Let $M_n(F)$ be the matrix space on the field $F$, $f$ be the linear map from $M_n(F)$ to $F$ such that $f(I)=n$, where $I$ is the identitiy matrix. Furthermore, $f(AB)=f(BA)$ for any $A,B\in M_n(F)$. Show that $f=tr$, where $tr(A)=\sum_{i=1}^n a_{ii}$ for $A=(a_{ij}).
I do not have any idea....
For this kind of problems, I would prefer a more elementary approprach. See Nathan Portland's answer, for instance. Here is an answer from a higher perspective.
By the given assumptions, we have $f(AB-BA)\equiv0$. However, the set of all commutators is exactly the space of all traceless matrices (see this question, for instance). So, we have $f(X)=0$ for all traceless matrices $X$. Thus, for a general $X$, $$ f(X)=\underbrace{f\left(X-\frac{\operatorname{trace}(X)}{n}I\right)}_{=0}+\frac{\operatorname{trace}(X)}{n}\underbrace{f(I)}_{=n} = \operatorname{trace}(X). $$
It is probably already on this forum, this is well known. Denote by $E_{i,j}$ the $n\times n$ matrix with a $1$ in position $(i,j)$ and zero elsewhere.
For $i,j$ in $\{1,\ldots,n\}$ we have $f(E_{i,i}) = f(E_{i,j}E_{j,i}) = f(E_{j,i}E_{i,j}) = f (E_{j,j}) = 1$.
If $i\neq j$, we have $f(E_{i,j}) = f (E_{i,j}E_{j,j}) = f (E_{j,j}E_{i,j}) = f (0) = 0$.
We conclude writing $A= \sum_{i,j=1}^n a_{i,j}E_{i,j}$.
