## 5 Interpolation Model with Scalar Subdivision Schemes  
Given the problem \(M_{n}\mathbf{P}^{0}=\mathbf{V}^{0}\) in (10), let us apply the tools available for circulant matrices to solve it. Indeed, by Remark 1 we deduce that the spectrum of \(M_{n}\) is  
\[\Lambda(M_{n})=\left\{b\Big{(}\frac{2\pi i}{n}\Big{)},j=0,\ldots,n-1\right\} \tag{16}\]  
and for the stencils with compact support in \([-p,q]\) the symbol can be written as  
\[b(\theta):=\sum_{k=-p}^{q}\beta_{k}\text{exp}(\text{i}k\theta),\]independent of \(n\). Thus, the singularity of \(M_{n}\) depends on the existence of roots for \(b(\theta)\) in the grid \(\frac{2\pi\mathbb{N}}{n}\cap[0,2\pi]\). For odd-symmetric schemes, \(\mathbf{\beta}_{n}:=\big{[}\beta_{0},\beta_{-1},\ldots,\beta_{-p},0,\ldots,0,\beta_{ p},\ldots,\beta_{1}\big{]}\in\mathbb{R}^{n}\) and thus  
\[b(\theta)=\sum_{k=-p}^{p}\beta_{k}\mathrm{exp}(\mathrm{i}k\theta)=\beta_{0}+2 \sum_{k=1}^{p}\beta_{k}\cos(k\theta). \tag{17}\]  
Conversely, for even-symmetric schemes \(\mathbf{\beta}_{n}:=\big{[}\beta_{0},\beta_{-1},\ldots,\beta_{-p+1},0,\ldots,0, \beta_{p},\ldots,\beta_{1}\big{]}\in\mathbb{R}^{n}\), hence  
\[b(\theta) =\sum_{k=-p+1}^{p}\beta_{k}\mathrm{exp}(\mathrm{i}k\theta)=\sum_ {k=1}^{p}\beta_{k}(\mathrm{exp}(\mathrm{i}(-k+1)\theta)+\mathrm{exp}(\mathrm{ i}k\theta))\] \[=\mathrm{exp}\bigg{(}\frac{\mathrm{i}\theta}{2}\bigg{)}\sum_{k=1 }^{p}\beta_{k}\bigg{(}\mathrm{exp}\bigg{(}\mathrm{i}\bigg{(}-k+\frac{1}{2} \bigg{)}\theta\bigg{)}+\mathrm{exp}\bigg{(}\mathrm{i}\bigg{(}k-\frac{1}{2} \bigg{)}\theta\bigg{)}\bigg{)}\] \[=2\mathrm{exp}\bigg{(}\frac{\mathrm{i}\theta}{2}\bigg{)}\sum_{k=1 }^{p}\beta_{k}\cos\Big{(}(2k-1)\frac{\theta}{2}\Big{)}. \tag{18}\]  
Either way, the first limit stencil satisfies \(b(0)=1\). Note that for odd-symmetric schemes it holds  
\[b(2\pi-\theta)=\beta_{0}+2\sum_{k=1}^{p}\beta_{k}\cos(k(2\pi-\theta))=\beta_{ 0}+2\sum_{k=1}^{p}\beta_{k}\cos(k\theta)=b(\theta),\]  
while for the even-symmetric schemes  
\[b(2\pi-\theta) =2\mathrm{exp}\bigg{(}\frac{\mathrm{i}(2\pi-\theta)}{2}\bigg{)} \sum_{k=0}^{p}\beta_{k}\cos\Big{(}(2k-1)\frac{(2\pi-\theta)}{2}\Big{)}\] \[=2\mathrm{exp}\bigg{(}\frac{-\mathrm{i}\theta}{2}\bigg{)}\sum_{k= 0}^{p}\beta_{k}\cos\Big{(}(2k-1)\frac{\theta}{2}\Big{)}\] \[=\mathrm{exp}(-\mathrm{i}\theta)b(\theta).\]  
Therefore, the study of the symbol can be restricted to the interval \([0,\pi]\) instead of \([0,2\pi]\). In particular, for both (17) and (18), \(b(\theta)=0\) implies \(b(2\pi-\theta)=0\). From (18), for even-symmetric schemes we find \(b(\pi)=0\) independently of \(n\) and \(\mathbf{\beta}\). As \(\pi\) belongs to the grid \(\frac{2\pi\mathbb{N}}{n}\cap[0,2\pi]\) for even values of \(n\), then from (16) we obtain the following result. Proposition 4: _For any even-symmetric subdivision scheme, if the amount of interpolated points \(n\) is even, then the interpolation matrix \(M_{n}\) is singular._  
On the other hand, in the context of odd-symmetric schemes the symbol may also vanish in the grid \(\frac{2\pi\mathbb{N}}{n}\cap[0,2\pi]\). As an example, consider the primal family of J-spline schemes [38] for the particular case that generates \(C^{3}\) subdivision curves  
\[ \begin{cases}P_{2j}^{k+1}&=\frac{1}{4}P_{j-1}^{k}+\frac{1}{2}P_{j}^{k}+\frac{ 1}{4}P_{j+1}^{k},\\ P_{2j+1}^{k+1}&=\frac{1}{16}P_{j-1}^{k}+\frac{7}{16}P_{j}^{k}+\frac{7}{16}P_{j+ 1}^{k}+\frac{1}{16}P_{j+2}^{k},\quad k\in\mathbb{N},j\in\mathbb{Z},\end{cases}  \tag{19}\]  
with first limit stencil  
\[\mathbf{\beta}=\frac{1}{48}\{1,\ 12,\ 22,\ 12,\ 1\}.\]The symbol  
\[b(\theta)=\tfrac{11}{24}+\tfrac{1}{2}\cos(\theta)+\tfrac{1}{24}\cos(2\theta)=( \cos(\theta)+1)\Big{(}\cos(\theta)+\tfrac{1}{5}\Big{)}\]  
has a root at \(\theta=\pi\), which belongs to the grid \(\tfrac{2\pi\mathbb{N}}{n}\cap[0,2\pi]\) when \(n\) is an even number. In the Hermite interpolation scenario, we can find a singular point and tangents interpolation operator even when the point interpolation operator is not singular.