If $T : F^{2 \times 2} \to F^{2\times 2}$ is $T(A) = PA$ for some fixed $2 \times 2$ matrix $P$, why is $\operatorname{tr} T = 2\operatorname{tr} P$? I am asked to prove that if $T$ is a linear operator on the space of $2 \times 2$ matrices over a field $F$ such that $T(A) = PA$ for some fixed $2 \times 2$ matrix $P$, then $\DeclareMathOperator{\tr}{tr} \tr T = 2 \tr P$.
At first I thought about considering the isomorphism between $F^{2 \times 2}$ and $F^4$ so that the matrix of $T$ as it would act on $F^4$ would be a $4 \times 4$ block matrix $(P,0;0,P)$ with trace $2\tr P$, but the trace is not an invariant under isomorphism, only under change of basis. Furthermore, the definition of the trace of a matrix is for a linear operator acting on an arbitrary vector space $V$ over a field $F$, not specifically on $F^n$ for some $n$, so taking an isomorphism shouldn't be necessary at all.
Taking a basis of $F^{2 \times 2}$ and using linearity of the trace as a linear functional gives 
$$
\begin{align}
\tr T &= P_{11} \tr \left[\begin{array}{cc}1&0\\0&0\end{array}\right] + P_{12} \tr \left[\begin{array}{cc}0&1\\0&0\end{array}\right] + P_{21} \tr \left[\begin{array}{cc}0&0\\1&0\end{array}\right] + P_{22} \tr \left[\begin{array}{cc}0&0\\0&1\end{array}\right] \\&= P_{11} + P_{22} \\&= \tr P,
\end{align}
$$
which is not the desired result. (Since the trace is invariant under change of basis, it should not matter which basis we choose for $F^{2 \times 2}$.)
What am I missing?
 A: If we represent $T$ as a $4 \times 4$ matrix with respect to your basis, we have
$$
[T]_B = 
\pmatrix{P_{11}&0&P_{12}&0\\0&P_{11}&0&P_{12}\\P_{21}&0&P_{22}&0\\0&P_{21}&0&P_{22}}
$$
A: Let $\{e_1, e_2, e_3, e_4\}$ be the standard ordered basis of $F^{2\times 2}$. Note that
$$T(e_1) = \left[\begin{array}\ P_{11} & P_{12}\\ P_{21} & P_{22}\end{array}\right]\left[\begin{array}\ 1 & 0\\ 0 & 0\end{array}\right] = \left[\begin{array}\ P_{11} & 0\\ P_{21} & 0\end{array}\right] = P_{11}e_1 + P_{21}e_3.$$
Therefore, under the isomorphism $F^{2\times 2} \cong F^4$, $[T(e_1)] = (P_{11}, 0, P_{21}, 0)^T$. If you compute $[T(e_2)]$, $[T(e_3)]$ and $[T(e_4)]$ in the same way, you will be able to construct the $4\times 4$ standard matrix of $T$ with respect to the given basis of $F^{2\times 2}$. Note, this $4\times 4$ matrix is not the block matrix you suggest in your question.
Also note that the trace is an invariant. In particular, choosing a basis in order to calculate it is perfectly fine (as the choice of basis does not change the value of the trace).
A: The linear map $L_P:F^n\to F^n$ defined by a $n\times n$ matrix $P$ is given by left multiplication by $P$ on column vectors: $L_P:v\mapsto P\cdot v$. Therefore $\def\tr{\operatorname{tr}}\tr L_P=\tr P$. Now left matrix multiplication on $n\times m~$matrices operates separately on each of the $m$ columns:
$$
   P\cdot(v_1~v_2~\cdots~v_m)=(P\cdot v_1~P\cdot v_2~\cdots~P\cdot v_m)
$$
Identifying in this way the space of $n\times m~$matrices with $(F^n)^m$, this linear map $T:(F^n)^m\to(F^n)^m$ is just $L_P$ acting on each component $(v_1,\ldots,v_m)\mapsto(L_P(v_1),\ldots,L_P(v_m))$. It follows by additivity of trace across direct sum decompositions that $\tr T=m(\tr L_P)=m(\tr P)$. Apply this for $n=m=2$.
