A Linear Operator with Trace Condition

In summary, a linear operator with trace condition refers to a specific type of linear operator that satisfies the trace condition, which requires the sum of diagonal elements in its matrix representation to be constant. The trace condition is significant as it simplifies calculations and aids in analyzing the properties and behavior of linear operators. It is also used in various applications in physics, engineering, and mathematics. While a trace class operator is a type of linear operator that satisfies the trace condition and has a finite trace, not all linear operators with a trace condition are trace class operators. In infinite-dimensional spaces, the trace condition is extended using the concept of a trace class operator, where the trace is defined as the sum of eigenvalues.
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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##.
 
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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.##
 
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