Timeline for Dimension of a subspace of $n\times n$ real symmetric matrices
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 8, 2023 at 14:41 | comment | added | Christian Remling | We'd then have to place $L(e)$ by a $2$-dimensional subspaces $L(e,a_e)$ in general. As I said, I'm not claiming this works, and maybe it doesn't, but it did feel as if it came close. | |
| Aug 8, 2023 at 14:40 | comment | added | Christian Remling | $T(x,a)$ is just short-hand for $a\otimes x +x\otimes a$. If we have $a_e=e$ for some $e$ (but there's no obvious reason why this would be the case), then we can try to project on $e^{\perp}$. If also $a_x\not=e$ for all $x\perp e$, then this will give the claim by induction (because $\dim PWP\ge n-1$ by induction hypothesis, and then $e\otimes e$ will be linearly independent of a basis of the first space). If not, we can try to pick a maximal linearly independent set $x_1,\ldots , x_k$ with $a_j=e$ and proceed similarly. | |
| Aug 8, 2023 at 9:44 | comment | added | mathew | @ChristianRemling : I couldn't understand the notation for $T(x,a)$ here. Can it be given a little more explanation? Thank you. | |
| Aug 8, 2023 at 9:38 | comment | added | mathew | @ChristianRemling: I was first taking an element $x\otimes a_x +a_x\otimes x$ and after that another $y\in \mathbb R^n\setminus \{0\}$ such that $y\in \{x,a_x\}^{\perp}$ and this way I was getting another element $y\otimes a_y+a_y\otimes y$ and I was able to show that these two elements are linearly independent. But constructing this way , it's becoming difficult to show the linear independence of the elements. But it's working upto $n=5$ to get $n-1$ linearly independent elements. | |
| Aug 5, 2023 at 19:10 | comment | added | Christian Remling | I think one can fight one's way through this by induction, but it probably won't be fun. I think it's best to try to prove the more general version: if for all $x\in M$, for some subspace $M$, ..., then $\dim W\ge \dim M$. If $T(x,a)$ denotes the rank $2$ operator from the OP and $P$ is a projection, then $P(T(x,a))P=T(Px,Pa)$, so one can restrict to subspaces as long as there are $a$'s available that are not orthogonal to this subspace. I would like to claim that I got close to a proof along these lines, but couldn't make the details work, and I'm giving up now. | |
| Aug 5, 2023 at 18:43 | comment | added | Nathaniel Johnston | @ChristianRemling: you’re right, we just get n/2. I was just doing naive row/column counting, but forgot to take into account duplicates properly. | |
| Aug 5, 2023 at 18:16 | comment | added | Christian Remling | @NathanielJohnston: Could you explain this in more detail, I don't see how to get $n-1$ easily. Just listing rank $2$ operators by going through a basis with $x$ might not be enough since the operators could come in pairs, with repetitions: for example $(x,a)=(e_1,e_2)$, then $(x,a)=(e_2,e_1)$ etc., so when we're done, we have $n/2$ linearly independent operators. | |
| Aug 4, 2023 at 15:59 | comment | added | Nathaniel Johnston | Your argument trivially gives a bound of $\dim(W) \geq n-1$. Not sure if that’s useful. It’s just the one extra dimension that is tricky. | |
| Aug 4, 2023 at 15:28 | history | edited | mathew | CC BY-SA 4.0 |
deleted 1 character in body
|
| S Aug 4, 2023 at 15:22 | review | First questions | |||
| Aug 4, 2023 at 16:38 | |||||
| S Aug 4, 2023 at 15:22 | history | asked | mathew | CC BY-SA 4.0 |