Timeline for An optimization problem in numerical linear algebra
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2011 at 19:35 | comment | added | Federico Poloni | The first guess was right --- it's the largest singular value of $M$, or equivalently the ''square root'' of the largest eigenvalue/singular value of $M^TM$. The two concepts coincide for $M^TM$ as it is symmetric positive definite. | |
| Apr 20, 2011 at 13:15 | vote | accept | bobye | ||
| Apr 20, 2011 at 11:24 | answer | added | Denis Serre | timeline score: 2 | |
| Apr 20, 2011 at 11:01 | comment | added | bobye | ... the largest singular value of $M^TM$. | |
| Apr 20, 2011 at 11:00 | history | edited | bobye | CC BY-SA 3.0 |
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| Apr 20, 2011 at 11:00 | comment | added | bobye | Sorry, I got it. The L_2 induced norm is given as the largest singular value of M. | |
| Apr 20, 2011 at 10:55 | comment | added | Denis Serre | The trace of $M^TM$ is the square of the Frobenius (=Schur, Hilbert-Schmidt) norm, but this norm is not an induced one. For instance because the norm of $I_n$ is $\sqrt n$ instead of $1$. | |
| Apr 20, 2011 at 10:35 | history | edited | bobye | CC BY-SA 3.0 |
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| Apr 20, 2011 at 10:28 | history | edited | bobye | CC BY-SA 3.0 |
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| Apr 20, 2011 at 9:56 | comment | added | Suvrit | Oops, i read $A$ to be a diagonal matrix, not $A^TA$ as such! | |
| Apr 20, 2011 at 8:41 | comment | added | drbobmeister | Maybe this comment is misguided, but bobye only stipulated $U$ and $V$ be diagonal. $A$ is apparently a scalar multiple of an orthogonal matrix, since $A^{T}A/a^{2} =I$. | |
| Apr 20, 2011 at 6:53 | history | asked | bobye | CC BY-SA 3.0 |