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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 is self adjoint and positive definite and the other 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 separable Hilbert space, $B_t$ is a cylindrical Brownian motion, while $a$ 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 $x_k(t)= \langle X_t,\phi_k\rangle$, $A\phi_k=\lambda_k\phi_k$ and $\langle \phi_k,\sqrt{a}\phi_j\rangle=\sqrt{a_k}\delta_{jk}$.

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In linear Algebra if the commutator of both matrices is zero. Maybe this can be generalized to the infinite dimensional case... –  Alex Mar 17 '13 at 13:09
    
What does it have to do with stochastic calculus? In general, positive operators need not be diagonalizable at all, unless they're compact. You need to supply more details. –  tomasz Mar 17 '13 at 13:21
    
Sorry, I should have made that more clear. $A$ can be unbounded and so is not compact in general. It may rely on separability of $H$ –  David Mar 17 '13 at 14:16
    
@tomasz you are correct. One needs an assumption like compact and self adjont for simultaneous diagonalisation. Otherwise you can have operators that don't even have eigenvalues. –  David Mar 18 '13 at 15:03
    
@David: Also, when asking about a specific paper, you should link to it. (Not that I'll help you, I have no idea about stochastic calculus, but that would probably make it easier for others...) –  tomasz Mar 18 '13 at 19:17
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