Timeline for Non-square multiplication operator matrix
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2020 at 19:27 | comment | added | Gustave | Thank you sur for the answer. Good chance. | |
| Sep 29, 2020 at 22:59 | comment | added | Nate Eldredge | In your question you used the open interval $(0,1)$. If you want the closed interval then simply modify it: $$A(x) = \begin{cases} xI, & x \ne 0 \\ I, & x = 0. \end{cases}$$ Of course if you intended for $A$ to be continuous then it will change things, and you can use compactness to argue that in this case the singular values (or some other measure of surjectivity) are indeed bounded away from zero. | |
| Sep 29, 2020 at 20:38 | comment | added | Gustave | I see that in your example $A(x)=xI$ is not invertible for any $x$ in $[0,1]$ (for example x=0). | |
| Sep 29, 2020 at 19:01 | comment | added | Nate Eldredge | It's not enough. Even take $n=2$ and $A(x) = xI$, you have exactly the same problem as in my previous example. Anyway, your new statement is equivalent to saying that every $A(x)$ has rank $2$, up to a change of basis. Unless you want $B$ to be independent of $x$, in which case that is sufficient but much too strong. | |
| Sep 29, 2020 at 18:16 | comment | added | Gustave | thank you for the interest. In fact, I believe that the right condition is that the matrix A must contains a mini matrix B 2×2 that it is invertible. What do you think sir? | |
| Sep 29, 2020 at 13:43 | comment | added | Nate Eldredge | This is not even true for $1 \times 1$ matrices, i.e. scalars; the scalar function $A(x)=x$ is "full rank" at every $x$, but $f \mapsto Af$ is not onto; the constant function $1$ is not in its image (because if $Af = 1$ then $f(x) = 1/x$ which is not in $L^2$). I think a more plausible condition would be something like "singular values bounded away from 0". | |
| Sep 29, 2020 at 7:15 | history | asked | Gustave | CC BY-SA 4.0 |