Timeline for finding positive vector that multiplied by matrix give positive vector
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2012 at 11:44 | comment | added | Emil Jeřábek | @iko: You’re welcome, I’m glad it helped. | |
| Jan 25, 2012 at 15:29 | comment | added | Federico Poloni | If your matrix is a Z-matrix, then such an $x$ can also be chosen as the eigenvector relative to the smallest eigenvalue. This might be faster, especially for large matrices. | |
| Jan 25, 2012 at 15:22 | comment | added | user20881 | Did not know before the magical abilities of Linear programming. solved my problems with tools from that field. @Emil - thanks. | |
| Jan 25, 2012 at 11:52 | comment | added | Emil Jeřábek | If you have $x$, you can scale it so that all coordinates of $x$ and $Ax$ are greater than $1$. Thus, the existence of $x$ can be expressed in terms of linear programming as feasibility of the system $\vec x\ge\vec1$, $Ax\ge\vec1$. In particular, the linear programming duality may shed some light on the existence of $x$. | |
| Jan 25, 2012 at 11:35 | comment | added | iko | I'm curios about the general case, but in the case i research now its a Z-matrix / L-matrix (the diagonal is positive and non-diagonal are negative or zero). @Fredrico thanks for the definition. it describes well the A matrix. | |
| Jan 25, 2012 at 11:33 | history | edited | Federico Poloni | CC BY-SA 3.0 |
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| Jan 25, 2012 at 11:32 | answer | added | Federico Poloni | timeline score: 2 | |
| Jan 25, 2012 at 11:19 | comment | added | Federico Poloni | I do not know if it has a name, but I have seen it used a couple of times in M-matrix literature. Is your $A$ perchance a Z-matrix? | |
| Jan 25, 2012 at 10:50 | history | asked | iko | CC BY-SA 3.0 |