Timeline for $P(Z)$ is matrix polynomials. Why is $s_n$ smooth in a neighbourhood of $Z$?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 20, 2016 at 22:07 | comment | added | R.T MAN | - Which reference say that, " the simple roots of a polynomial are smooth functions with respect to the coefficients of the polynomial."? Please, send me a god reference for this text. | |
| Mar 12, 2016 at 4:03 | history | edited | Richard Zhang | CC BY-SA 3.0 |
Hermetian transpose for complex-coefficient matrices...
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| Mar 12, 2016 at 4:01 | vote | accept | R.T MAN | ||
| Mar 12, 2016 at 3:59 | vote | accept | R.T MAN | ||
| Mar 12, 2016 at 4:01 | |||||
| Mar 12, 2016 at 3:58 | comment | added | Richard Zhang | Yes, assuming you mean $s_n(z)$ and $s_{n-1}(z)$. And that statement even generalizes to the eigenvalues of the possibly nonsymmetric matrix $P(z)$. | |
| Mar 12, 2016 at 2:24 | comment | added | R.T MAN | Can we say that, the study of points where differentiability is lost, is confined to the points of the plane where, $s_n(λ)$ meets the one corresponding to $s_{n−1}(λ)$? | |
| Mar 11, 2016 at 18:03 | history | answered | Richard Zhang | CC BY-SA 3.0 |