According to Exercise 3, similar matrices have the same trace. Thus we can define the trace of a linear operator on a finite-dimensional space to be the trace of any matrix which represents the operator in an ordered basis. This is well-defined since all such representing matrices for one operator are similar.
Now let \(V\) be the space of all \(2 \times 2\) matrices over the field \(F\) and let \(P\) be a fixed \(2 \times 2\) matrix. Let \(T\) be the linear operator on \(V\) defined by \(T(A) = PA\). Prove that \(\operatorname{trace}(T) = 2\operatorname{trace}(P)\).
Solution
Write \[ P=\begin{bmatrix}p_{11}&p_{12}\\p_{21}&p_{22}\end{bmatrix}. \] Use the ordered basis \[ \mathcal B=(E_{11},E_{21},E_{12},E_{22}) \] of \(V\), grouped by columns. Left multiplication by \(P\) acts on each column independently by the same matrix \(P\). Therefore \[ [T]_{\mathcal B} = \begin{bmatrix} P&0\\ 0&P \end{bmatrix}. \] Hence \[ \operatorname{trace}(T) =\operatorname{trace}([T]_{\mathcal B}) =2\operatorname{trace}(P). \]