Timeline for Resolvent of a triangular matrix
Current License: CC BY-SA 3.0
7 events
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| Nov 28, 2014 at 19:35 | comment | added | loup blanc | In my mind "no kidding"="is it a joke ?", that is about the Michele's sentence: " where $p_A$ is the characteristic polynomial of $A$, which is easy to compute once we know an eigendecomposition of A"; that is funny when one knows that $A$ is a triangular matrix. This seems to me absolutely innocuous. Compare with "Récoltes et semailles", the book written by Grothendieck; at least, read the introduction (chapter 0). | |
| Nov 28, 2014 at 17:39 | comment | added | S. Carnahan♦ | What do you mean when you say "no kidding!"? | |
| Nov 28, 2014 at 12:52 | history | edited | loup blanc | CC BY-SA 3.0 |
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| Nov 27, 2014 at 19:00 | comment | added | Michele | concerning comment 1. You don't need to make Leverrier-Faddeev to work in $K[x]$ to recover $N(x)=adj(xI-A)$. In fact, L-F recovers $n$ (constant) matrices $N_0,...,N_{n-1}$ which are the coefficients of $N(x)$ seen as a polynomial matrix, $N(x)=N_0+N_1 x+...+N_{n-1}x^{n-1}$, and does so working in $K$ (plus, it also computes the characteristic polynomial $p_A(x)$. See e.g.jstor.org/discover/10.2307/…) So the overall complexity is still $O(n^4)$. | |
| Nov 27, 2014 at 18:10 | history | edited | loup blanc | CC BY-SA 3.0 |
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| Nov 27, 2014 at 10:05 | comment | added | Michele | But a multiply in $K[x]$ costs $O(n)$, leading us again to an overall complexity of $O(n^4)$. Not to speak of numerical instability problems arising when working in $K[x]$ with finite-precision arithmetic. | |
| Nov 26, 2014 at 19:22 | history | answered | loup blanc | CC BY-SA 3.0 |