How to claim the subspace $AB-BA$ has dimension $n^2-1$? There is a problem given in LINEAR ALGEBRA by  HOFFMAN & kUNZE PAGE $107$, 
Show that the trace functional on squre matrices of order $n$ is characterized in the following sense:
If $W$ is the space of size $n$ matrices over the field $F$ and if $f$ is a linear functional on $W$ such that  $f(AB)=f(BA)$ for each $A$ and $B$ in $W$, then $f$ is a scalar multiple of the trace function.
MY TRY: Put $C=AB-BA$. Clearly $\operatorname{Tr} C=0$.
I claim that dimension of the subspace of such $C$ is $n^2-1$. So by the rank-nullity theorem $$\dim \ker f +\dim\operatorname{Im} f=n^2$$ i.e $n^2-1+\dim\operatorname{Im}=n^2$ ( as $f(C)=0$).
Hence $\operatorname{rank} f=1$.
Now since trace functional satisfies such $f$'s and it's scalar multiple is also a subspace of dimension $1$, $f$ needs to be scalar multiple of $\operatorname{Tr}$.
MY PROBLEM: I claimed that subspace (the terrible fact is I'm not sure whether it's a subspace or not, can't prove it) of the matrices of the form $AB-BA$ has dimension with $n^2-1$. It means that there is no restriction on the entries of $AB-BA$ except the fact that their trace is $0$. How can I prove it?
Any other approach or hint to solve that problem will also help me. Thanks for reading! 
 A: 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 $\ker \operatorname{Tr} \subseteq \{AB - BA : A,B \in M_n(\mathbb{C})\}$.
