If $A$ is a semisimple algebra over $\mathbb{C}$, then the codimension of $[A,A]$ in $A$ is the number of isomorphism classes of simple $A$-modules. Claim: (p. 145 in Groups and Representations by Alperin & Bell) 
If $A$ is a semisimple algebra over $\mathbb{C}$, then the codimension of $[A,A]$ in $A$ is the number of isomorphism classes of simple $A$-modules, where $[A,A]:=\{ab-ba:a,b\in A\}$.
Pf: What I have thus far is the following: 
Consider the matrices $e_{ij}=\delta_{ij}$.
If $i\neq j$, $e_{ij}=e_{ii}e_{ij}-e_{ij}e_{ii}\in [A,A]$
Also, $e_{ii}-e_{jj}=e_{ij}e_{ji}-e_{ji}e_{ij}\in[A,A]$
So, 
$[A,A]= \left\{ 
\begin{bmatrix}
    x_{11}       & x_{12} & x_{13} & \dots & x_{1n} \\
    x_{21}       & x_{22} & x_{23} & \dots & x_{2n} \\
    \vdots &\vdots&\vdots& &\vdots \\
    x_{n1}       & x_{n2} & x_{n3} & \dots & x_{nn}
\end{bmatrix}: x_{11}+\ldots +x_{nn}=0\right\}$
I do not understand which set $\{e_{ij}\}$ belong to. 
 A: If $A$ is a semisimple algebra over $\Bbb C$ then it is isomorphic to
$\prod_{i=1}^n M_{a_i}(\Bbb C)$ for some $a_i$. The number of isomorphism
classes of simple modules in $n$. For $A=\prod_{i=1}^n M_{a_i}(\Bbb C)$,
$[A,A]=\prod_{i=1}^n M_{a_i}^0(\Bbb C)$ where $M_a^0(\Bbb C)$ is the
space of trace-zero matrices. This has codimension one in $M_a(\Bbb C)$,
so that $[A,A]$ has codimension $n$ in $A$.
A: The technological answer here is the following one:
If $A$ is an algebra, the vector space $A/[A,A]$ is called the $0$th Hochschild homology of $A$. 


*

*One of the key properties of Hochschild homology is that it is invariant under Morita equivalence: if $A$ and $B$ are algebras which are Morita equivalent, then the vector spaces $A/[A,A]$ and $B/[B,B]$ are isomorphic. In particular, since for every algebra $A$ the algebras $A$ and $M_n(A)$ are Morita equivalent (this is the most basic example of such an equivalence), we always have $$\frac{A}{[A,A]}\cong\frac{ M_n(A)}{[M_n(A),M_n(A)}.$$

*On the other hand, Hochschild homology playes well with direct product: if $A$ and $B$ are algebras, then $$\frac{A\times B}{[A\times B,A\times B]}\cong\frac{A}{[A,A]}\times\frac{B}{[B,B]}$$.
What you want follows from these two facts directly, but in fact it follows form the first one alone. That an algebra is semisimple over an algebraically closed field implies easily that its category of modules is equivalent to that of a product of copies of the field —this is a strong form of Schur's lemma—  so one does not really need Wedderburn's theorem.
