A Linear Operator with Trace Condition

[Euge]

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 $L = 0$.

 

[fresh_42]

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

Spoiler

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 $F.$