Polar Decomposition: Ranges

Question

This is just a note.

Given Hilbert spaces $\mathcal{H}$, $\mathcal{K}$.

Consider a closed operator: $$A:\mathcal{D}(A)\to\mathcal{K}:\quad A=A^{**}$$

Construct its modulus: $$|A|:=\sqrt{A^*A}:\quad|A|^*=|A|$$

Regard a decomposition: $$A=J|A|:\quad JJ^*J=J$$

Then for its kernel: $$\mathcal{N}J=(\mathcal{R}|A|)^\perp\implies\mathcal{N}J=(\mathcal{R}A^*)^\perp$$

How can i check this?

(The latter is better to work with.)

Polar decompositions are not necessarily unique. You should state your particular definition. — Disintegrating By Parts, Feb 23 '15 at 22:19
@T.A.E.: Yep, I haven taken as granted that this is well-known. (I just didn't want to spoil the thread with basics.) — freishahiri, Feb 23 '15 at 22:22
@T.A.E.: Remember there was a great answer on cardinality of infinite Hamel dimension which got terribly spoiled by linear algebra lemmas. — freishahiri, Feb 23 '15 at 22:26
But the result changes if you choose a different polar part. — Disintegrating By Parts, Feb 23 '15 at 22:29
@T.A.E.: But isn't it unique when requiring to act only the range of the modulus?? — freishahiri, Feb 23 '15 at 22:32
@T.A.E.: Besides I found this great answer given by Arturro Magadin which got spoiled: Algebraic Dual — freishahiri, Feb 23 '15 at 22:35
$J$ is not uniquely defined on $\mathcal{R}(|T|)^{\perp}$ if this subspace is non-zero. Some people prefer minimal partial isometries, and others like full isometries. I think Rudin likes maximal. — Disintegrating By Parts, Feb 23 '15 at 22:37
@T.A.E.: Ok I've changed "its polar" to "the polar" and specified its null space. Do you think it is ok this way? — freishahiri, Feb 23 '15 at 22:48

freishahiri · Accepted Answer · 2015-07-15T19:17:30.213

1

This answer is community wiki.

By square root lemma: $$\|A\varphi\|=\||A|\varphi\|\implies\mathcal{N}A=\mathcal{N}|A|$$

Exploit selfadjointness: $$\mathcal{N}J=(\mathcal{R}|A|)^\perp=\mathcal{N}|A|=\mathcal{N}A=(\mathcal{R}A^*)^\perp$$

Concluding the assertion.

edited Jul 15 '15 at 19:17

answered Feb 23 '15 at 17:00

freishahiri

1 Answers1