- #1

- 12

- 0

## Main Question or Discussion Point

Is it true that an nxn symmetric matrix has n linearly independent eigenvectors even for non-distinct eigenvalues? How can we show it rigorously? Basically, I want to prove that if an nxn symmetric matrix has eigenvalue 0 with multiplicity k, then its rank is (n - k). If we can prove that there exist k linearly independent eigenvectors which solve Ax = 0, then we can use the rank-nullity theorem to directly show the result, right?