1. ## Uniqueness of Trace

Here's the problem:

Let $f$ be a function defined on the set of $n \times n$ matrices with entries from the field $F$ such that

$f(A + B) = f(A) + f(B)$,

$f(\lambda A) = \lambda f(A)$,

$f(AB) = f(BA)$.

Prove that there is an element $\alpha_0 \in F$ such that $f(A) = \alpha_0 \text{trace} (A)$.

I know that it can be shown that $f$ has the same value for all matrices in a similarity class, but I have no idea where to go from there. Any insight is appreciated!

2. Originally Posted by syme.gabriel
Let $f$ be a function defined on the set of $n \times n$ matrices with entries from the field $F$ such that

$f(A + B) = f(A) + f(B)$,

$f(\lambda A) = \lambda f(A)$,

$f(AB) = f(BA)$.

Prove that there is an element $\alpha_0 \in F$ such that $f(A) = \alpha_0 \text{trace} (A)$.
Write $E_{ij}$ for the matrix with a 1 (the identity element of F) in the (i,j)-position and zeros everywhere else. If i≠j then $E_{ii}E_{ij} = E_{ij}$ and $E_{ij}E_{ii} = 0$, from which it follows that $f(E_{ij})=0$. Also, $E_{ij}E_{ji} = E_{ii}$ and $E_{ji}E_{ij} = E_{jj}$, so that $f(E_{ii}) = f(E_{jj})$. So if $f(E_{11}) = \alpha_0$ then $f(E_{ii}) = \alpha_0$ for all i. Thus $f(E_{ij}) = \alpha_0\text{tr}(E_{ij})$ for each matrix unit $E_{ij}$. Since these elements linearly generate the whole of $M_n(F)$, the result follows.