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First take $k\ge 2$, so the simple-knot B-splines are continuous, vanish at both support endpoints, and are strictly positive inside their supports. Thus the diagonal condition is $t_i<x_i<t_{i+k}$. We prove both directions of the Schoenberg–Whitney theorem, deriving the weighted zero-count fact needed for sufficiency rather than relying on an ambiguous coefficient-free hint.

For necessity, suppose $N_i(x_i)=0$. If $x_i\le t_i$, then for every row $r\le i$ and column $j\ge i$ we have $x_r\le x_i\le t_i\le t_j$, hence $N_j(x_r)=0$. The first $i$ rows have nonzero entries in at most $i-1$ columns and are linearly dependent. If instead $x_i\ge t_{i+k}$, then for $r\ge i$ and $j\le i$ we have $x_r\ge x_i\ge t_{i+k}\ge t_{j+k}$, so those entries vanish. The last $n-i+1$ rows have nonzero entries in at most $n-i$ columns and are again dependent. In either case the B-spline collocation matrix is singular. Thus invertibility implies the diagonal positivity condition.

We establish two elementary spline facts for the converse. For local linear independence of B-splines, extend the strictly increasing knots in both directions. On a cell $(t_j,t_{j+1})$ the $k$ active basis functions have indices $j-k+1,\dots ,j$. Their knot polynomials are ${\psi }_i(y)={\prod }_{\ell =1}^{k-1}(y-t_{i+\ell })$. The Cox-de Boor recurrence proves Marsden's identityHere is the induction behind that identity. For order one the indicator functions sum to one. On passing from order $k-1$ to $k$, collect the coefficient of each lower-order basis function. With $u=t_i$, $v=t_{i+k-1}$, its extra factor isThus the identity at one order multiplies by $y-x$ at the next.

The $k$ polynomials ${\psi }_i$ are independent: evaluate successively at $y=t_j,t_{j-1},\dots ,t_{j-k+1}$. The first evaluation isolates ${\psi }_j$; after its coefficient has been eliminated the next isolates ${\psi }_{j-1}$, and so on, with nonzero diagonal factors because the knots are distinct. Comparing powers of $y$ in the identity then shows that the active $N_i$ span all polynomials of degree at most $k-1$ on the cell. There are $k$ of them, so they are independent. In particular, a spline vanishes on a whole open cell exactly when all its active coefficients vanish. Padding the finite coefficient vector with zeros makes this fact apply also near the outer knots.

Next let $s={\sum }_{i=p}^q{c}_i{N}_i$ be a real spline on $I=(t_p,t_{q+k})$, with no identically zero knot cell there. The simple-knot B-spline regularity gives $s\in C^{k-2}$, and its derivatives through order $k-2$ vanish at both endpoints because it is supported on the closed interval. Suppose it has $Z$ distinct interior zeros. Its zero set initially has $Z+2$ connected components, counting the endpoints; the zeros are isolated because no cell vanishes identically.

A useful form of Rolle's theorem counts zero components, including any zero intervals that arise after differentiation. Between two successive zero components a real differentiable function is nonzero and has an interior extremum, supplying a zero of its derivative. These derivative-zero components are distinct: a zero interval connecting two such extrema would make the original function constant across an intervening zero and nonzero gap, which is impossible. If its derivative also vanishes at both endpoints, those supply two further distinct components. Thus differentiation increases the zero-component count by at least one. Apply this successively through derivative order $k-2$; every required endpoint derivative vanishes. No derivative in this process can vanish identically: that would make $s$ a global polynomial of degree below $k-1$, with all its endpoint derivatives through that degree zero, forcing $s=0$. The final derivative has at least $Z+k$ zero components.

That final derivative is continuous and piecewise affine on $q-p+k$ knot cells. Each closed affine cell can meet at most one zero component: two different zeros would force the whole cell to vanish, joining them into one component. Every zero component meets a cell, so there are at most $q-p+k$ of them. For $k=2$ this argument applies directly without differentiation. We have proved the compact-support spline zero countThis proof also accommodates zero intervals of the derivatives; counting them as infinitely many distinct zeros would invalidate a naive repeated-Rolle argument.

Now assume $N_i(x_i)>0$ for all $i$ and suppose $A_{x}c=0$ for a nonzero coefficient vector. Since $A_x$ is real, a nonzero real null vector can be chosen. Set $s={\sum }_i{c}_i{N}_i$, so $s(x_i)=0$ for all $i$. Consider one connected component of the union of the open supports of the terms with $c_i\ne 0$. If its first and last indices are $p,q$, the component is exactly $I=(t_p,t_{q+k})$. On this interval $s$ equals $s_0={\sum }_{i=p}^q{c}_i{N}_i$. Each knot cell in $I$ has at least one active nonzero coefficient, so local independence ensures $s_0$ does not vanish on any whole cell.

For every $p\le i\le q$, diagonal positivity gives $t_i<x_i<t_{i+k}$, and this places $x_i$ inside $I$. Thus $s_0$ has the $q-p+1$ distinct zeros $x_p,\dots ,x_q$, contradicting the upper bound $q-p$. There is no nonzero null vector, proving sufficiency. Together the two directions giveFor completeness, order $k=1$ consists of disjoint interval indicators, with the usual left-closed, right-open convention. An invertible ordered collocation requires exactly one site in each interval; ordering makes this equivalent to $N_i(x_i)>0$, and the matrix is then the identity. This handles possible sites at left knots without incorrectly replacing positivity by a strict interior inequality for order one.

Use the usual strictly increasing spline knot sequence, and initially assume $k\ge 2$. The explicit formula for the order-$k$ divided difference givesEach summand is a truncated power function in $t$, with its junction at $t_j$. It is a piecewise polynomial function of degree at most $k-1$ and has $k-2$ continuous derivatives. Finite summation preserves these properties.

For $t\ge t_{i+k}$ all summands vanish. For $t\le t_i$, all spline knot data come from the polynomial $u\mapsto (u-t{)}^{k-1}$. An order-$k$ divided difference annihilates a polynomial of smaller degree, so $M_i(t)=0$ there too. Thus the support is contained in $[t_i,t_{i+k}]$.

On the last open spline knot interval only the $j=i+k$ term survives; it is a nonzero multiple of $(t_{i+k}-t{)}^{k-1}$. On the first open spline knot interval, compare the truncated spline knot data with the untruncated polynomial data: only the $j=i$ value differs. Since the untruncated divided difference is zero, the result is a nonzero multiple of $(t_i-t{)}^{k-1}$. Consequently both outer pieces are nonzero, giving exact closed support. The Cox-de Boor recurrence used below also proves positivity throughout the open support. ThereforeThis is simple-knot B-spline regularity. For $k=1$ the B-splines are interval indicators; they are piecewise constant and need not be continuous. The notation $C^{k-2}$ should then be read as no continuity requirement, not as a classical negative-order differentiability space.

Write the strictly increasing local spline knots as $u_j=t_{i+j}$, $0\le j\le k$. The explicit formula for a divided difference givesEach summand is a truncated power function of $t$: it is a polynomial on either side of its knot $u_j$, and for $k\ge 2$ is globally $C^{k-2}$. Hence $M_i$ is a piecewise polynomial function of degree at most $k-1$, with these spline knots and at least this global smoothness.

For $t\ge u_k$, all summands vanish. For $t<u_0$, all knot values agree with those of the ordinary polynomial $(s-t{)}^{k-1}$ in the divided-difference variable $s$. Its order-$k$ divided difference is zero, because its degree is less than $k$. Thus $M_i$ also vanishes to the left of $u_0$.

The closed support is exactly the indicated interval, rather than merely contained in it. For $u_0<t<u_1$, subtracting the omitted $j=0$ term from the zero divided difference of $(s-t{)}^{k-1}$ givesFor $u_{k-1}<t<u_k$, only the last truncated-power term survives, givingThere are therefore nonzero values arbitrarily close to either endpoint, provingAt each simple knot, the $(k-1)$st derivative has a nonzero jump from exactly one truncated-power summand, so this is also the exact global smoothness. For order $k=1$, the function is instead the normalized interval indicator; there is no assertion of classical continuity, and the notation $C^{k-2}$ is only the customary formal spline smoothness notation. This distinction is part of simple-knot B-spline regularity.