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It is true that for orthogonal matrices, all eigenvectors of a single eigenvalue are orthogonal to all other vectors of different eigenvalues(distinct eigenbases are orthogonal). My question is are these eigenbases also orthonormal to eachother, or simply orthogonal?

fool_of_a_took
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    Yes, it is true: $\alpha_2 u^Tv=u^T (Av)=(Au)^T v=\alpha_1 u^T v$, where $\alpha_1\ne \alpha_2$ are eigenvalues of eigenvectors $u, v$. Hence $u^Tv=0$. – markvs Apr 14 '22 at 20:01

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If you're working over $\mathbb{R}$, then this is true, but for kind of trivial reasons. First, if $A$ is orthogonal and $u,v$ are arbitrary vectors (of the correct dimension), then $$\langle Au, Av \rangle = \langle A^TAu,v\rangle = \langle u,v \rangle,$$ or in other words, $A$ preserves inner products. This implies that the only real eigenvalues that $A$ can have are $1$ and $-1$: indeed, if $Av = \lambda v$ for some $v \neq 0$, then $$\langle v,v\rangle = \langle Av, Av \rangle = \lambda^2\langle v, v \rangle,$$ so $\lambda^2 = 1$.

This means that if you have two eigenvectors $u$ and $v$ corresponding to different eigenvalues, then one of the eigenvalues must be $-1$ (say, for $v$), and the other $1$. Then $$\langle u,v\rangle = \langle Au,Av\rangle = -\langle u,v \rangle,$$ so $u$ and $v$ are indeed orthogonal.

If you are considering orthogonal matrices (not unitary!) over $\mathbb{C}$, then the statement is unlikely to be true. In particular, in this case you can have a pair of eigenvalues $\lambda$ and $\lambda^{-1}$, and for such a pair the above reasoning gives nothing. For a concrete example, you can consider the matrix of the $90$ degree rotation in the plane. This has eigenvalues $i$ and $-i$ and the corresponding eigenvectors are not orthogonal.

Dániel G.
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