Timeline for A strange matrix equality
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2013 at 9:19 | vote | accept | driss-alamilouati | ||
| Jun 16, 2013 at 10:41 | comment | added | Denis Serre | This sounds like a polynomial identity. There are many over $M_n(k)$, the simplest being that $S_{2n}(A^1,\ldots,A^{2n})=0_n$, with $S_m$ the standard non-commutative polynomial in $m$ variables. See the MO question mathoverflow.net/questions/38698 . | |
| Jun 16, 2013 at 10:31 | answer | added | Vladimir Dotsenko | timeline score: 13 | |
| Jun 16, 2013 at 4:40 | answer | added | Steven Landsburg | timeline score: 5 | |
| Jun 15, 2013 at 22:40 | comment | added | Gerry Myerson | Ralph, when $n=2$ and $A$ has trace zero, its square is a scalar matrix, whence $A^2B=BA^2$. | |
| Jun 15, 2013 at 22:08 | comment | added | Ralph Furman | Is this an actual identity? If you take A to have zero trace and B to have nonzero trace then the above implies AAB=BAA | |
| Jun 15, 2013 at 22:03 | history | asked | driss-alamilouati | CC BY-SA 3.0 |