Corollary 3 (« Hadamard’s inequalities »)

— Let $(a_{ij})_{1 \leq i, j \leq n}$ be a matrix with complex elements, and with determinant D. Put

and $m = \sup_{i,j} |a_{ij}|$. Then we have

If $D \neq 0$, in order that $|D| = c_1 ... c_n$ it is necessary and sufficient that the rows $y_i = (a_{ij})_{1 \leq j \leq n}$ of the matrix $(a_{ij})_{1 \leq i, j \leq n}$ are two by two orthogonal vectors.

Let the space $\mathbf{C}^n$ be assigned a scalar product defined by

Let $(\mathbf{x}_1, ..., \mathbf{x}_n)$ be the canonical basis of $\mathbf{C}^n$ and $\mathbf{y}_i$ the vector with components $a_{ij}$ for $1 \leq j \leq n$. We have $\| \mathbf{x}_i \| = 1$ and $\| \mathbf{y}_i \| = c_i$ for $1 \leq i \leq n$; also $\langle \mathbf{x}_i|\mathbf{y}_j \rangle = a_{ji}$. The inequality $|D| \leq c_1 ... c_n$ and the condition of equality are then particular cases of V, p. 35, cor. 1. Obviously we have $c_i \leq m \cdot n^{1/2}$, hence $c_1 ... c_n \leq m^n \cdot n^{n/2}$.