Spent hours trying to prove this after encountering it in Lax's discussion of the spectral theorem, but no luck. Here's the problem (it is Theorem 18 in Lax 2ed, Chapter 6):
A mapping $A$ has distinct eigenvalues $a_1, …, a_n$. We know that a matrix has the same eigenvalues as its transpose, so these eigenvalues correspond to the eigenvectors $x_1, …, x_n$ of $A$, as well as the eigenvectors $y_1, …, y_n$ of $A^T$.
(1) Show that, for any fixed $i$ in $1, …, n$, $\langle x_i, y_i\rangle\ne 0$.
(2) Now let $x = \sum{k_jx_j}$ be the expansion of a vector as the sum of the eigenvectors of $A$. Show that $k_i = \dfrac{\langle l_i, x\rangle}{\langle l_i, x_i\rangle}$ for $i=1, …, n$.
I'm missing something fundamental here. The previous theorem discusses orthogonality of eigenvectors when they do not have the same eigenvalue, but I don't believe it is relevant here.
I also wanted to think about this in terms of the null space of the characteristic polynomials for $A$ and $A^T$ (they obviously have the same roots), but didn't get anywhere there.
Best Answer
Either there should be an additional hypothesis, or you should be allowed to choose the eigenvectors cleverly in the case where several eigenvalues are equal. As it stands, there are easy counterexamples: Let $A$ be the zero matrix or the identity matrix. then the $x_i$'s and $y_j$'s could be anything, and in particular you could have unwanted orthogonality.
EDIT, after the question was edited to say the eigenvalues are distinct: Now that the necessary hypotheses are in place, the result can be proved as follows. First show that the eigenvectors $x_i$ and $y_j$ for different eigenvalues are orthogonal. Indeed, (taking the vectors to be column vectors) $$ a_jy_j^Tx_i=(y_j^TA)x_i=y_j^T(Ax_i)=a_iy_j^Tx_i, $$ and $a_j\neq a_i$, so $y_j^Tx_i=0$. Now if $x_i$ were also orthogonal to $y_i$, then it would be orthogonal to all the $y$'s. But these form a basis for the space, so $x_i$ would be orthogonal to everything, including itself. That would make $x_i=0$, which is false because $x_i$ is an eigenvector.