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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 '14 at 12:07
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The answer is yes (in the finite dimensional case). By definition, the eigenvectors of $A$ with eigenvalue $\lambda$ must be linearly independent. So use the Gram-Schmidt algorithm on them.

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