Let \(M_{2\times2}(\mathbb R)\) be the space of real \(2\times2\) matrices and \[ A=\begin{pmatrix}1&2\\-1&3\end{pmatrix},\qquad B=\begin{pmatrix}2&1\\0&4\end{pmatrix}. \] Define \(L(X)=AXB\). Compute the trace and determinant of \(L\).
Solution
Write \[ L=L_A\circ R_B, \] where \[ L_A(X)=AX, \qquad R_B(X)=XB. \]
1. Compute the trace of \(L\).
Let \(E_{ij}\) be the standard basis of \(M_{2\times2}(\mathbb R)\). For a general matrix \(X=(x_{ij})\), \[ (AXB)_{ij}=\sum_{k,\ell}a_{ik}x_{k\ell}b_{\ell j}. \] The coefficient of \(x_{ij}\) in the same \((i,j)\) output coordinate is \[ a_{ii}b_{jj}. \] Therefore the trace of the linear transformation is \[ \operatorname{tr}L =\sum_{i,j}a_{ii}b_{jj} =(\operatorname{tr}A)(\operatorname{tr}B). \] Here \[ \operatorname{tr}A=1+3=4, \qquad \operatorname{tr}B=2+4=6, \] so \[ \boxed{\operatorname{tr}L=24.} \]
2. Compute the determinant of left multiplication by \(A\).
Left multiplication acts independently on the two columns of \(X\), applying \(A\) to each column. Hence, in a basis grouped by columns, the matrix of \(L_A\) is block diagonal with two copies of \(A\). Thus \[ \det L_A=(\det A)^2. \] Since \[ \det A=1\cdot3-2(-1)=5, \] we get \[ \det L_A=25. \]
3. Compute the determinant of right multiplication by \(B\) and conclude.
Right multiplication acts independently on the two rows of \(X\). On each row it has determinant \(\det B\), so \[ \det R_B=(\det B)^2. \] Here \[ \det B=2\cdot4=8, \] and therefore \[ \det R_B=64. \]
Since \(L=L_A\circ R_B\), \[ \det L =(\det L_A)(\det R_B) =25\cdot64 =1600. \] Hence \[ \boxed{\det L=1600.} \]