TITLE: Linear functional vanishing on matrices of square zero is a scalar multiple of the trace
QUESTION [1 upvotes]: Let $f: \mathcal{M}_n(F)\to F$ be a linear functional, satisfying 
$$ f(A)=0\text{ whenever }A^2=0$$Show that $f(A)=c\operatorname{tr}(A)$ for some $c\in F$.
Can someone help me out here? I am out of ideas in this one. Thanks.

REPLY [1 votes]: Every rank$~1$ endomorphism $\def\tr{\operatorname{tr}}\phi$ of an $n$-dimensional space with trace$~0$ satisfies $\phi^2=0$. This can be obtained from the more general fact that every rank$~1$ matrix$~A$ has minimal polynomial $X(X-\tr A)$, but also simply be expressing $\phi$ on a basis obtained by completing a basis of the $n-1$-dimensional subspace $\ker\phi$, for which the final diagonal entry equals $\tr\phi=0$.
One easily sees that the space of matrices of trace$~0$ is spanned by its subset of (traceless) matrices of rank$~1$: it suffices to note that every elementary matrix $E_{i,j}$ (with $i\neq j$) has rank$~1$, as does every nonzero (traceless) matrix with equal entries along each of its rows. Then the fact that $f$ vanishes on matrices$~A$ with $A^2=0$ implies it vanishes on all matrices of trace$~0$, and therefore it must be a multiple of the trace function.
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