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Given $X\in \operatorname{GL}(n, \mathbb{C}) $, let $X_r\in \operatorname{GL}(2n, \mathbb{R}) $ be the real matrix obtained by substituting to each complex entry $a+ib$ the matrix $\begin{pmatrix} a & & - b \\ b & & a \end{pmatrix}$. Does someone can give me the details of the proof that $\det(X_r) =|\det(X)|^2$? The only thing I have tried is to use the fact that the determinant is a wedge product but it seems not work.

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3 Answers 3

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One may view this as a determinant of a block matrix. In general (cf. John Silvester, Determinants of Block Matrices), if $R$ is a commutative ring and $C$ is a commutative subring of $M_m(R)$, each matrix $Z\in M_n(C)$ can be viewed as a matrix $Y\in M_{mn}(R)$ and $$ \det{}_RY=\det{}_R\left(\det{}_CZ\right).\tag{1} $$ In your case, let $m=2,\, R=\mathbb R$ and $C=\left\{\pmatrix{a&-b\\ b&a}:a,b\in\mathbb R\right\}$. Since $f:a+ib\mapsto\pmatrix{a&-b\\ b&a}$ is a ring isomorphism between $\mathbb C$ and $C$, when it is applied entrywise to a matrix $X\in M_n(\mathbb C)$, we have $$ \det{}_Cf(X)=f(\det{}_{\mathbb C}X).\tag{2} $$ Now let $Z=f(X)\in M_n(C)$ and $Y=X_r\in M_{2n}(\mathbb R)$. By $(1)$ and $(2)$, $$ \det{}_{\mathbb R}Y =\det{}_{\mathbb R}\left(\det{}_CZ\right) =\det{}_{\mathbb R}\left(\det{}_Cf(X)\right) =\det{}_{\mathbb R}\left(f(\det{}_{\mathbb C}X)\right) =|\det{}_{\mathbb C}X|^2. $$

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  • $\begingroup$ Don't have clear why $det_C(f(X)) =f(det_{\mathbb{C}} (X)) $. Please can you add details? $\endgroup$
    – yo yo
    Jan 14, 2021 at 10:19
  • $\begingroup$ @yoyo $f$ is a ring homomorphism and the determinant of a matrix is a polynomial in the matrix entries. Hence $$f(\det_{\mathbb C}X) =f\left(\sum_\sigma\operatorname{sgn}(\sigma)\prod_ix_{i\sigma(i)}\right) =\sum_\sigma\operatorname{sgn}(\sigma)\prod_if(x_{i\sigma(i)}) =\det_Cf(X).$$ $\endgroup$
    – user1551
    Jan 14, 2021 at 11:37
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Since $|a+ib|^2=\det\begin{pmatrix} a & & - b \\ b & & a \end{pmatrix}$,$$\begin{align}\det X\cdot\overline{\det X}&=\sum_{\sigma,\,\sigma^\prime\in S_n}\varepsilon_\sigma\varepsilon_{\sigma^\prime}\prod_{j=1}^n\prod_{k=1}^nX_{j\sigma(j)}\overline{X_{k\sigma^\prime(k)}}\\&=\sum_{\sigma,\,\sigma^\prime}\varepsilon_{\sigma\circ\sigma^\prime}\prod_j(XX^\dagger)_{j(\sigma\circ\sigma^\prime)}\\&=\sum_{\sigma\in S_{2n}}\varepsilon_\sigma\prod_{j=1}^{2n}(X_r)_{j\sigma(j)}\\&=\det X_r.\end{align}$$

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  • $\begingroup$ Impressive knowledges...+1 $\endgroup$
    – Sebastiano
    Jan 13, 2021 at 21:00
  • $\begingroup$ Unfortunately i can't understand the second and third equalities. Can you add details please? $\endgroup$
    – yo yo
    Jan 14, 2021 at 10:05
  • $\begingroup$ @yoyo Do you count the line starting with $\det X\cdot\overline{\det X}$ as the second equation due to including the one starting with $|a+ib|^2$, or as the first due to excluding it? $\endgroup$
    – J.G.
    Jan 14, 2021 at 10:10
  • $\begingroup$ To prove that $det(X)\overline{det(X)}=det(X_r)$ you used 4 equalities; i refer to the second and third. $\endgroup$
    – yo yo
    Jan 14, 2021 at 10:27
  • $\begingroup$ @yoyo The second uses the fact that permutation compositions sum parities. The third expresses such compositions as elements of $S_{2n}$. $\endgroup$
    – J.G.
    Jan 14, 2021 at 12:13
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Another option is to think about the matrix $X_r$ as a $2n \times 2n$ complex matrix. Motivated by the $2 \times 2$ case, over the complex numbers, we have the identity

$$ \begin{pmatrix} a - ib & 0 \\ 0 & a + ib \end{pmatrix} = P^{-1} \begin{pmatrix} a & -b \\ b & a \end{pmatrix} P$$

where

$$ P = \frac{1}{\sqrt{2}} \begin{pmatrix} -iI_n & iI_n \\ I_n & I_n \end{pmatrix}, \,\,\, P^{-1} = \frac{1}{\sqrt{2}} \begin{pmatrix} iI_n & I_n \\ -iI_n & I_n \end{pmatrix}. $$

Since similar matrices have the same determinant, we have

$$ \det(X_r) = \det \begin{pmatrix} a - ib & 0 \\ 0 & a + ib \end{pmatrix} = \det(a - ib) \det(a + ib) = \\ \det(X) \cdot \det(\overline{X}) = \det(X) \cdot \overline{\det(X)} = \left| \det(X) \right|^2. $$

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  • $\begingroup$ So if i understand you take $X$ as comples matrix, write as $A+iB$ real matrices and than view it as $\begin{pmatrix} A && -B \\ B && A \end{pmatrix}$ use $P$ to get the form \begin{pmatrix} A-iB && 0 \\ 0 && A+iB \end{pmatrix} and than conclude.. Right? $\endgroup$
    – yo yo
    Jan 14, 2021 at 11:03
  • $\begingroup$ This is ok, but i am not sure it is what i asked. Becouse if you consider $\mathbb{C}$ you have that the condition of $\mathbb{C}$ linearity is equivalent to be of the form \begin{pmatrix} a && -b \\ b && a \end{pmatrix} but when the dimension is higher requairing $\mathbb{C}$ linearity it is not equivalent to take the complex blocks matrix.. I think $\endgroup$
    – yo yo
    Jan 14, 2021 at 11:07
  • $\begingroup$ @yoyo: It actually has nothing to do with $\mathbb{C}$-linearity. You can check that if you multiply $X_r$ by $P$ and $P^{-1}$ given by what I wrote using the rules of block multiplication, you get the block diagonal matrix with two $n \times n$ blocks given by $A \pm iB$. Since you know the determinant of a block diagonal matrix and similar matrices have the same determinant, you get what you want. $\endgroup$
    – levap
    Jan 14, 2021 at 13:27

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