 Determinant of the minimum-index matrix

Let $L$ be the triangular matrix with $L_{ij}=1$ for $j\le i$ and zero otherwise. Then $(L{L}^T{)}_{ij}=\min (i,j)$ and $\det L=1$, so the determinant of $L{L}^T$ is one. More generally, for $0=t_0<t_1<\cdots <t_n$, the matrix with entries $\min (t_i,t_j)$ factors as $L\mathrm{diag}(t_k-t_{k-1})L^T$ and has determinant ${\prod }_k(t_k-t_{k-1})$. This is the finite-dimensional structure behind the Brownian covariance kernel.