prove that $\det T_{B} = |\det B|^{2n}$ I'm doing an exercise in Kunze Hoffman book and be stucked in this exercise about calculating the determinant of a linear operator. Can anyone help me?

Suppose $H$ is the vector space of all $n \times n$ Hermitan matrices over the field of real number. Prove that $T_B (A) = BAB^{*}$ is a lineat operator over the real vector space $H$ (I can prove this). Moreover, $\det T_B = |\det B|^{2n}$

Thanks every one.
 A: Remarks. If $T_B$ is defined over the space of all $n\times n$ complex matrices, and provided you know Kronecker product and vectorization, the proof is just a one-liner:
$$\det T_B=\det\left((B^\ast)^T\otimes B\right)=\det(\bar{B}\otimes B)=\det(\bar{B})^n\det(B)^n=\overline{\det(B)}^n\det(B)^n=|\det(B)|^{2n}.$$
However, the $T_B$ in this question is defined over the space of all $n\times n$ (complex) Hermitian matrices over the field $\mathbb{R}$. In this case, it is not obvious why $\det T_B$ is still equal to $|\det(B)|^{2n}$, as the determinant of a linear map is in general different from the determinant of the restriction of this linear map to an invariant subspace. For instance, the restriction of $f:\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}a&b\\0&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ to the span of $(1,0)^T$ is the map $g:(x,0)^T\mapsto(ax,0)^T$, so $\det f=ad\not=a=\det g$ in general.
Sketch of proof of the problem statement. Since $B$ is a complex square matrix, it can be unitarily triangularized. As unitary conjugation $A\mapsto U^\ast AU$ preserves Hermitian matrices and the determinant of a linear operator is invariant under change of basis, we may assume WLOG that $B$ is upper triangular. Now, let $E_{ij}$ denotes the $(i,j)$-th elementary matrix and let $\omega=\sqrt{-1}$. For any $i<j$, let $F_{ij}=E_{ij}+E_{ji}$ and $G_{ij}=\omega(E_{ij}-E_{ji})$. Then
$${\cal B}=\{E_{ii}:i=1,\ldots,n\}\cup\{F_{ij}:i<j\}\cup\{G_{ij}:i<j\}$$
is a basis of $H$.
We can order this basis as follows.


*

*$F_{ij}<E_{II}$ if $i+j\le 2I$.

*$F_{ij}<F_{IJ}$ if (a) $i+j<I+J$ or (b) $i+j=I+J$ and $i<I$.

*$F_{ij}<G_{ij}$.

*$G_{ij}<F_{IJ}$ if $F_{ij}<F_{IJ}$.

*$G_{ij}<E_{II}$ if $F_{ij}<E_{II}$.


For example, when $n=4$, the $16$ basis matrices of $H$ are ordered as
$$E_{11},F_{12},G_{12},F_{13},G_{13},E_{22},F_{14},G_{14},F_{23},G_{23},F_{24},G_{24},E_{33},F_{34},G_{34},E_{44}.$$
The matrix representation $[T_B]_{\cal B}$ of $T_B$ w.r.t. the basis ${\cal B}$ is a block lower triangular matrix, where the size of each diagonal block is either a 1-by-1 or 2-by-2 block. It can be shown that each diagonal entry $b_{ii}$ of $B$ corresponds uniquely to a 1-by-1 block of $[T_B]_{\cal B}$ containing the entry $|b_{ii}|^2$, and each pair of diagonal entries $(b_{ii},b_{jj})$ with $i<j$ corresponds uniquely to a 2-by-2 block
$$
\begin{pmatrix}
\operatorname{Re}(b_{ii}\overline{b_{jj}})&\operatorname{Im}(b_{ii}\overline{b_{jj}})\\
\operatorname{Re}(\omega b_{ii}\overline{b_{jj}})&\operatorname{Im}(\omega b_{ii}\overline{b_{jj}})
\end{pmatrix}
$$
of $[T_B]_{\cal B}$. Multiplying the determinants of these 1-by-1 blocks and 2-by-2 blocks together, the result follows.
