Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H$ be a separable Hilbert space and let $T:H \rightarrow H$ be a symmetric bound linear map.

a) Show that for every orthogonal projection $P$ on $H$ ($P' = P$, $P^2 = P$) PTP is symmetric.

b) Prove the existens of a sequence $C_n$ of compact symmetric linear maps such that $C_nx\rightarrow Tx$ for $x\in H$.

My try:

a) I don't see that $P^* = P$ for all orthogonal projection, why is it so? but if that is true $(PTP)' = P'T'P' = PTP$

b) I do not really know how to start here, I know that the limit of compact maps are compact if they converge uniformly. Any help or hint would be greatfull

share|improve this question
For (a): Didn't you write in your assumption on $P$ that you want to have $P' = P$? Where is your point? For (b): Let $(e_n)_{n\in\mathbb N}$ an orthogonal basis for $H$. Such a thing exists, as $H$ is seperable. Now let $P_n$ denote the orthogonal projection onto $\operatorname{span} \{e_1, \ldots, e_n\}$ and $C_n := P_nTP_n$. Then $C_n$ is symmetric by (a) and compact by finite-dimensionality. –  martini Dec 10 '12 at 14:30

1 Answer 1

Here's one way to see (a): write $\langle Px, y \rangle = \langle Px, Py \rangle + \langle Px, y-Py \rangle$. Since $P$ is orthogonal projection, $y - Py$ is orthogonal to the range of $P$, so the second term vanishes. Thus $\langle Px, y \rangle = \langle Px, Py \rangle$. By the same argument, $\langle x, Py \rangle = \langle Px, Py \rangle$. Since $\langle Px, y \rangle = \langle x, Py \rangle$, $P$ is symmetric.

For (b): fix an orthonormal basis $\{e_i\}$ for $H$, and let $P_n$ be orthogonal projection onto the span of $e_1, \dots, e_n$. Show that for any $x$, $P_n x \to x$. Now try taking $C_n = P_n T P_n$. (Note that $P_n \to I$ pointwise, but not uniformly, so this doesn't contradict the fact about compact operators that you state.)

share|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.