# Do Real Symmetric Matrices have 'n' linearly independent eigenvectors? [duplicate]

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?

Thanks!

• For each eigenvalue $\lambda$, you can use gram-schmidt process to find an orthonormal basis for the eigenspaces with respect to $\lambda$.
– user99914
May 12, 2018 at 4:21
• 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? May 12, 2018 at 4:24
• Basically you are asking why a real symmetric matrix is diagonalizable. This answer addressed this, but some details are skipped.
– user99914
May 12, 2018 at 4:32
• What you are seeking is the spectral theorem
– user169852
May 12, 2018 at 4:32
• Does this answer your question? Eigenvectors of real symmetric matrices are orthogonal
– user53259
Jul 25, 2020 at 22:53

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