Timeline for Can we claim that all the terms in a matrix are less than equal to 1 if spectral radius is less than 1?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Mar 29, 2016 at 8:11 | comment | added | reuns | @RohitShukla : but when the matrix is not symmetric, it is not true anymore, take $M = \left(\begin{array}{ll}1&2000000\\0&0\end{array}\right)$ whose eigenvalues are $1$ and $0$, hence its spectral radius is $1$ ... | |
| Mar 29, 2016 at 4:10 | comment | added | Nate Eldredge | To say it another way, for a symmetric matrix, the spectral radius equals the operator norm. And by Cauchy-Schwarz $|u^T H v| \le |u| |Hv| \le |u| |v| \|H\|$. Then take $u,v$ to be unit basis vectors. | |
| Mar 29, 2016 at 3:03 | vote | accept | Rohit Shukla | ||
| Mar 29, 2016 at 2:59 | history | answered | David E Speyer | CC BY-SA 3.0 |