# Showing that real symmetric matrices are diagonalizable

Summary:: Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.

Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.

We know that if ##A## is diagonalizable then ##g(\lambda)=a(\lambda)##. So all we have to show is that ##A## is diagonalizable.

I found a proof by contradiction: Assuming ##A## is not diagonalizable we have

$$(A- \lambda_i I)^2 v=0, \ (A- \lambda_i I) v \neq 0$$

Where ##\lambda_i## is some repeated eigenvalue. Then

$$0=v^{\dagger}(A-\lambda_i I)^2v=v^{\dagger}(A-\lambda_i I)(A-\lambda_i I) \neq 0$$

Which is a contradiction (where ##\dagger## stands for conjugate transpose).

But I do not really understand it; what is meant by 'generalized eigenvalues of order ##2## or higher'?

I asked about this proof a while ago but the answer I got did not really convince me...

We could go over an alternative proof of course, the aim is understand how to show that real symmetric matrices are diagonalizable.

Thank you!

## Answers and Replies

Well, there is a quite elegant analysis argument for that using compactness of a certain group of matrices (and other machinery), but I'll treat it as a linear algebra problem.

Let $E\neq 0$ be a real Euclidean space. It is well known that a transformation $\varphi$ on $E$ is symmetric if and only if $E$ admits an orthonormal basis of eigenvectors of $\varphi$.

Now let $A$ be a symmetric real matrix and fix an orthonormal basis $e$ on $E$ (take any basis, give it the Gram-Schmidt treatment and normalise it, say). The corresponding transformation $\varphi$ is then symmetric. Let $f$ be an orthonormal basis of eigenvectors of $\varphi$.

The matrix of $\varphi$ with respect to $f$, call it $B$, is diagonal.

Let $T$ be the transition matrix from $e$ to $f$. Then $B=T^{-1}AT$, thus $A$ is similar to a diagonal matrix i.e is diagonalisable. Transition matrix between orthonormed bases is orthogonal, so can replace $T^{-1}=T^t$.

Remark: in the complex case, this is not true. However, Hermitian matrices are diagonalisable.

Last edited:
JD_PM
You have two separate questions, I feel. So I'll briefly address the second one. Yes, diagonalisability is equivalent to $g(\lambda) = a(\lambda)$. This is because equality of multiplicities is equivalent to the space admitting a basis of eigenvectors. Also, similar matrices have the same characteristic polynomial.