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 $\langle \phi\_k,\sqrt{a}\phi\_j\rangle=\sqrt{a\_k}\delta\_{jk}$.



When are two operators simultaneously diagonalisable?
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.


