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Probably, this question has already been answered, but I did not find an answer.

If a matrix $A$ is hermitian and an eigenvalue $\lambda$ has multiplicity $k$, are there always $k$ pairwise orthogonal vectors $x_1,...,x_k$ with $Ax_i=\lambda x_i$ for all $i=1,...,k$ ? If yes, how can this be proven ?

I know that a hermitian matrix has real eigenvalues, and that the eigenvectors to different eigenvalues have to be orthogonal. But I am not quite sure about the situation for multiple eigenvalues.

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The answer to this depends on whether you choose an orthogonal basis for the eigenspaces or not. But if you do, you can do some actually scaling on the vectors to diagonalize $A$ with an orthonormal basis of $\mathbb{C}^n$. This is called unitary diagonalization, and this is a notable property: the unitarily diagonalizable matrices are a proper subset of all diagonalizable matrices.

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    $\begingroup$ Note that the spectral theorem guarantees that $A$ is always diagonalizable period. This shows that $A$ is always unitarily diagonalizable. $\endgroup$
    – user137769
    Jun 22, 2014 at 12:07
  • $\begingroup$ This is a misleading answer. You can always choose an orthonormal basis for any (subspace of an) inner product space, in particular for any eigenspace. The problem is whether the eigenspaces are mutually orthogonal; if they are not, then no matter how you choose bases for each of them, they will never combine to an orthonormal basis of the whole space$~V$. But if they are mutually orthogonal, than any choice of orthonormal basis for each eigenspace will combine to an orthonormal basis of$~V$. Eigenspaces for a Hermitian operator are mutually orthogonal. Also this is beside the OP. $\endgroup$ Jul 19, 2022 at 9:52

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