Timeline for Square root of normal positive operators over real Hilbert spaces
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8 at 4:37 | comment | added | Jianing Song | Thanks for your reply! | |
| Dec 29, 2023 at 1:53 | comment | added | J. van Dobben de Bruyn | I don't see how one would ensure that the obtained square root is quasi-positive. We get $\sigma(S^{1/2}) \subseteq \{\text{Re}(z) \geq 0\}$, but for non-normal operators this doesn't necessarily imply $\langle S^{1/2}x,x \rangle \geq 0$ for all $x$. Maybe a more careful analysis, using the series expansion of ${\scriptstyle\surd}$, can somehow show that $S^{1/2}$ is also quasi-positive? I'm not sure. | |
| Dec 29, 2023 at 1:48 | comment | added | J. van Dobben de Bruyn | In this setting, it might even be possible to get a real (instead of complex) square root of the operator by working with a power series expansion of ${\scriptstyle\surd}$ on $U$ with only real coefficients. For instance, if $\sigma(S) \subseteq \{\text{Re}(z) > 0\}$, then a series expansion of ${\scriptstyle\surd}$ like the one given here shows that $S$ has a real square root. | |
| Dec 29, 2023 at 1:44 | comment | added | J. van Dobben de Bruyn | @JianingSong Come to think of it, it is still possible to obtain a square root of certain non-normal operators by using the holomorphic functional calculus instead of the continuous one. This requires ${\scriptstyle\surd}$ to be holomorphic in an open neighbourhood $U$ of $\sigma(S)$, so this does not work for all operators (indeed, not all operators have a square root). In particular, for this approach to work, we must have $0 \notin \sigma(S)$ (that is, $S$ must be invertible). | |
| Dec 28, 2023 at 1:51 | comment | added | J. van Dobben de Bruyn | Here, continuous functional calculus refers to the common technique of applying a continuous function $f : \sigma(S) \to \mathbb{C}$ to a normal operator $S$, by approximating $f$ by polynomials in $z$ and $\overline{z}$ that converge to $f$ uniformly on $\sigma(S)$. This can only be done if $S$ and $S^*$ commute — otherwise, one can only meaningfully apply polynomials in either $z$ or $\overline{z}$, not both. | |
| Dec 28, 2023 at 1:50 | comment | added | J. van Dobben de Bruyn | @JianingSong The assumption is definitely needed — some square matrices do not have a square root at all, let alone a quasi-positive one (see e.g. here). The assumption is used when we apply continuous functional calculus to approximate ${\scriptstyle\surd}$ by polynomials in $\lambda$ and $\overline\lambda$ — this is only defined when $S$ and $S^*$ commute (i.e. $S$ is normal). | |
| Dec 15, 2023 at 23:01 | comment | added | Jianing Song | Just out of curiosity: where do we use the fact that $T$ is normal? Do existence and uniqueness of positive square root still hold if we drop that condition? Thanks! | |
| Nov 12, 2022 at 16:52 | history | edited | J. van Dobben de Bruyn | CC BY-SA 4.0 |
language
|
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
| Feb 1, 2018 at 17:05 | review | First posts | |||
| Feb 1, 2018 at 17:14 | |||||
| Feb 1, 2018 at 17:03 | history | answered | J. van Dobben de Bruyn | CC BY-SA 3.0 |