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I know that a symmetric matrix has a set of orthogonal eigenvectors if all its eigenvalues are distinct. As far as I know, their eigenvalues might not be distinct. However, some sources claim that symmetric matrices always have eigenbasis.

I know that we can get orthonormal basis using Gram Schmidt on the eigenvectors of each basis. However, I don't think that it guarantees the resultant vectors are still eigenvectors. Shouldn't the Gram Schmidt process turns the eigenvectors into non eigenvectors? If not how to prove that the resultant vectors are still eigenvectors?

So, my question then is if symmetric matrices always have eigenbasis regardless of whether its eigenvalues are distinct? Best if a calculus-free proof for whether or not symmetric matrices have eigenbasis can be given.

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  • $\begingroup$ This is true over reals. The proof is simple: find some eigenvector (it exists) v and continue by induction on $v^\perp$ (which is $A$-invariant by the symmetry of $A$) $\endgroup$ – Peter Franek Nov 16 '17 at 17:47
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Yes, each real symmetric matrices is orthogonaly equivalent to a diagonal matrix. In other words, you can find an orthonormal basis of eigenvectors. This is the spectral theorem for symmetric matrices.

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Yes, a symmetric matrix always has an eigenbasis: as an $n \times n$ symmetric matrix always has $n$ eigenvectors (spectral theorem), which can be made orthogonal by the Gram-Schmidt theorem. This proves that the eigenbasis doesn't depend on repeated or distinct eigenvalues.

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