Eigenvectors of a symmetric matrix.

[chocolatefrog]

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?

 

[Bacle2]

For an nxn matrix to be diagonalizable, it must have n L.I eigenvectors , as you said:

here:

http://en.wikipedia.org/wiki/Diagonalizable_matrix

They show that symmetric matrices are diagonalizable.