# geometric multiplicity= algebraic multiplicity for a symmetric matrix

Could any one tell me how to prove:

$\lambda$ be an eigen value of a symmetric matrix $A$ then how to show that the geometric multiplicity and algebraic multiplicity are equal?

-
What facts do you know so far? If you know the fact that all symmetric matrices are orthogonally diagonalizable then this is easy. If not then it gets a bit more complicated. – EuYu May 16 '13 at 4:57
see the thing is was reading that proof from my local writer book, and he says "let A be orthogonally diagonalizable..." I understand that that then $A$ is symmetric, but for the converse part he uses this lemma which he has not proved in his books so I asked. – La Belle Noiseuse May 16 '13 at 5:02
I think the standard way to prove that symmetric matrices are orthogonally diagonalizable uses the fact that matrices with real eigenvalues are orthogonally triangularizable. This is sometimes referred to as Schur's theorem. Have you heard of that before? – EuYu May 16 '13 at 5:05
No Dear Sir EUYU – La Belle Noiseuse May 16 '13 at 6:14

If $A$ is real symmetric then it is diagonalizable. That is, there is some orthogonal $P$ such that $AP=PD$, where $D$ is diagonal. The columns of $P$ are each eigenvectors, and form a basis. Hence all geometric multiplicities equal algebraic multiplicities.