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Hmm, aren't all eigenvectors orthogonal (not just those of a symmetric matrix -- nevermind, see my edit below)?

Say $\vec{v}$ and $\vec{u}$ are eigenvectors of a matrix $A$ with eigenvalues $\lambda_{1}$ and $\lambda_{2}$ respectively (where $\lambda_{1} \neq \lambda_{2}$). Then:

$$ \lambda_{1}<v, u>\ =\ <\lambda_{1}v, u>\ =\ <Av, u>\ =\ <v, A^{T}u>\ =\ <v, \lambda_{2}u>\ =\ \lambda_{2}<v, u> $$

Because $\lambda_{1} \neq \lambda_{2}$, it follows that $(\lambda_{1} - \lambda_{2})<v, u>\ =\ 0\ \Leftrightarrow\ <v, u>\ =\ 0$.

Edit: Extreme derp. I just realized that $<Av, u>\ =\ <v, A^{T}u>$ is only true if $A$ is symmetric.