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The real Jacobian determinant of a complex-linear automorphism is the squared modulus of its complex determinant

Statement

Let $m \geq 1$ , let $L : C^{m} \to C^{m}$ be $C$ -linear, and let $A$ be its matrix in the standard complex basis. Regard $L$ as an $R$ -linear map on $R^{2 m}$ through the usual identification $C^{m} ≅ R^{2 m}$ . Then

$\det_{R} L = {∣ \det_{C} A ∣}^{2} .$

In particular $\det_{R} L \neq 0$ if and only if $\det_{C} A \neq 0$ .

Facts & Assumptions

Given: The complex-linear map $L$ and its matrix $A = B + i C$ with real matrices $B$ and $C$ .

[L1]

Determinants multiply under matrix products, and the determinant of a triangular block matrix is the product of its diagonal-block determinants (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , The determinant of a triangular matrix is the product of its diagonal entries).

[A1]

In the real basis $(e_{1}, \dots, e_{m}, i e_{1}, \dots, i e_{m})$ , the real matrix of $L$ is

$R = (\begin{array}{cc} B & - C \\ C & B \end{array}) .$

The complex matrices

$P = \frac{1}{2} (\begin{array}{cc} I & I \\ - i I & i I \end{array}), Q = (\begin{array}{cc} I & i I \\ I & - i I \end{array})$

are inverse to one another.

Proof

technique · direct

1.1A1algebra

Using [A1], direct block multiplication gives $Q R P = (\begin{array}{cc} A & 0 \\ 0 & \overline{A} \end{array}) .$ Indeed the two columns of $P$ are the $(z, \overline{z})$ coordinates of a real vector, and the $C$ -linearity of $L$ makes the transformed action split into $A$ on the $z$ block and $\overline{A}$ on the $\overline{z}$ block.

2.1step 1.1L1algebra

By [L1], step 1.1, and the identity $Q P = I$ , one has $\det_{C} R = \det_{C} (Q R P) = \det_{C} A \cdot \det_{C} \overline{A} = \det_{C} A \cdot \overline{\det_{C} A} = {∣ \det_{C} A ∣}^{2} .$ Since $R$ has real entries, the determinant polynomial gives the same real number whether computed over $R$ or over $C$ , so $\det_{R} L = \det_{C} R$ .

3.1step 2.1∎

The displayed formula immediately makes $\det_{R} L$ nonzero exactly when $\det_{C} A$ is nonzero.

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jiří Lebl, Tasty Bits of Several Complex Variables, Section 5.2 (standard reference, not scraped)
Jaap Korevaar and Jan Wiegerinck, Several Complex Variables, Section 5.2 (standard reference, not scraped)