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It is true that if $T:H \to H$ is a compact operator ($H$ Hilbert space) then $T^\ast T$ is algo compact and indeed self-adjoint.

Conversely, is it true that every compact and self-adjoint operator $S$ can be decomposed like $S=A^\ast A$ with $A$ compact?

Thanks!

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You also need $S$ to be positive semidefinite, i.e. $\langle x, S x \rangle \ge 0$ for all $x \in H$. Then you can take $A = \sqrt{S}$ using the continuous functional calculus. Note that any continuous function $f$ on $\sigma(S)$ with $f(0)=0$ is the uniform limit on $\sigma(S)$ of a sequence of polynomials $p_n$ with $p_n(0)=0$, and so $f(S)$ is the norm limit of $p_n(S)$. If $S$ is compact, then so are $p_n(S)$ and therefore $f(S)$.

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  • $\begingroup$ Of course! :) Thanks a lot!! $\endgroup$ – Daniel Oct 18 '12 at 21:06
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No, an operator of the form $A^*A$ is selfadjoint and has non-negative spectrum. So any compact selfadjoint operator with a negative eigenvalue cannot be written in that form.

For the simplest example, fix a vector $x\in H$ and let $S$ be the operator $$ Sy=-\langle y,x\rangle x. $$

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  • $\begingroup$ Thank you very very much for the example. $\endgroup$ – Daniel Oct 18 '12 at 21:06

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