Showing a linear functional satisfies $f(A) = \lambda tr(A)$ I have a linear functional from the space of nxn matrices over a field F. The functional satisfies $f(A) = f(PAP^{-1})$ for all invertible $P$ and $A$ an nxn matrix. I'm trying to show that $f(A) = \lambda tr(A)$ for some constant $\lambda$.
So far I have:


*

*The linear functionals have basis $f_{E_{i,j}}(A) = tr({E_{i,j}}^t A)$ where $E_{i,j}$ is zero everywhere except in the $i,j$ position.

*If $PAP^{-1} = A$ for all invertible $P$, then A is a multiple of the identity matrix.


Thanks for any help 
 A: I write this as a separate answer, because the method of solution is different from my previous post.
By your first observation, or simply using the Riesz Representation Theorem, one can deduce that there exists a matrix $B$ such that $f(X)=\text{tr }(BX)$. Since $f(X)=f(PXP^{-1})$ and $\text{ tr} AB=\text{tr } BA$, one concludes that
$$f(X)=\text{tr } (BPXP^{-1})=\text{tr }(P^{-1}BP X),$$
and consequently without loss of generality $B$ we can (and we will) assume that $B$ is an upper triangular matrix. The diagonal entries of $B$ are equal because otherwise they can be permuted by means of conjugation by some $P$ and this changes the value of $\text{tr }BX$ for an appropriate choice of diagonal matrix $X$. 
The upper diagonal entries of $B$ have to be zero, because otherwise we can fix some $k\ne l$ such that $b_{kl}\ne 0$. Now let the nonzero entries of the $n\times n$ matrix $X=[x_{ij}]$ be as follows $x_{ii}=i$ and $x_{lk}=1$. This leads to a contradiction, because 
$$f(X)=BX\ne f(\text{diag} (1, \dots, n))=B\text{ diag} (1, \dots, n)$$
whereas $X$ and the diagonal matrix $\text{diag} (1, \dots, n)$ are similar.
Putting these together, we conclude that $B$ is a scalar matrix.
A: From the given condition it is clear that $f$ satisfies $f(AB)=f(BA)$ for any two square matrices $A$ and $B$. Next recall that the kernel of the trace function on $M_n(\mathbb{R})$ is $n^2-1$ dimensional. Thus any square matrix $X$ can be written in the form $X=Z+cA$ where $Z$ is a matrix with zero trace and $A$ is any fixed matrix with nonzero trace. Hence 
$$f(X)=cf(A).$$
This shows that $f$ depends only on the value of $c$. On the other hand tr $X=c\text{tr }A$ and hence $f$ has to be a multiple of the trace function.
