Is a family of commuting self adjoint operators simultaneously diagonalizable?

Let $V$ be a finite-dimensional inner product space over $\mathbb{R}$.

Let $\mathscr{A}$ be a family of self-adjoint operators on $V$ such that $ST=TS$ for all $S,T\in \mathscr{A}$.

Then, does there exists an orthonormal basis $\beta$ for $V$ such that $[T]_{\beta}$ is diagonal for every $T\in\mathscr{A}$?

I'm going to prove this by induction: Decopmose $V$ into the direct sum of eigenspaces, then find an orthonormal basis for each eigenspace and take a union.

Before I really write down to prove this, I want to clarify whether this is true or not. (I don't want to waste time)..

Is this $\mathscr{A}$ simultaneously diagonalizable by an orthonormal basis?

• Commuting diagonalizable matrices are simultaneously diagonalizable, then use Gram-Schmidt process to make an orthonormal basis as the process preserves diagonalizability. – Ehsan M. Kermani Jan 7 '15 at 2:53
• Hint: Try proving they share the same eigenvectors. Now, use the fact they are both diagonalizable. – dineshdileep Jan 7 '15 at 6:41