Show that a real symmetric matrix is always diagonalizable

Question

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).

OK but isn't there a better proof? I see it could be approached by the Spectral theorem or Gram Schmidt Prove that real symmetric matrix is diagonalizable. A hint for how to do so would be appreciated.

Answer 1

The proof with the spectral theorem is trivial: the spectral theorem tells you that every symmetric matrix is diagonalizable (more specifically, orthogonally diagonalizable). As you say in your proof, "all we have to show is that $A$ is diagonalizable", so this completes the proof.

The Gram Schmidt process does not seem relevant to this question at all.

Honestly, I prefer your proof. If you like, here is my attempt at making it look "cleaner":

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We are given that $A$ is real and symmetric. For any $\lambda$, we note that the algebraic and geometric multiplicities disagree if and only if $\dim \ker (A - \lambda I) \neq \dim \ker (A - \lambda I)^2$. With that in mind, we note the following:

Claim: All eigenvalues of $A$ are real.

Proof of claim: If $\lambda$ is an eigenvalue of $A$ and $x$ an associated unit eigenvector, then we have $$ Ax = \lambda x \implies x^\dagger Ax = x^\dagger (\lambda x) = \lambda. $$ However, $$ \bar \lambda = \overline{x^\dagger Ax} = (x^\dagger A x)^\dagger = x^{ \dagger } A^\dagger x^{\dagger\dagger} = x^\dagger A x = \lambda. $$ That is, $\lambda = \bar \lambda$, which is to say that $\lambda$ is real. $\square$

With that in mind, it suffices to note that for any matrix $M$, we have $\ker M = \ker M^\dagger M$. Indeed, it is clear that $\ker M \subseteq \ker M^\dagger M$, and we have $$ x \in \ker M^\dagger M \implies M^\dagger Mx = 0 \implies x^\dagger M^\dagger M x = 0 \\\implies (Mx)^\dagger (Mx) = 0 \implies Mx = 0 \implies x \in \ker M. $$ Now, taking $M = A - \lambda I$, we see that for any eigenvalue $\lambda$ of $A$, we have $$ \dim \ker(A - \lambda I)^2 = \dim \ker M^\dagger M = \dim \ker M, $$ which mean that the algebraic and geometric multiplicities are indeed the same for each eigenvalue $\lambda$.