Proving that dimension of two vector spaces are the same.

Question

Let $V=\{A\in \mathbb{F^{n \times n}} \ \ | \ \mathrm{Trace}(A)=0\}$ and $W=\{B\in \mathbb{F^{n \times n}} \ \ | \ \ B=CD-DC, C \ \text{and} \ D \in \mathbb{F^{n \times n}} \}$. Prove that $V=W$.

It is easy to prove that $W \subset V$. To prove the other containment, I need to show that $\dim W=\dim V$. It is clear to see that $\dim V=n^2-1$. I am stuck in finding the dimension of $W$...

We can observe that $$E^{kl}E^{rs}-E^{rs}E^{kl}=\delta_{lr}E^{ks}-\delta_{ks}E^{rl}$$($E^{kl}$ is a matrix having all entries as zero except the one in the $k$-th row and $l$-th column) Hence if $i \neq j$, $E^{ij} $ can be written as $$E^{ij} =E^{il}E^{rj}-E^{rj}E^{il}$$ Therefore $E^{ij} \in W$ for $i\neq j$.

If $i=j$, then we have to choose $l=r$. then $$E^{ir}E^{ri}-E^{ri}E^{ir}=(E^{ii}-E^{rr})\in W$$ Now let $i=1$, then we can find n-1 linearly independent matrices in W : $$E^{11}-E^{22},E^{11}-E^{33}, \ldots, E^{11}-E^{nn} \ \ \ (*)$$ So far we have $(n^2-n)+(n-1)=n^2-1$ linearly independent matrices in $W$. But How can I show that $\dim W\neq n^2$, equivalently how can I show that any $E^{jj}$ can be written as a linear combination of elements (*) ?

Thanks !

Answer 1

Well, obviously $\dim W\ne n^2$, because it would imply that $W$ includes all $n\times n$ matrices, and it does not. The only choice left is $\dim W=n^2-1$.

Answer 2

I think you're better off proving that $tr(CD-DC)=0$ directly. The fact that commutators are trace-free is actually a fairly standard fact that comes up frequently in physics. Anyway, we have

$$(CD)_{ii} = \sum_{j=1}^n c_{ij} d_{ji}$$

and

$$(DC)_{ii} = \sum_{j=1}^n d_{ij} c_{ji}$$

So

$$tr(CD-DC)=\sum_{i=1}^n \sum_{j=1}^n c_{ij} d_{ji} - d_{ij} c_{ji}$$

Now you just pair off terms. That is, if we let $f_{ij} = c_{ij} d_{ji} - d_{ij} c_{ji}$, then $f_{ij} = -f_{ji}$, so the $f_{ii}$ are zero and all the other $f_{ij}$ are canceled by the corresponding $f_{ji}$.