# 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!

• Please see the link given in the (solution) for the following post. math.stackexchange.com/questions/1007473/… – Anurag A Jul 17 '18 at 16:54
• Don't you mean $C = AB - BA$ after the words "MY TRY:"? Cheers! – Robert Lewis Jul 17 '18 at 16:58
• Sorry1let me edit – Subhajit Saha Jul 17 '18 at 17:42
• Buried in this past thread we also have a proof for the fact that all trace zero matrices are linear combinations of commutators. – Jyrki Lahtonen Jul 17 '18 at 17:55

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 $\ker \operatorname{Tr} \subseteq \{AB - BA : A,B \in M_n(\mathbb{C})\}$.

• I am trying to understand your answer.How did you space in my question in nice way?I haven't learn Latex properly. – Subhajit Saha Jul 17 '18 at 18:20
• @SubhajitSaha You can see the Latex code if you try to edit my post, or you can mouse over the formulas and right click, and then select Show Math As $\to$ TeX Commands. – mechanodroid Jul 17 '18 at 18:22
• It isn't obvious that matrices of the form $AB - BA$ form a subspace. – Christian Sykes Jul 17 '18 at 18:41
• @ChristianSykes True, this proof only works if we know that it is a subspace. – mechanodroid Jul 17 '18 at 18:58
• @mechanodroid I can't understand why your are saing the matrices of the form $AB-BA$ may not be a subspace ? You have proved it as a subspace $KerTr$. So obviously a subspace. – Subhajit Saha Jul 17 '18 at 22:23