Let \(V\) be the vector space of all \(n \times n\) matrices over the field of complex numbers, and let \(B\) be a fixed \(n \times n\) matrix over \(C\). Define a linear operator \(M_B\) on \(V\) by \(M_B(A) = BAB^*\), where \(B^* = \overline{B^t}\). Show that \[ \det M_B = |\det B|^{2n}. \] Now let \(H\) be the set of all Hermitian matrices in \(V\), \(A\) being Hermitian if \(A = A^*\). Then \(H\) is a vector space over the field of real numbers. Show that the function \(T_B\) defined by \(T_B(A) = BAB^*\) is a linear operator on the real vector space \(H\), and then show that \(\det T_B = |\det B|^{2n}\). (Hint: In computing \(\det T_B\), show that \(V\) has a basis consisting of Hermitian matrices and then show that \(\det T_B = \det M_B\).)
Solution
Goal. Show \(\det M_B = |\det B|^{2n}\) on \(V\), and \(\det T_B = |\det B|^{2n}\) on the real space \(H\) of Hermitian matrices.
1. \(\det M_B = |\det B|^{2n}\).
1.1. \(M_B = L_B \circ R_{B^*}\), where \(L_B(A) = BA\) and \(R_{B^*}(A) = AB^*\).
\(M_B(A) = BAB^* = L_B(R_{B^*}(A))\).
1.2. \(\det L_B = (\det B)^n\) and \(\det R_{B^*} = (\det B^*)^n = \overline{\det B}^n\).
\(L_B\) acts as \(B\) on each of the \(n\) columns, so \(\det L_B = (\det B)^n\); similarly \(R_{B^*}\) acts as \(B^*\) on each row, so \(\det R_{B^*} = (\det B^*)^n\).
1.3. Hence \(\det M_B = \det L_B \cdot \det R_{B^*} = (\det B)^n \overline{\det B}^n = |\det B|^{2n}\).
determinant is multiplicative.
2. \(H\) is a real vector space, and \(T_B\) is a linear operator on it.
2.1. \(H = \theset{A : A = A^*}\) is closed under real linear combinations.
if \(A, C\) are Hermitian and \(r \in \RR\), then \((A + C)^* = A + C\) and \((rA)^* = rA\).
2.2. \(T_B(A) = BAB^*\) is Hermitian when \(A\) is.
\((BAB^*)^* = B A^* B^* = B A B^*\) (since \(A^* = A\)).
2.3. Hence \(T_B\) is a linear operator on the real vector space \(H\).
2.1 and 2.2.
3. \(\det T_B = |\det B|^{2n}\).
3.1. \(V\) has a basis consisting of Hermitian matrices.
the matrices \(E_{jj}\) (diagonal) and \(E_{jk} + E_{kj}\), \(i(E_{jk} - E_{kj})\) (off-diagonal) form a basis of \(V\) over \(\CC\), and each is Hermitian; this gives a real basis of \(H\) and a complex basis of \(V\).
3.2. \(M_B\) and \(T_B\) have the same matrix (in the appropriate bases), so \(\det T_B = \det M_B\).
\(M_B\) and \(T_B\) are the same map \(A \mapsto BAB^*\); on the Hermitian basis, the real matrix of \(T_B\) and the complex matrix of \(M_B\) agree, so their determinants agree.
3.3. Hence \(\det T_B = \det M_B = |\det B|^{2n}\).
3.2 and 1.3.
4. Q.E.D.
1.3 and 3.3 give the two results.