# Determinant of associated real matrix from a complex one

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.

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 $$(2)$$, $$\det{}_{\mathbb R}Y =\det{}_{\mathbb R}\left(\det{}_CZ\right) =\det{}_{\mathbb R}\left(\det{}_Cf(X)\right) =\det{}_{\mathbb R}\left(f(\det{}_{\mathbb C}X)\right) =|\det{}_{\mathbb C}X|^2.$$

• Don't have clear why $det_C(f(X)) =f(det_{\mathbb{C}} (X))$. Please can you add details? Jan 14, 2021 at 10:19
• @yoyo $f$ is a ring homomorphism and the determinant of a matrix is a polynomial in the matrix entries. Hence $$f(\det_{\mathbb C}X) =f\left(\sum_\sigma\operatorname{sgn}(\sigma)\prod_ix_{i\sigma(i)}\right) =\sum_\sigma\operatorname{sgn}(\sigma)\prod_if(x_{i\sigma(i)}) =\det_Cf(X).$$ Jan 14, 2021 at 11:37

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}

• Impressive knowledges...+1 Jan 13, 2021 at 21:00
• Unfortunately i can't understand the second and third equalities. Can you add details please? Jan 14, 2021 at 10:05
• @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?
– J.G.
Jan 14, 2021 at 10:10
• To prove that $det(X)\overline{det(X)}=det(X_r)$ you used 4 equalities; i refer to the second and third. Jan 14, 2021 at 10:27
• @yoyo The second uses the fact that permutation compositions sum parities. The third expresses such compositions as elements of $S_{2n}$.
– J.G.
Jan 14, 2021 at 12:13

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

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

• 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? Jan 14, 2021 at 11:03
• 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 Jan 14, 2021 at 11:07
• @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. Jan 14, 2021 at 13:27