Timeline for Matrix decomposition the other way
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2017 at 9:27 | history | edited | Mark | CC BY-SA 3.0 |
removed duplicate sentence.
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| May 19, 2011 at 15:36 | comment | added | Mark | Hauke: In this case it's as simple as it can be: your matrix (which I shall denote $A$) has just one eigenvalue $c_1 = 1$ with multiplicity 6, so the formula for $E_1$ ($T_1$ in your post) is to be interpreted as an empty product and gives the identity operator/matrix $I=I_{6 \times 6}$. Obviously $A \ne I$ but $A-I$ is a nilpotent matrix (and a very simple one, one might add) and you get the decomposition I mentioned. One can also arrive at this by using the uniqueness of the decomposition, since $I$ is clearly diagonalizable, $A-I$ is clearly nilpotent and of course they commute. | |
| May 19, 2011 at 14:46 | comment | added | Hauke Reddmann | @Charles - I know I could LaTeX here...but I can't LaTeX :-) THX. @Geoff - "Looks like Langrange interpolation" was my thought also. @Darij - using the same "n" for dimension and sum was unfortunate, since if I have an (not defective!) eigenvalue of multiplicity m>1, I simply throw them together in one of the T_i. @Mark - could you compute an actual example (the horror of any true mathematician :-) with a defective eigenvalue involved? Lets say, to keep it simple and already in Jordan form 110000 011000 001000 000110 000010 000001 - how does the decomposition look? | |
| May 19, 2011 at 14:31 | vote | accept | Hauke Reddmann | ||
| May 18, 2011 at 19:52 | history | answered | Mark | CC BY-SA 3.0 |