 Solution

If $A$ is a Hermitian matrix, then $(BA{B}^{\ast }{)}^{\ast }=BA{B}^{\ast }$, so $M_B$ preserves $W$. Addition and multiplication by real scalars commute with this operation, making $T_B$ a real linear map.

A real basis of $W$ consists of $E_{ii}$ and, for $i<j$, $E_{ij}+E_{ji}$ and $i(E_{ij}-E_{ji})$. It has $n^2$ elements and is also a complex basis of all matrices. Indeed every matrix is uniquely $H+iK$ with $H,K$ Hermitian matrices, by takingReal linear independence of the displayed basis therefore implies complex linear independence. Since its images under $M_B$ are Hermitian matrices, the matrix of $M_B$ in this complex basis has exactly the same real entries as the matrix of $T_B$ in the real basis. Their determinants are equal. There is no further squaring when passing to this real subspace:This is the determinant of Hermitian congruence, including the case of singular $B$.