Unique solution comes out to be trace Let $f: \mathcal{M}_n(F)\to F$ be a linear functional satisfying $f(\mathbf{AB})=f(\mathbf{BA})\;\forall \mathbf{A,B}\in \mathcal{M}_n(F)$. Also it satisfies $f(\mathbf{I})=n$. Prove that $f(\mathbf{A})=\operatorname{tr}(\mathbf{A})$.
Can someone tell me how to prove this? I have no idea about how to start it. My first idea was to showing this for the canonical basis but I failed. Can someone help me? Thanks.
 A: You can indeed use the canonical basis fruitfully here:
Let $E_{ij}$ that has $1$ in the $(i, j)$ entry and $0$ elsewhere. Then, for $i \leq k \leq j$ but $i < j$ we have $E_{ik} E_{kj} = E_{ij}$ but $E_{kj} E_{ik} = 0$, and so for every element $E_{ij}$ with $i \neq j$ we have
$$f(E_{ij}) = f(E_{ik} E_{kj}) = f(E_{kj} E_{ik}) = f(0) = 0.$$
What can we conclude from this?
Spolier/additional hint:

 Thus, $f$ is a linear functional on the diagonal elements of $\mathcal{M}_n(F)$. A similar argument shows that $f(E_{ii}) = f(E_{jj})$ for all $i, j$, and so the normalization condition $f({\bf I}) = n$ forces the functional to be the trace.

Alternate solution using a fact about the Lie algebra $\mathfrak{gl}(n, F)$.
If you know a little about the Lie algebra $\mathfrak{gl}(n, F) \cong \mathcal{M}_n(F)$ we can proceed even more quickly. For any $A, B \in \mathcal{M}_n(F)$, we have
$$f([A, B]) = f(AB - BA) = f(AB) - f(BA) = 0.$$ But, the derived Lie algebra of $\mathcal{M}_n(F)$, that is, the Lie subalgebra of all elements that can be written as the bracket of two elements of $\mathcal{M}_n(F)$ is $\mathfrak{sl}(n, F)$, and so, $f(A) = 0$ for any tracefree element $A \in \mathcal{M}_n(F)$. Thus, $f$ depends only on the trace of the argument, and hence we may think of it as a linear map $F \to F$. But the space of such maps is $1$-dimensional, and the trace is such a map (and nonzero), so $f$ must be a multiple of the trace operator. Then, the condition $f({\bf I}) = n$ forces it again to be the trace itself.
A: Consider a linear functional $f$. There exists a matrix of coefficients $C = (c_{ij})$ so that we have for any $A$
$$f(A) = \sum c_{ji} a_{ij} $$
This can also be written as $f(A) = \text{tr}(AC)$. We'll show that $C = I_n$. 
From $f(AB) = f(BA)$ we get 
$$\text{tr}(ABC)= \text{tr}(BAC)$$
However, by the properties of trace we have $\text{tr}(BAC)=\text{tr}(B(AC))=\text{tr}(ACB) $. Therefore
$$\text{tr}(ABC)= \text{tr}(ACB)$$
for any matrices $A$, $B$. Fix a matrix $B$ and let $A$ vary in the equality above. $A$ provides the coefficients like before $C$. It follows that $CB = BC$. This happens for all $B$ and this implies $C$ is a scalar matrix $C = \lambda I_n$. Now use $n=f(I) = \text{tr}(C) = \text{tr}(\lambda I_n) = n \lambda$ and therefore $\lambda = 1$, $C= I_n$ and $f(A)= \text{tr}(A \cdot I_n) = \text{tr}(A )$ for all $A$.
