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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.

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  • $\begingroup$ 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
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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 $$ f(A) = tr(A)\underbrace{n^{-1} f(I_n)}_{=:c}. $$

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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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