Hadamard's inequality

Remarks

If the columns are linearly dependent, the left side is $0$. Otherwise the Gram–Schmidt process gives $A = QR$ with $Q$ unitary and $R$ upper triangular with diagonal entries $r_{ii}$. Then $\abs{\det A} = \prod_i \abs{r_{ii}}$, and $\abs{r_{ii}}$ is the distance from $v_i$ to the span of $v_1, \dots, v_{i-1}$, so $\abs{r_{ii}} \le \norm{v_i}$.

Example: for $A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, the columns are orthogonal with norm $\sqrt{2}$, and indeed $\abs{\det A} = 2 = \sqrt{2} \cdot \sqrt{2}$. For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, the bound is $1 \cdot \sqrt{2}$, while $\det A = 1$.

An equivalent form: if $P$ is an $n \times n$ positive semidefinite matrix, then $\det P \le \prod_i p_{ii}$. Apply it to $P = A^* A$, whose diagonal entries are $\norm{v_i}^2$. If all entries of $A$ satisfy $\abs{a_{ij}} \le 1$, the inequality gives $\abs{\det A} \le n^{n/2}$, which is the starting point of Hadamard's maximal determinant problem.

Going deeper

For a student

The inequality bounds the size of a determinant without computing it. It is used to bound the size of integer determinants in algorithms, and it gives the bound $n^{n/2}$ for matrices with entries of absolute value at most $1$. The positive semidefinite form is a special case of Fischer's inequality for block matrices.