How do i directly show that $\text{rng} (T)$ is $F$ where $T$ is a trace over a field $F$? Let $T:M_{n\times n} (F) \rightarrow F : A\mapsto \text{tr}(A)$.
I can explicitly find a basis for $\ker (T)$, so that $\text{nullity}(T)=n^2 -1$.
Then, by dimension theorem, $\text{rng}(T)$ must be equal to $F$, but how do i show this directly?
If $F$ is infinite, this is trivial, but what if $F$ is finite? Say $F$ is of characteristic $n$ and fix $a\in F$ such that $a$ is nonzero. Then $\frac{1}{n}$ does not exist. How do i show that $F\subset \text{rng}(T)$ directly?

Show that $T$ is linear, and has at least one non-zero value. Then deduce from linearity that $T$ is surjective.
Or, show that for every $x\in F$ the matrix which has $x$ in the top-left corner and zero elsewhere, has trace $x$.
