# Uniqueness of Trace

• Mar 27th 2008, 10:39 AM
syme.gabriel
Uniqueness of Trace
Here's the problem:

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

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

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

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

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

I know that it can be shown that $\displaystyle f$ 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!
• Mar 28th 2008, 03:12 AM
Opalg
Quote:

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

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

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

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

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

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