Linear-Algebra-Done-Right · Unit 29 · Video 6 · Interactive Practice

Hadamard's Inequality: Column Lengths Cap |det A|

IKey Formulas

Formula	Name	Hypotheses
$\|\det A\| \le \\|v_1\\| \cdots \\|v_n\\|$	Hadamard's inequality	$A \in \mathbf{F}^{n,n}$ with columns $v_1, \dots, v_n$ ; Euclidean norm on $\mathbf{F}^n$
$A = QR$ , $\ \|\det A\| = R_{1,1} \cdots R_{n,n}$	QR factorization	$A$ invertible; $Q$ unitary, $R$ upper triangular with positive diagonal
$R_{k,k} \le \\|R_{\cdot,k}\\| = \\|v_k\\|$	The one inequality	$Q$ preserves norms; $\\|R_{\cdot,k}\\|^2 = \sum_j \|R_{j,k}\|^2$

Key Insight: $|\det A|$ is the volume of the parallelepiped spanned by the columns; for given edge lengths, the box with orthogonal edges has the largest volume.

IISwinging an Edge

With edge lengths fixed, the area $L_1 L_2 \sin\theta$ of a parallelogram is largest when the edges are perpendicular.

L₁ 4 L₂ 3 θ 40°

IIIWhere the Gap Comes From

Gram–Schmidt splits $v_2$ into a part along $v_1$ ( $R_{1,2}$ ) and a perpendicular part ( $R_{2,2}$ ); only $R_{2,2}$ counts toward the area.

IVQuiz Questions

Problem 1 · Compute the Bound

Given: $A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 3 & 0 \\ 2 & 4 & 0 \end{pmatrix}$ . Find the upper bound Hadamard's inequality gives for $|\det A|$ .

✅ Correct! The columns have lengths

3

5

and

2

, so

|\det A| \le 30

❌ Not quite.

9 + 25 + 4

adds squared lengths; the bound multiplies the lengths themselves.

❌ Not quite.

3 + 5 + 2

adds the lengths; the bound is their product.

❌ Not quite.

4

|\det A|

itself; the question asks for the bound.

Show solution

\|(1, 2, 2)\| = 3, \quad \|(0, 3, 4)\| = 5, \quad \|(2, 0, 0)\| = 2, \qquad |\det A| \le 3 \cdot 5 \cdot 2 = 30.

The actual value is $\det A = 4$ (for instance by row operations), well below the bound: the columns are far from orthogonal.

Problem 2 · Equality Case

Given: $A = \begin{pmatrix} 2 & -3 \\ 3 & 2 \end{pmatrix}$ with columns $v_1, v_2$ . Compare $|\det A|$ with $\|v_1\|\,\|v_2\|$ .

✅ Correct!

\langle v_1, v_2 \rangle = -6 + 6 = 0

: orthogonal columns give equality.

❌ Check the sign.

ad - bc = 2 \cdot 2 - (-3)(3) = 4 + 9

❌ Not quite. Multiply the lengths:

\sqrt{13} \cdot \sqrt{13} = 13

, not

13 + 13

❌ Not quite. Hadamard's bound is the product of both lengths,

\sqrt{13}\cdot\sqrt{13}

; it can never be exceeded.

Show solution

\det A = 2 \cdot 2 - (-3) \cdot 3 = 13, \qquad \|v_1\| = \|v_2\| = \sqrt{4 + 9} = \sqrt{13}, \qquad \|v_1\|\,\|v_2\| = 13.

Equality holds; indeed $\langle v_1, v_2 \rangle = 2(-3) + 3 \cdot 2 = 0$ , so the parallelogram is a square of side $\sqrt{13}$ .

Problem 3 · An Impossible Determinant

Given: $A \in \mathbf{R}^{3,3}$ whose columns have lengths $1$ , $2$ and $3$ . Which value can $\det A$ not take?

✅ Correct!

|\det A| \le 1 \cdot 2 \cdot 3 = 6 < 7

❌ Not quite.

6

is attained:

A = \operatorname{diag}(1, 2, 3)

❌ Not quite.

-6

is attained:

A = \operatorname{diag}(-1, 2, 3)

❌ Not quite.

0

is attained: take two parallel columns, e.g.

e_1

2e_1

3e_2

Show solution

Hadamard's inequality gives $|\det A| \le 1 \cdot 2 \cdot 3 = 6$ , so $7$ is impossible. The values $6$ , $-6$ and $0$ occur for $\operatorname{diag}(1, 2, 3)$ , $\operatorname{diag}(-1, 2, 3)$ and the matrix with columns $e_1, 2e_1, 3e_2$ .

Problem 4 · Recovering the Angle

Given: $A \in \mathbf{R}^{2,2}$ with columns of lengths $5$ and $2$ , and $|\det A| = 5$ . Find the angle $\theta \in [0\degree, 180\degree]$ between the columns.

✅ Correct!

5 = 5 \cdot 2 \sin\theta

gives

\sin\theta = \tfrac{1}{2}

❌ Not quite. Area uses

\sin\theta

, not

\cos\theta

❌ Not quite. At

90\degree

the area would be the full

5 \cdot 2 = 10

❌ Almost.

\sin 150\degree = \tfrac{1}{2}

as well; the absolute value cannot tell the two apart.

Show solution

$|\det A|$ is the area of the parallelogram spanned by the columns, $L_1 L_2 \sin\theta$ :

5 = 5 \cdot 2 \cdot \sin\theta \implies \sin\theta = \tfrac{1}{2} \implies \theta = 30\degree \text{ or } 150\degree.

Solved: 0 / 4