Timeline for The multiplicity of the max eigenvalue in matrix multiplication
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 5, 2012 at 23:01 | comment | added | Aaron Meyerowitz | In this variation you want to consider $C=kA+X$ for the matrix $X=AB$. The eigenvalues are simply $k$ times the eigenvalues of $C'=A+X/k. $ So you wonder when $C'$ has all eigenvalue real and the largest one of multiplicity $1$ (given that these things are true of $A$) It seems a sure thing that whatever $X$ is, for and $\epsilon \gt 0$ there is a $K$ such that for all $k \gt K,$ $\max_j(|\lambda^A_j-\lambda^{C'}_j|) \lt \epsilon.$ How to find $K$ given $A$ and $X$, or just their spectra, I can't say. | |
| Mar 5, 2012 at 20:06 | comment | added | David | So $A$ satisfies these constrains, $A^{-1}$ must not satisfies. The question is not reasonable. If matrices $A$ and $B$ are given, does there always exists $k \in Z^+$ such that $C=A(kI + B)$ has the property 2 (i.e. the max eigenvalue of $C$ is real and has multiplicity of 1)? If there always exist $k$, how can we find it? Thanks, David | |
| Mar 5, 2012 at 19:50 | comment | added | David | Thank you Igor. Can we add more constrains (properties) on $A$ and $B$ such that the max eigenvalue of $AB$ is real and has multiplicity of 1? If so, can you suggest some kind of constrains that I can use? | |
| Mar 5, 2012 at 16:55 | comment | added | Igor Rivin | Sigh, reading comprehension is not my friend :( | |
| Mar 5, 2012 at 16:23 | comment | added | user6976 | You cannot take $B=A^{-1}$ because its highest eigenvalue will be $>1$. I have deleted my answer. | |
| Mar 5, 2012 at 16:10 | history | answered | Igor Rivin | CC BY-SA 3.0 |