If $\mathrm{tr}(A)=0$, then we have $A=BC-CB?$

Question

For any matrix $A_{n\times n}$ with $\mathrm{tr}(A)=0$ show that there exist two matrices $B$ and $C$ such that $$A=BC-CB.$$

I know to prove this: if $A=BC-CB$, then we have $\mathrm{tr}(A)=0$ because $$\mathrm{tr}(BC)=\mathrm{tr}(CB)$$ so $$\mathrm{tr}(A)=\mathrm{tr}(BC-CB)=\mathrm{tr}(BC)-\mathrm{tr}(CB)=0.$$ But my problem is that I can't prove it.

The trace operator is linear form, so $tr(A+B)=tr(A)+tr(B)$. Moreover the elements among the diagonal are the same even if you invert the order in the product. — 7raiden7, Apr 16 '14 at 10:36
Note that the trace operation is commutative: trace(AB)=trace(BA) — Alijah Ahmed, Apr 16 '14 at 11:02

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

I have shown, in an answer to a related question, that over a field $\mathbb F=\mathbb R$ or $\mathbb C$, every traceless matrix is a commutator (i.e. $\operatorname{trace}(C)=0\Rightarrow C=AB-BA$ for some square matrices $A$ and $B$ over $\mathbb F$). For a general ground field, a reference is also given in the answer.

edited Apr 13 '17 at 12:21

Community

1

answered Apr 16 '14 at 11:02

user1551

139,064

If $\mathrm{tr}(A)=0$, then we have $A=BC-CB?$

1 Answers1

Linked