= Solution
A B-spline $N_j$ is positive exactly in the interior of its support $(t_j,t_{j+k})$. If the <B-spline collocation matrix> $A_{\mathbf x}$ is invertible, its <determinant> contains a nonzero permutation term. Hence there is a permutation $\pi$ such that
$$
N_{\pi(r)}(x_r)\gt0,
\qquad r=1,\ldots,n.
$$
If $x_i\leq t_i$, then every one of the first $i$ points satisfies $x_r\lt t_i$. Any support containing such a point must have left endpoint $t_j\lt x_r$, hence $j\lt i$. Only $i-1$ B-splines are available to match these $i$ rows, contradicting that $\pi$ is a permutation. Thus $x_i\gt t_i$. Similarly, if $x_i\geq t_{i+k}$, each of the last $n-i+1$ points can only be matched to an index $j\gt i$, but only $n-i$ such indices exist. Therefore $x_i\lt t_{i+k}$. We have proved
$$
\boxed{t_i\lt x_i\lt t_{i+k}},
\qquad
\boxed{N_i(x_i)\gt0}.
$$
The <Schoenberg–Whitney theorem> states, for strictly increasing knots and interpolation sites, that
$$
\boxed{A_{\mathbf x}=(N_j(x_i))_{i,j=1}^n
\text{ is invertible}
\iff t_i\lt x_i\lt t_{i+k}
\iff N_i(x_i)\gt0\quad\forall i}.
$$
Indeed its determinant is positive under these inequalities.
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