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.

  • $\begingroup$ I don't think that your condition alone is sufficient for $f$ to be a scalar multiple of a trace- see my answer below. However, I might have missed something. One needs to show (or have) that $f(e_{i,i})$ have the same value for each $i$. $\endgroup$ – voldemort Oct 14 '14 at 13:48
  • $\begingroup$ These comments are enough. One just has to find a basis of the space of traceless matrices, such that $B^2=0$ for all the elements in the basis. $\endgroup$ – daw Oct 14 '14 at 13:51
  • $\begingroup$ @daw: I might be missing something obvious: but why is this counter example wrong? take $n=2$, and let $f(E_{1,1})=1$, $f(E_{2,2})=0=f(E_{1,2})=f(E_{2,1})$. Then $f$ is not a scalar multiple of the trace. $\endgroup$ – voldemort Oct 14 '14 at 13:53
  • $\begingroup$ Of course I meant extend $f$ linearly in my example above.. $\endgroup$ – voldemort Oct 14 '14 at 13:54
  • $\begingroup$ @voldemort see the matrix $C$ in my example below: $C^2=0$, but $f(C)\ne0$ for your $f$. $\endgroup$ – daw Oct 14 '14 at 14:05

Denote by $E_{ij}$ the matrices with all entries zero with the exception of the entry $(i,j)$ that is $1$.

Then $E_{ij}^2 = 0 $ for all $i\ne j$.

The matrices $\{E_{ij}, i \ne j\}$ are linear independent. The set $$ \{ E_{ij}, i\ne j\} \cup \{ E_{ii}-E_{jj}, i< j\} $$ is a basis of the space of all matrices with zero trace. However, $(E_{ii}+E_{jj})^2 \ne 0$.

Notice that $$ C:=\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix} $$ satisfies $C^2=0$. Moreover, $tr(C)=0$.

Set $F_{ij}:= E_{ii} + E_{ij} - E_{ji} - E_{jj}$, $i<j$. Then with the matrix $C$ in mind, we find $F_{ij}^2 =0$. Moreover, $tr(F_{ij})=0$. And the set $$ \{ E_{ij}, i\ne j\} \cup \{ F_{ij}, i< j\} $$ is a basis of the set of matrices with zero trace. Hence it follows $f(A)=0$ for all matrices with trace zero.

Since the set of all matrices with zero trace has dimension $n^2-1$, which is one less than the dimension of the space $\mathcal M_n$, it follows $$ f(A) = c \ tr(A). $$ To see this, extend the basis above by $n^{-1}I_n$ to a basis of $\mathcal M_n$. Then $f$ is completely determined by $f(I_n)$, as $f(B)=0$ for $B$ with $tr(B)=0$.

Since any matrix $A$ can be written as $A= tr(A)n^{-1}I_n + B$ with $tr(B)=0$, it follows $$ f(A) = tr(A)\underbrace{n^{-1} f(I_n)}_{=:c}. $$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.