The trace functional and its scalar multiples I am trying to solve the following problem:
Show that the trace functional on $n \times n$ matrices is unique in the following sense. If $W$ is the space of $n \times n$ 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.
I know that one way to prove this is to show that the null space of $f$ equals the null space of the trace function, but I found this solution rather complicated; any other suggestions?
 A: The space $M_n(F)$ is the span of the matrix units $\{E_{kj}\}$, where $E_{kj}$ is the matrix with $1$ in the $k,j$ coordinate and zero elsewhere. These matrix units satisfy 
$$
E_{kj}E_{st}=\delta_{j,s}\,E_{kt},
$$
and $\sum_kE_{kk}=I$. Now let $f$ be a tracial functional. Then, if $k\ne j$,
$$
f(E_{kj})=f(E_{k1}E_{1j})=f(E_{1j}E_{k1})=f(0)=0.
$$
So $f(E_{kj})=\text{Tr}(E_{kj})=0$ when $k\ne j$. We have
$$
f(E_{kk})=f(E_{k1}E_{1k})=f(E_{1k}E_{k1})=f(E_{11}).
$$
Then 
$$
f(I)=\sum_kf(E_{kk})=nf(E_{11}).
$$
Then $f(E_{kk})=f(I)/n$ for all $k$. We have shown that, for arbitrary $k,j$
$$
f(E_{kj})=\frac{f(I)}{n}\,\text{Tr}\,(E_{kj}).
$$
As the matrix units span all the matrices, $$f(A)=\frac{f(I)}{n}\,\text{Tr}\,(A)$$ for all matrices $A$.
A: Hint: let $sl(N)$ be the matrices whose trace is zero, it is semi simple, this implies that is $sl(N)=[SL(N),SL(N)]$, remark that the $f(AB)=f(BA)$ implies that $f$ vanishes on the commutator of two matrices$ [A,B]=AB-BA$ so it vanishes on $sl(N)$. Now the codimension of $sl(N)$ is 1, thus write $f(I_n)=ctrace(I_n)$, you obtain $f= c(trace)$ since you can write every matrix as asum of an element of $sl(N)$ and an element of the form $xI_n$.
