MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Have they been studied? In particular, what is the analogue of the Schmidt theorem for compact operators in Hilbert spaces?

Helemskii A. Ya., Lectures and Exercises on Functional Analysis, Ch. 3, $\S4$,

Theorem (Schmidt). Let $H$ and $K$ be Hilbert spaces and $T:H\to K$ a compact operator. Then there exist

a) a finite or countable orthonormal system $e'_1$, $e'_2,\ldots$, in $H$,

b) an orthonormal system $e''_1$, $e''_2,\ldots$, in $K$ of the same cardinality,

c) a (finite or infinite) sequence $s_1\ge s_2\ge s_3\ge\ldots$ of positive numbers with the index set of the same cardinality, tending to zero if it is infinite,

such that

$$ Tx=\sum_n s_n\langle x,e'_n\rangle e''_n. $$ (In other words, $T(e'_n)=s_n e''_n$ for all $n$ and $T$ takes every vector orthogonal to all $e_n'$ to zero.)

An equivalent statement: a compact operator in Hilbert spaces is weakly unitary equivalent to a diagonal operator.

The numbers $\{s_n\}$ are called the $s$-numbers of the operator $T$. What would they be in the quaternioniс case? Is where a variant of the theorem in which they are complex?

share|cite|improve this question
Unless there's something further... restricting scalars to $\mathbb C$ gives a compact operator on the associated complex Hilbert space, so the usual spectral things apply. (I do not know what you mean by "Schmidt's theorem".) The fact that the operator on the restricted-scalars Hilbert space commutes with the operators inherited from the quaternions obviously has some impact, but, lacking more specifics, it's not clear what one gains by having had a quaternionic structure. – paul garrett Jun 20 '13 at 23:06
@paul garrett I included Schmidt's theorem in the question. – Andrew Jun 21 '13 at 9:38
I think that the spectral theorem for compact operators is less ambiguous. The spectral theorem directly generalizes in your case as Paul Garrett explains. – Marc Palm Jun 21 '13 at 13:22
Schmidt's theorem would be for me that the Kernel function is square integrable iff the Kernel transform is Hilbert Schmidt. – Marc Palm Jun 21 '13 at 13:23
up vote 2 down vote accepted

Quaternionic functional analysis is reviewed in (Chi-Keung Ng, On quaternionic functional analysis). In fact "Hilbert spaces of any one of the three kinds - real, complex and quaternionic - can be seen as Hilbert spaces of the other kinds, equipped with extra structure" (John C. Baez, Division Algebras and Quantum Theory, ).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.