I know that Real Symmetric Matrices have real eigenvalues and the vectors corresponding to each distinct eigenvalue is orthogonal from this answer. But what if the matrix has repeated eigenvalues? Does it have linearly independent (and orthogonal) eigenvectors? How to prove that?

PS: In the answer I referred to has another answer which might have answered this question. I'm not sure if it answered my question since I didn't understand it. If it did answer my question, can anyone please explain it?


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    $\begingroup$ For each eigenvalue $\lambda$, you can use gram-schmidt process to find an orthonormal basis for the eigenspaces with respect to $\lambda$. $\endgroup$ – user99914 May 12 '18 at 4:21
  • $\begingroup$ In general for a square matrix, if there are repeated eigenvalues, there may not be n distinct eigenvectors right? Is it true for Symmetric Matrices as well or do symmetric matrices have distinct eigenvectors even with repeated eigenvalues? $\endgroup$ – Nagabhushan S N May 12 '18 at 4:24
  • $\begingroup$ Basically you are asking why a real symmetric matrix is diagonalizable. This answer addressed this, but some details are skipped. $\endgroup$ – user99914 May 12 '18 at 4:32
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    $\begingroup$ What you are seeking is the spectral theorem $\endgroup$ – Bungo May 12 '18 at 4:32
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    $\begingroup$ Does this answer your question? Eigenvectors of real symmetric matrices are orthogonal $\endgroup$ – Nai Jul 25 at 22:53

Real Symmetric Matrices have $n$ linearly independent and orthogonal eigenvectors.

There are two parts here.
1. The eigenvectors corresponding to distinct eigenvalues are orthogonal which is proved here.
2. If some of the eigenvalues are repeated, since the matrix is Real Symmetric, there will exist so many independent eigenvectors. (Proof here and here.) As John Ma pointed out, in this case we can use Gram–Schmidt orthogonalization to get orthogonal vectors.

So, all $n$ eigenvectors of a Real Symmetric matrix are linearly independent and orthogonal

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