If \(V\) is the vector space of \(n \times n\) matrices over \(F\) and \(B\) is a fixed \(n \times n\) matrix over \(F\), let \(L_B\) and \(R_B\) be the linear operators on \(V\) defined by \(L_B(A) = BA\) and \(R_B(A) = AB\). Show that
\(\det L_B = (\det B)^n\);
\(\det R_B = (\det B)^n\).
Solution
1. Part (a): Determinant of left multiplication \(L_B\):
1.1. Decompose the vector space \(V = M_n(F)\) as a direct sum of its \(n\) column subspaces:\[ V = V_1 \oplus V_2 \oplus \cdots \oplus V_n, \]where \(V_j = \{A \in M_n(F) \mid \text{column } k \text{ of } A \text{ is } 0 \text{ for all } k \neq j\} \cong F^n\).
every matrix is uniquely the sum of its single-column components.
1.2. For any \(A = [A_1 \mid A_2 \mid \cdots \mid A_n] \in V\):\[ L_B(A) = BA = [B A_1 \mid B A_2 \mid \cdots \mid B A_n]. \]Thus each subspace \(V_j\) is an invariant subspace under \(L_B\), and the restriction \(L_B|_{V_j}: V_j \to V_j\) has matrix representation \(B\) with respect to the standard basis of \(V_j \cong F^n\).
definition of matrix multiplication on column vectors.
1.3. With respect to the ordered basis \(\mathcal{B} = (E_{11}, E_{21}, \dots, E_{n1}, E_{12}, \dots, E_{nn})\), the matrix of \(L_B\) is block diagonal:\[ [L_B]_\mathcal{B} = \begin{pmatrix} B & 0 & \cdots & 0 \\ 0 & B & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & B \end{pmatrix} = I_n \otimes B. \]
invariant direct sum decomposition into \(n\) identical blocks.
1.4. The determinant of a block diagonal matrix is the product of the determinants of the diagonal blocks:\[ \det L_B = \prod_{j=1}^n \det B = (\det B)^n. \]
determinant of block diagonal matrix.
2. Part (b): Determinant of right multiplication \(R_B\):
2.1. Let \(\tau: V \to V\) be the matrix transposition operator \(\tau(A) = A^T\).\(\tau\) is an invertible linear operator on \(V\) with \(\tau^2 = \operatorname{id}_V\).
\((A^T)^T = A\) and linearity of transpose.
2.2. For any \(A \in V\):\[ (\tau \circ R_B \circ \tau)(A) = \tau\big(R_B(A^T)\big) = \tau(A^T B) = (A^T B)^T = B^T A = L_{B^T}(A). \]Thus \(R_B = \tau^{-1} \circ L_{B^T} \circ \tau\).
anti-automorphism property of transpose \((XY)^T = Y^T X^T\).
2.3. Taking the determinant of both sides:\[ \det R_B = \det(\tau^{-1} \circ L_{B^T} \circ \tau) = \det(L_{B^T}) = (\det B^T)^n = (\det B)^n. \]
multiplicative property of determinants, Part (a) applied to \(B^T\), and \(\det B^T = \det B\).
3. Conclusion:\(\det L_B = (\det B)^n\) and \(\det R_B = (\det B)^n\). Q.E.D.
1 and 2.