# 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.

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It disturbs me a little to call a real matrix Hermitian instead of simply symmetric...is there some meaning I'm missing here? – DonAntonio Jan 18 '13 at 3:55
It follows from the fact that $T_B$ can be represented by the Kronecker product $B^*\otimes B$. Unfortunately if Kronecker products have not been treated by this point in the book, this will not be very helpful :-( – Chris Godsil Jan 18 '13 at 16:06
@DonAntonio: Not really, the matrix still has complex entries, but the vector space is defined over real number set, meaning that the scalar multiplication of this vector space must be real number multiplication. – le duc quang Jan 21 '13 at 1:37
@leducquang, I'm baffled by your comment: as far as I know, when we talk of "matrices over the field of real numbers", it means all the matrix's entries are real, of course. Why would anyone want to restrict himself to the reals field if he's working with complex matrices? Unless you meant that the definition field of the complex vector space (or algebra) of complex matrices is the reals...? – DonAntonio Jan 21 '13 at 3:29
@DonAntonio: I copy exactly the same line in the book. Here, the phrase "over the field of real number" is not for matrix. It is for vector space. – le duc quang Jan 21 '13 at 9:39

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.

1. $F_{ij}<E_{II}$ if $i+j\le 2I$.
2. $F_{ij}<F_{IJ}$ if (a) $i+j<I+J$ or (b) $i+j=I+J$ and $i<I$.
3. $F_{ij}<G_{ij}$.
4. $G_{ij}<F_{IJ}$ if $F_{ij}<F_{IJ}$.
5. $G_{ij}<E_{II}$ 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.

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