5
$\begingroup$

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 $\langle \phi_k,\sqrt{a}\phi_j\rangle=\sqrt{a_k}\delta_{jk}$.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ They would certainly have to commute. Commuting diagonalizable operators are simultaneously diagonalizable, since they share all of their eigenvectors. $\endgroup$
    – Ben McKay
    Commented Mar 17, 2013 at 13:20
  • $\begingroup$ @Ben: I find your last statement a little unclear. Each operator acts on each eigenspace of the other, I agree, and I agree that if they commute and each have an ONB of eignvectors, there does exist an ONB of common eigenvectors. $\endgroup$
    – Geoff Robinson
    Commented Mar 17, 2013 at 14:29
  • $\begingroup$ Is commutativity a necessary condition? There is a result that says if they are compact and self adjoint and commute, then they are simultaneously diagonalisable. Is the compact/self adjoint assumption just to ensure diagonalisability? I am not sure of any reason or assumption that the operators commute or are compact. Both operators are positive and so they are diagonalisable. Also, please can you recommend a reference for this? Thanks. $\endgroup$
    – David
    Commented Mar 17, 2013 at 14:42
  • $\begingroup$ @Geoff: In the paper it is assumed that $A$ has a complete ONB $\phi_k$ and this satisfies the conditions above (in the edit) $\endgroup$
    – David
    Commented Mar 17, 2013 at 14:47
  • 1
    $\begingroup$ @David: Two diagonal operators definitely commute, and two matrices commuting is basis-independent. $\endgroup$
    – Ryan Reich
    Commented Mar 17, 2013 at 14:48

2 Answers 2

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for your response. That is a good observation. Is it still false if the Hilbert space is separable? If so is it impossible in general to assume that we can find $\{\phi_k\}$ such that $\langle \phi_i,\sqrt{a}\phi_j\rangle =\sqrt{a_i}\delta{ij}$? $\endgroup$
    – David
    Commented Mar 17, 2013 at 22:04
5
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ That is a helpful explanation. So does this mean that you only need $B$ to be compact and self adjoint, say, if we know that $A$ is diagonalisable with respect to some ONB? Also, are there any invariants in the case of $a$ above. Would positivity be an invariant? Thanks. $\endgroup$
    – David
    Commented Mar 17, 2013 at 18:05
  • $\begingroup$ @David: This is where my understanding of topological linear algebra (e.g. analysis) fails, but the answer to your first question is yes for purely formal reasons. For the second question I refer you to Robert Israel's answer concerning whether positivity is at all useful. The "invariant" thing is just this: suppose $P$ is a property of linear operators implying diagonalizability and preserved under restriction to invariant subspaces. Then $P$ works in place of "compact and self-adjoint" in the above argument. $\endgroup$
    – Ryan Reich
    Commented Mar 17, 2013 at 18:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .