Hadamard’s Determinant Inequality

Abstract

This note is devoted to a short, but elementary, proof of Hadamard’s determinant inequality.

Let X be an n × n matrix with complex entries and columns x₁, …, x_n. Hadamard’s determinant inequality [2] reads

$$\mid det\mathit{X}\mid \le \prod _{i=1}^n{\left|\right|{\mathit{x}}_i\left|\right|}_2.$$

Horn and Johnson [1] supply at least a half dozen proofs for this classical result. Hadamard’s inequality is obviously true if any x_i = 0. If we assume otherwise and divide both sides by the right-hand side, then Hadamard’s inequality reduces to the inequality |detX| ≤ 1 subject to the Euclidean length constraints ||x_i||₂ = 1. Equality is attained by taking X to be the identity matrix I or any unitary matrix. Recall that a unitary matrix X has orthonormal columns and consequently satisfies

$${\mid det\mathit{X}\mid }^2=det{\mathit{X}}^{\ast }det\mathit{X}=det({\mathit{X}}^{\ast }\mathit{X})=det\mathit{I}=1.$$

In view of Weierstrass’ theorem, the necessarily positive maximum of |detX| is attained on the compact set defined by the constraints. Suppose X has columns of unit length and yields the maximum value, but two columns of X, say the first two, are not orthogonal. Let us demonstrate that |detX| can be increased by replacing the first column by the linear combination y₁ = ax₁+bx₂ for carefully chosen complex scalars a and b. The determinant of the new matrix Y satisfies

$$det\mathit{Y}=det(a{\mathit{x}}_1+b{\mathit{x}}_2,{\mathit{x}}_2,{\mathit{x}}_3,\dots ,{\mathit{x}}_n)=adet\mathit{X}.$$

We force y₁ to be a unit vector by imposing the constraint

$${\mid a\mid }^2+{\mid b\mid }^2+2Re(\overline{a}bs)=1,s={\mathit{x}}_1^{\ast }{\mathit{x}}_2.$$

The Cauchy-Schwarz inequality implies |s| ≤ 1. Furthermore, s ≠ 0 by assumption, and |s| ≠ 1, for otherwise x₁ and x₂ are collinear and detX vanishes. The reader can check that we may choose

$$\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}\frac{1}{\sqrt{1-{\mid s\mid }^2}}\\ -\frac{\overline{s}}{\sqrt{1-{\mid s\mid }^2}}\end{array}\right).$$

Since a > 1, the identity detY = a detX shows that |detY| > |detX|. This contradiction proves that X has orthonormal columns, so it is unitary and |detX| = 1.