Timeline for Behavior of matrix rank under thresholding of its elements
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 2, 2017 at 3:00 | comment | added | gradstudent | Yes. I meant if that rank 2 decomposition you wrote somehow helps understand why a max of that is rank 5 instead of looking at the full matrix. | |
| Aug 2, 2017 at 2:37 | comment | added | Robert Israel | Lower triangular matrix with $1$'s on the diagonal. | |
| Aug 2, 2017 at 1:08 | comment | added | gradstudent | How does it make it obvious that the rank is 5 after the max-0 operation? | |
| Aug 2, 2017 at 0:58 | comment | added | Robert Israel | Or maybe a bit easier to see: $$ \pmatrix{1 & 0 & -1 & -2 & -3\cr 2 & 1 & 0 & -1 & -2\cr 3 & 2 & 1 & 0 & -1\cr 4 & 3 & 2 & 1 & 0\cr 5 & 4 & 3 & 2 & 1\cr} = \pmatrix{1\cr 2\cr 3\cr 4\cr 5} \pmatrix{1 & 1 & 1 & 1 & 1} - \pmatrix{1\cr 1\cr 1\cr 1\cr 1\cr} \pmatrix{0 & 1 & 2 & 3 & 4}$$ | |
| Aug 2, 2017 at 0:08 | comment | added | gradstudent | Okay. So for every n you can have a rank 2 matrix whose rank after max-0 will be n. | |
| Aug 2, 2017 at 0:05 | comment | added | Gerry Myerson | I leave it to you, gradstudent, to show that 5 is a variable here, that is, the construction works for every $n$. | |
| Aug 2, 2017 at 0:02 | comment | added | gradstudent | Maybe this is an effect of the fact that here the rank i.e $2$ is already pretty close to the dimension i.e 5? | |
| Aug 1, 2017 at 23:46 | history | answered | Gerry Myerson | CC BY-SA 3.0 |