When are two operators simultaneously diagonalisable? I am reading a paper and they have diagonalised both operators in an equation, on a separable Hilbert space, with respect to the same basis. My question is, when can two operators be simultaneously diagonalised? In the paper, one operator, $A$, is self adjoint and positive definite and the other, $a$ is bounded and positive.
Thanks.
Context: The problem comes from the generalised Ornstein-Uhlenbeck equation,$dX_t=-AX_t dt + \sqrt{2a}dB_t$ where $A$ is a constant self-adjoint positive definite operator on $H$, a seperable Hilbert space, and $B_t$ is a cylindrical Brownian motion. $a$ is a constant, positive operator . The author then diagonalises the system to become $dX_k(t)=-\lambda_kX_k(t) dt + \sqrt{2a_k}dB_k(t)$ where $x_k(t)= \langle X_t,\phi_k\rangle$ and with $A\phi_k=\lambda_k\phi_k$ and $\langle \phi_k,\sqrt{a}\phi_j\rangle=\sqrt{a_k}\delta_{jk}$.
 A: Even one positive definite operator on an infinite-dimensional Hilbert space need not have any eigenvectors at all: it might have continuous spectrum.  The more general statement is the Spectral Theorem.  Commuting bounded normal linear operators generate a closed commutative
$*$-subalgebra $A$ of ${\mathcal B}(H)$, and a resolution of the identity $E$ on the Borel subsets of the maximal ideal space $\Delta$ of $A$ by which each of these operators can be represented as $$ T = \int_{\Delta} \widehat{T}\ dE$$
where $\widehat{T}$ is the Gelfand transform of $T$.  See Rudin, "Functional Analysis", theorem 12.22.
A: The abstract linear algebra argument that two commuting, diagonalizable linear endomorphisms are simultaneously diagonalizable is as follows.  If $AB = BA$, and if $\vec{v}$ is an eigenvector of $A$, then
$$A(B\vec{v}) = B(A \vec{v}) = B(\lambda \vec{v}) = \lambda(B \vec{v})$$
so $B\vec{v}$ is also an eigenvector of $A$, with the same eigenvalue.  Therefore, $B$ acts on the eigenspaces of $A$.  If $A$ is diagonalizable, then the ambient vector space $V$ is the direct sum of its eigenspaces, so we only have to show that $B$ is diagonalizable acting on each eigenspace of $A$.  This furnishes an eigenbasis of $B$ that is the union of subsets each contained in eigenspaces of $A$, so is a fortiori an eigenbasis of $A$ as well.
In the case that $V$ is finite-dimensional, this can be proven sort of non-constructively by observing that since $B$ is globally diagonalizable, its minimal polynomial is split and separable (factors completely into distinct linear factors).  The minimal polynomial of its restriction is a factor of the big polynomial, and this property of its factorization implies that $B$ is diagonalizable.
If $V$ is infinite-dimensional, you use a different invariant of restriction.  For example, being compact and self-adjoint is such an invariant, and implies diagonalizability.
