Hadamard's Inequality

| (A)| ≤ _i |r_i|_2 — equality if and only if rows are orthogonal.

The idea

Theorem (Hadamard's inequality).

Let $A$ be an $n \times n$ real matrix with rows $r_1, \ldots, r_n$. Then $|\det(A)| \leq \prod_{i=1}^{n} \|r_i\|_2,$ with equality if and only if the rows are mutually orthogonal, or one of them is zero.

The inequality compares two solids. $|\det(A)|$ is the volume of the solid whose edges are the rows of $A$, and $\|r_i\|_2$ is the length of the $i$-th edge, so the right-hand side is the volume of a rectangular box built from edges of the same lengths. Tilting the edges away from perpendicular while holding their lengths fixed flattens the solid and reduces its volume, so the rectangular box has the largest volume of any solid with those edge lengths. The two volumes agree only when no edge was tilted, which is the equality case.

The bound is useful because its right-hand side costs little: a row length takes a handful of multiplications, while a determinant takes far more. Hadamard's inequality therefore bounds a determinant from above without computing one, and the bound is nearly exact when the rows are nearly orthogonal.

Ways to work on it

• Walkthrough. Verify Hadamard's inequality on a small matrix and identify the equality condition.
• Practice. Verify Hadamard on a random 2 × 2 matrix — sometimes tight, sometimes loose.
• Hardest. Positive definite form: (A) ≤ a_ii on a 3 × 3 matrix.

Not sure where to start? Take the ten-question placement test.