Timeline for Square root of normal positive operators over real Hilbert spaces
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 11, 2016 at 5:43 | comment | added | Nik Weaver | Yes, I noticed that too. Indeed, it is the unique "positive" square root. | |
| Nov 11, 2016 at 1:24 | comment | added | Nate Eldredge | Incidentally, though, my operator $A$ does have a "positive" square root, namely $\frac{1}{\sqrt{2}} (A+I)$. (This is "positive" since $A+I$ is the sum of two "positive" operators.) | |
| Nov 11, 2016 at 0:11 | comment | added | Nik Weaver | Oh, I didn't notice that his definition of "positive" was nonstandard! | |
| Nov 10, 2016 at 21:10 | comment | added | Nate Eldredge | I don't think it's that easy. With $H = \mathbb{R}^2$, the operator $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is normal and positive by this definition, but it isn't self-adjoint. It's not diagonalizable over the reals, so no isomorphism of real Hilbert spaces can make it into a multiplication operator. | |
| Nov 10, 2016 at 19:21 | history | answered | Nik Weaver | CC BY-SA 3.0 |