Timeline for Nth root of a matrix as an analytic function?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 8, 2013 at 8:54 | comment | added | Piotr Migdal | @Federico So, you have my thanks in arxiv.org/abs/1305.1506. | |
| Mar 6, 2013 at 22:53 | vote | accept | Piotr Migdal | ||
| Feb 24, 2013 at 21:11 | comment | added | Federico Poloni | Sorry - I meant to add Chapter 1, but I pressed "enter" too quickly. | |
| Feb 24, 2013 at 21:08 | comment | added | Federico Poloni | The multiplicity of the eigenvalue $1$ is $2$, so you need a polynomial that matches $g(z)$ and $g'(z)$ in $z=1$, where $g(x)=\sqrt[n]{x}$. For some more detail on this approach, you can check Higham's Functions of matrices, SIAM Press 2008. | |
| Feb 24, 2013 at 20:30 | comment | added | Piotr Migdal | However, it is not as simple - I cannot assume that the matrix is diagonalizable; so any function which just maps eigenvalues to their roots won't work. Take as a counterexample $A = [[1, 1], [0, 1]]$ and $f(z)=z$ (sure, another polynomial works for this $A$). | |
| Feb 24, 2013 at 20:23 | vote | accept | Piotr Migdal | ||
| Feb 24, 2013 at 20:26 | |||||
| Feb 24, 2013 at 20:19 | comment | added | Piotr Migdal | Yes, I'm fine with coefficients depending on the matrix. I don't know why I overlooked this solution. | |
| Feb 24, 2013 at 20:08 | history | answered | Federico Poloni | CC BY-SA 3.0 |