \end{array} \right].$$
$A$ has eigenvectors $v_1 = \left[ \begin{array}{c} -1 \\ 1 \end{array} \right]$ and $v_2 = \left[ \begin{array}{c} -2 \\ 1 \end{array} \right]$ with eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 1$. Its transpose
$$A^T = \left[ \begin{array}{cc} 0 & 1 \\ -2 & 3 \end{array} \right]$$
has the same eigenvalues but eigenvectors $w_1 = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right], w_2 = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$.
You'll notice in the above example that $v_1$ is orthogonal to $w_2$ and vice versa. This generalizes: if $v_i$ is an eigenvector of $A$ with eigenvalue $\lambda_i$ and $w_j$ is an eigenvector of $A^T$ with eigenvalue $\lambda_j \neq \lambda_i$, then
$$w_j^T A v_i = w_j^T (\lambda v_i) = \lambda_i w_j^T v_i$$
but we also have
$$w_j^T A v_i = (A w_j)^T v_i = (\lambda_j w_j)^T v_i = \lambda_j w_j^T v_i$$
which gives $w_j^T v_i = 0$. This says that if $A$ has distinct real eigenvalues then the eigenvectors of $A$ and $A^T$ are dual bases with respect to the inner product (up to scale), which among other things is one way to prove the above result: if the eigenvectors of $A$ aren't orthogonal then they can't be their own dual basis.
