Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216). I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me. Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation. Part (b) asks the reader to use part (a) to show that similar matrices have the same trace. If $A$ and $B$ are similar, then

$\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$

$= \operatorname{Tr}(BSS^{-1})$

$= \operatorname{Tr}(B)$,

which completes part (b). Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 - 1$. Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation. From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space. Part (d), however, I'm stuck on. It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB - BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices. I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 - 1$ matrixes. I also tried to think of a linear transformation whose image would have the form of $AB - BA$, but this also didn't help me. I'm kind of stuck...

Many thanks in advance!

share|cite|improve this question
Wikipedia says: any square matrix with zero trace is the commutator of some pair of matrices. | Proof: $\mathcal{sl}_n$ is a semisimple Lie algebra and thus every element in it is the commutator of some pair of elements, otherwise the derived algebra would be a proper ideal. Not that I understand this. – anon Jun 14 '12 at 0:46
That doesn't sound very suitable for someone reading an introductory linear algebra text... – Ben Millwood Jun 14 '12 at 0:49
@Yuki: why is that a comment rather than an answer? – Ben Millwood Jun 14 '12 at 0:49
@benmachine: "corrected"! =p – Yuki Jun 14 '12 at 0:54
As Yuki's answer shows, it is far easier to show that the commutators span the set of matrices of trace zero, than it is to show that every matrix of trace zero is a commutator (I think this also requires some hypothesis on the field $F$; I am not sure it is true for all fields). But, one reasonably elementary proof of this second result, which should work for $F = \mathbb{R}$ or $\mathbb{C}$ anyway, is given in William Kahan's short note "Only commutators have trace zero": – leslie townes Jun 14 '12 at 1:00
up vote 6 down vote accepted

One way of proving this: Note that for all $A,B$ matrices, $AB−BA$ has trace equal to zero. Denote by $E_{ij}$ the matrix with entry $1$ in row $i$, column $j$ and $H_{ij}:=E_{ii}−E_{jj}$. Then $\{E_{ij}:i\ne j\}∪\{H_{i,i+1}:1\le i\le n-1\}$ form a basis for the space. Also, $H_{ij}=E_{ij}E_{ji}−E_{ji}E_{ij}$ and $2E_{ij}=H_{ij}E_{ij}−E_{ij}H_{ij}$. So, you have a basis formed by elements of the form $AB−BA$.

share|cite|improve this answer
Thanks Yuki, that's a huge help. I noticed that the basis you mentioned has some extra vectors in it. It suffices to use $\{E_{i,j}:i\ne j\}∪\{H_{i,i+1}:1 \le i \le n-1\}$. Great answer! – Andrew Jun 14 '12 at 1:44
@Andrew: oops... Sorry! And thanks for correcting! =p – Yuki Jun 14 '12 at 1:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.