If $C = M_{n \times n}(F)$ and $\text{trace}(C)=0$ then $\exists{A,B}$ s.t $C=AB-BA$.
Asked
Active
Viewed 315 times
0
-
I think you might want $C \in M_{n \times n}(F)$ instead of $C = M_{n \times n}(F)$. Cheers! – Robert Lewis Apr 06 '19 at 17:25
-
Hello and welcome to math.stackexchange. Please tell us what your thoughts are for this problem. Can you prove this in the $2 \times 2$ case? When $F$ is the field of real numbers? Does the reverse implication hold? – Hans Engler Apr 06 '19 at 17:25
-
(Note that all of the discussion besides the Albert-Muckenhoupt paper seems to be beating around the bush.) – darij grinberg Apr 06 '19 at 17:28