Timeline for Use of traces in physics
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Aug 27, 2019 at 13:24 | comment | added | LSpice | @MarcPalm, aren't the statement "[the] trace is coordinate independent" and "[it is] invariant under conjugation by invertible matrices" the same thing? (Your comment seems to imply that the latter is a stronger statement.) | |
| Mar 29, 2011 at 0:54 | comment | added | Logan M | I'll attempt to justify some of why traces are important, though not solely from a physics perspective. The trace of a matrix $M$ comes naturally (as does the determinant) from the characteristic polynomial $p(\lambda) = \det(\lambda I-M)$. Namely, for any algebraically closed field, it is the sum of the roots of $p(\lambda)$ counted with multiplicity (the determinant is the product). This is clearly coordinate-independent and a rather fundamental quantity. The usual definition lacks any intuition, but is more useful for generalizing to arbitrary matrix rings and for efficient computation. | |
| Mar 28, 2011 at 17:49 | answer | added | Russell May | timeline score: 4 | |
| Mar 28, 2011 at 16:01 | answer | added | Simon Wadsley | timeline score: 0 | |
| Mar 28, 2011 at 13:56 | answer | added | José Figueroa-O'Farrill | timeline score: 7 | |
| Mar 28, 2011 at 12:42 | answer | added | Jon Bannon | timeline score: 6 | |
| Mar 28, 2011 at 12:38 | answer | added | Pieter Naaijkens | timeline score: 4 | |
| Mar 28, 2011 at 12:32 | comment | added | Marc Palm | What kind of traces do you consider? The usual matrice trace is coordinate indepenent and even invariant under conjugation by invertible matrices, since $tr(ab) = tr(ba)$. | |
| Mar 28, 2011 at 12:29 | history | asked | John R Ramsden | CC BY-SA 2.5 |