# A Linear Operator with Trace Condition

• POTW
• Euge

#### Euge

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Let ##V## be a finite dimensional vector space over a field ##F##. If ##L## is a linear operator on ##V## such that the trace of ##L\circ T## is zero for all linear operators ##T## on ##V##, show that ##L = 0##.

• topsquark

I hope I didn't overlook something. It is quite late here.

Set ##L=\sum_\mu \lambda_\mu \otimes X_\mu ## and ##T=\sum_\nu \tau_\nu \otimes X_\nu\,.## Then
\begin{align*}
(L\circ T)(v)&=\sum_\mu \lambda_\mu\left(\sum_\nu \tau_\nu (v) X_\nu \right)X_\mu =\sum_\mu \sum_\nu \tau_\nu(v) \lambda_\mu(X_\nu) X_\mu\\
&=\left(\sum_\mu \left(\sum_\nu \lambda_\mu(X_\nu)\tau_\nu\right)\otimes X_\mu\right)(v)
\end{align*}
Therefore
$$0=\operatorname{trace}\left(\sum_\mu \left(\sum_\nu \lambda_\mu(X_\nu)\tau_\nu\right)\otimes X_\mu \right)= \sum_{\mu,\nu} \lambda_\mu(X_\nu) \operatorname{trace}\left(\tau_\nu \otimes X_\mu\right)$$
Now we can choose ##T## as ##T(v)=\alpha v+\beta v_\nu## with scalasr ##\alpha,\beta## such that
$$\operatorname{trace}\left(\tau_\nu \otimes X_\mu\right)=1$$
consecutively, and making all ##\lambda_\mu(X_\nu)=0## and thus ##L\equiv 0.## ##\alpha## and ##\beta## are chosen in a way that norms the trace and avoids conflicts with the characteristic of ##F.##

• topsquark