= Solution
To prove that <trace is the unique normalized cyclic linear functional>, let $E_{ij}$ be the <matrix units>. For $i\ne j$, $E_{ii}E_{ij}=E_{ij}$ whereas $E_{ij}E_{ii}=0$, so the assumed identity for the <linear functional> gives $f(E_{ij})=0$. Also, $E_{ij}E_{ji}=E_{ii}$ and $E_{ji}E_{ij}=E_{jj}$, giving $f(E_{ii})=f(E_{jj})$. Write this common value as $c$. Since the <matrix units> form a <basis>,
$$
\boxed{f(A)=\sum_{i,j}a_{ij}f(E_{ij})=c\sum_i a_{ii}=c\operatorname{Tr}(A).}
$$
For $n=1$ the same conclusion follows immediately from the one-dimensional <basis> $E_{11}$. Finally, $f(I)=nc$; if this equals $n$, then $c=1$ over either specified field. \b[The normalization makes $f$ exactly the trace.]
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