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when toggle format what by license comment
Mar 2, 2023 at 1:31 history edited LSpice CC BY-SA 4.0
Capitalise title
Feb 27, 2023 at 16:41 answer added Bewol timeline score: 3
S Feb 22, 2023 at 19:17 history suggested CommunityBot CC BY-SA 4.0
expanded acronym "psd" in title
Feb 22, 2023 at 18:35 review Suggested edits
S Feb 22, 2023 at 19:17
Feb 22, 2023 at 17:44 history edited Dima Pasechnik CC BY-SA 4.0
added example and a better question
Feb 22, 2023 at 14:43 history edited Dima Pasechnik CC BY-SA 4.0
added extra condition
Feb 22, 2023 at 14:25 comment added Dima Pasechnik @ChrisWuthrich - your example is not psd
Feb 22, 2023 at 14:23 comment added Dima Pasechnik sorry, edited the question so that it makes sense, indeed, the condition on $\det P$ is needed.
Feb 22, 2023 at 14:16 comment added Chris Wuthrich .. and how do you diagonalise $(\begin{smallmatrix} 0 & 2\\1& 0 \end{smallmatrix})$ staying over $\mathbb{Q}$?
Feb 22, 2023 at 14:04 comment added YCor @DimaPasechnik but $\det(P)^{1/n}$ is not necessary in $\mathbf{Q}$...
Feb 22, 2023 at 14:02 comment added Dima Pasechnik $P$ is p.s.d., its determinant is in $\mathbb{Q}$, anyway
Feb 22, 2023 at 14:01 comment added YCor @DimaPasechnik Tom's point is that the possibility writing the matrix in this way implies that $\det(P)^{1/n}$ belongs to $\mathbf{Q}[i]$. And this is not always true (take $n\ge 2$ and $P$ be diagonal($2,1,\dots,1$)).
Feb 22, 2023 at 13:58 comment added Dima Pasechnik no, why? The $1/n$ is there cause for a scalar $t$ one has $\det (tI)=t^n$, and I'd like $\det P$ to cancel each other on both sides of $P=tAA^*$.
Feb 22, 2023 at 13:53 comment added Tom De Medts Do you assume $\det P$ to be an $n$-th power of a rational number?
Feb 22, 2023 at 13:53 answer added marco timeline score: 1
Feb 22, 2023 at 13:19 history asked Dima Pasechnik CC BY-SA 4.0