Hadamard's inequality for determinants

Idea. The determinant is the signed volume of the parallelepiped spanned by the columns. This volume is at most the product of the edge lengths, with equality for a rectangular box. The proof makes this precise by orthogonalising.

Step 1: trivial case. If $a_1,\dots,a_n$ are linearly dependent, then $\det A=0\le\prod|a_j|$.

Step 2: Gram-Schmidt. Otherwise define orthonormal vectors $q_1,\dots,q_n$ by Gram-Schmidt: $q_j$ is the normalised component of $a_j$ orthogonal to $\operatorname{span}(a_1,\dots,a_{j-1})$. Then $a_j=\sum_{i\le j}R_{ij}q_i$, i.e. $A=QR$ with $Q=(q_1|\dots|q_n)\in O(n)$ and $R$ upper triangular with
$$R_{jj}=\langle q_j,a_j\rangle,\qquad|R_{jj}|=\operatorname{dist}\big(a_j,\operatorname{span}(a_1,\dots,a_{j-1})\big)>0.$$

Step 3: bound the diagonal. The distance from $a_j$ to a subspace is at most the distance to $0$ (which belongs to the subspace), so $|R_{jj}|\le|a_j|$.

Step 4: determinant. $\det Q=\pm1$ and the determinant of the triangular matrix $R$ is $\prod R_{jj}$, so
$$|\det A|=\prod_{j=1}^n|R_{jj}|\le\prod_{j=1}^n|a_j|.$$

Equality case. If $\det A\ne0$, equality holds iff $|R_{jj}|=|a_j|$ for every $j$, i.e. iff $a_j\perp\operatorname{span}(a_1,\dots,a_{j-1})$ for every $j$: the columns are pairwise orthogonal. If $\det A=0$, equality holds iff some column is zero.

Example. For $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, $\det A=1\le\sqrt1\cdot\sqrt2=1.41$, strict since the columns are not orthogonal.

Remark. Applying the inequality to $A^T$ gives the same bound with the rows. It is the key tool in the next two problems (an upper bound on $|\det A|$ in terms of the Frobenius norm, and the Lipschitz estimate for $\det$ on bounded sets).