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Sampling consecutive Cardinal cubic B-splines at the centers of their supports gives diagonal values $2/3$ and adjacent values $1/6$, with all other values zero. Finite truncation gives the displayed B-spline collocation matrix, including its endpoint rows without wrapping or adding boundary conditions. Its inverse has alternating signs. The exact inverse norm of midpoint cubic spline collocation is less than three for each finite dimension and approaches three as the dimension grows.

The B-spline collocation matrix in the cubic case is the finite tridiagonal matrixFor $n=1$, this is simply the one-entry matrix $(2/3)$.

To evaluate the inverse norm exactly, let $D=\mathrm{diag}((-1{)}^1,\dots ,(-1{)}^n)$. Then $B_n=D{A}_{n}D=(4I-C)/6$, where $C$ has ones immediately above and below the diagonal and zeros elsewhere. Since $\Vert C/4{\Vert }_{{\ell }^{\infty }}\le 1/2$, the convergent Neumann serieshas nonnegative entries and proves invertibility. Since $A_n^{-1}=D{B}_n^{-1}D$, its absolute entries equal those of $B_n^{-1}$. The absolute row sums are therefore the entries of $u=B_n^{-1}1$.

The equation $B_{n}u=1$ becomesIts constant particular solution is $3$; the two characteristic roots of the homogeneous recurrence are $\rho =2-\sqrt{3}$ and ${\rho }^{-1}=2+\sqrt{3}$. Imposing the two boundary values givesThis can also be checked by direct substitution, since $\rho +{\rho }^{-1}=4$. The numerator is smallest at the central index or central pair of indices: its successive difference is $(1-\rho )({\rho }^{n-i}-{\rho }^i)$. Thus, with $j=\lfloor (n+1)/2\rfloor$, the exact inverse norm of midpoint cubic spline collocation isEquivalently, if $n=2m-1$ the fraction is $2{\rho }^m/(1+{\rho }^{2m})$, and if $n=2m$ it is $({\rho }^m+{\rho }^{m+1})/(1+{\rho }^{2m+1})$. The finite norm is not exactly three: it equals $3/2$ for $n=1$, two for $n=2$, and tends to three as $n$ grows. Hence three is the sharp dimension-independent bound for these inverse matrix norms.