Careful: when you say "multiplicative closure", you have to be clear, since when dealing with $n\times n$ matrices there is a "multiplication" that has nothing to do with the vector space structure (the multiplication of matrices). It is clearer if you refer to it as the "scalar multiplication".
So, fix the matrix $B$. You need to show that:
- There is at least one matrix $A$ such that $AB=BA$;
- If $A_1$ and $A_2$ are two matrices such that each of them commutes with $B$, then $A_1+A_2$ also commutes with $B$ (closure of your set under vector addition).
- If $A$ is a matrix that commutes with $B$, and $k$ is any scalar, then $kA$ also commutes with $B$ (closure of your set under scalar multiplication).
So, with that in mind:
Is there a matrix that necessarily commutes with $B$? (If this thing is really going to be a subspace, it better have what "vector" [i.e., matrix] in it for sure? Try that matrix).
Suppose $A_1$ and $A_2$ both commute with $B$. That is, $A_1B=BA_1$ and $A_2B=BA_2$. You want to show that $(A_1+A_2)$ also commutes with $B$. That is, you want to show that
$$(A_1+A_2)B = B(A_1+A_2).$$
Of course, you'll want to use the fact that each of $A_1$ and $A_2$ commutes with $B$, and perhaps some properties you know about matrix multiplication. Is there some property of matrix multiplication that would let you relate $(A_1+A_2)B$ with the products you do know something about, namely $A_1B$ and $A_2B$?
Suppose $A$ commutes with $B$, and $AB=BA$. Let $k$ be a scalar. You want to show that $kA$ also commutes with $B$: that is, you need to prove that
$$(kA)B = B(kA).$$
Again, is there some property of matrix multiplication that you know and that might help here?
And if you establish these three, you're done: the set in question is a subspace!
We claim that $\{AB - BA : A,B \in M_n(\mathbb{C})\} = \{A \in M_n(\mathbb{C}), \operatorname{Tr }A = 0\} = \ker \operatorname{Tr}$.
We already know that $\{AB - BA : A,B \in M_n(\mathbb{C})\} \subseteq \ker \operatorname{Tr}$.
Let $E_{ij}$ denote the matrix with $1$ at the position $(i,j)$ and $0$ elsewhere.
Check that $B = \{E_{ij} : 1 \le i, j \le n, i\ne j\} \cup \{E_{ii} - E_{nn} : 1 \le i \le n-1 \}$ is a basis for $\ker \operatorname{Tr}$.
For $1 \le i, j \le n, i\ne j$ we have
$$E_{ij} = E_{ik}E_{kj} - E_{kj}E_{ik}$$
where $k$ is some index $\ne i,j$. To see this, let $\{e_1, \ldots, e_n\}$ be the standard basis for $\mathbb{C}^n$ and note that $E_{ij}e_j = e_i$ and $E_{ij}e_r = 0$ for $r \ne j$. Now verify that
$$(E_{ik}E_{kj} - E_{kj}E_{ik})e_r =
\begin{cases}
0, &\text{if } r \ne j,k\\
E_{ik}E_{kj}e_j = E_{ik}e_k = e_i, &\text{if }r = j\\
-E_{kj}E_{ik}e_k = -E_{kj}e_i = 0, &\text{if }r = k\\
\end{cases}$$
For $1 \le i \le n-1$ we have
$$E_{ii} - E_{nn} = E_{in}E_{ni} - E_{ni}E_{in}$$
Indeed
$$(E_{in}E_{ni} - E_{ni}E_{in})e_r =
\begin{cases}
0, &\text{if } r \ne i,n\\
E_{in}E_{ni}e_i = E_{in}e_n = e_i, &\text{if }r = i\\
- E_{ni}E_{in}e_n = -E_{ni}e_i = -e_n, &\text{if }r = n\\
\end{cases}$$
Therefore $B \subseteq \{AB - BA : A,B \in M_n(\mathbb{C})\}$ so we conclude $\ker \operatorname{Tr} \subseteq \{AB - BA : A,B \in M_n(\mathbb{C})\}$.
Best Answer
Take X, Y two elements of W. Then they satisfy: XAB=BAX, YAB=BAY. So, for the sum we have (X+Y)AB=XAB+YAB=BAX+BAY=BA(X+Y). That means that the sum X+Y belongs to W. Now for the scalar multiply, (λΧ)(ΑΒ)=λ(ΧΑΒ)=λ(ΒΑΧ)=ΒΑ(λΧ). That means that λΧ belongs to W.