Timeline for Question on density of certain set of matrices
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23, 2023 at 3:38 | vote | accept | Kanghun Kim | ||
| Mar 21, 2023 at 17:54 | comment | added | Denis Serre | @KanghunKim Feel free to accept the answer. | |
| Mar 21, 2023 at 14:14 | comment | added | Kanghun Kim | Now I get the idea. Thank you :) | |
| Mar 20, 2023 at 14:44 | comment | added | Denis Serre | @WillieWong Thanks a lot for the details :) . | |
| Mar 20, 2023 at 13:57 | comment | added | Willie Wong | @KanghunKim: (c) When $S = Q$, the next task is to characterize those $B$ that leads to this. In this case, the "cubic polynomial" that is involved must be identically zero. This means all of its coefficients must be zero. In particular, the cubic term and the linear term must be separately equal to zero. Hence $B \Sigma = \Sigma B^T$ for every $\Sigma$. (d) Now let $v$ be an arbitrary vector, let $\Sigma$ be the symmetric matrix $v v^T$. Evaluating $B(v) (v^Tv) = B\Sigma v = \Sigma B^T v = v (v^T B^T v)$ we see that every $v$ is an eigenvalue of $B$, hence $B$ must be a multiple of $I_n$. | |
| Mar 20, 2023 at 13:49 | comment | added | Willie Wong | @KanghunKim: (b) your $Q$ is a vector space. Your requirement "$A + A (B^{-1}A)^2$ is symmetric" can be expressed as "a cubic polynomial vanishes". So your set $S$ is an algebraic subvariety (in fact affine) of $Q$. It is a general fact that $S$ therefore either has measure zero, or $S = Q$. (Here we use that if it doesn't have measure zero, then $S$ must contain an open subset of $Q$, but if a polynomial vanishes on an open set it must be identically zero.) | |
| Mar 20, 2023 at 13:41 | comment | added | Willie Wong | @KanghunKim: quick explanations. (a) The use of $Q'$ and $S'$ is to reparametrize stuff to make them easier to analyze, using the fact that when $B$ is invertible, the mapping $M\mapsto B^{-1} M$ is a invertible linear mapping of the space of all matrices, so do not change measurability properties. | |
| Mar 20, 2023 at 6:18 | comment | added | Denis Serre | @WillieWong If $AB^T=\Sigma$ is symmetric, then $B^{-1}A=B^{-1}\Sigma B^{-T}$ is congruent to $\Sigma$, thus symmetric. And conversely. | |
| Mar 20, 2023 at 5:20 | comment | added | Willie Wong | This must be something simple, but I am not seeing why $Q'$ is the space of symmetric matrices. $Q$ is defined by $AB^T$ being symmetric. | |
| Mar 17, 2023 at 9:04 | comment | added | Kanghun Kim | Hmmm.... any simplifications may I ask? I'm only a junior in maths | |
| Mar 17, 2023 at 8:27 | history | answered | Denis Serre | CC BY-SA 4.0 |