# On the Total Positivity and Accurate Computations of \(r\)-Bell Polynomial Bases
## 3 Bidiagonal Factorization of \(r\)-Stirling Matrices
\right]_{r}=i+r-j-1,\quad 1\leq j<i\leq n+1.\]  
Clearly, \(p_{i,i}>0\), for \(i=1,\ldots,n+1\), and \(m_{i,j}>0\), for \(1\leq j<i\leq n+1\). Then, using Theorem 1, we deduce that \(L_{n}^{r}\) is totally positive. Finally, taking into account the bidiagonal factorization (7) for the inverse of a nonsingular totally positive matrix, the factorization (26) for \((L_{n}^{r})^{-1}\) can be deduced easily. The provided bidiagonal decomposition (5) of the first kind of \(r\)-Stirling matrix \(L_{n}^{r}\) can be stored in a compact form through \(BD(L_{n}^{r})=\left(BD(L_{n}^{r})_{i,j}\right)_{1\leq i,j\leq n+1}\) with  
\[BD(L_{n}^{r})_{i,j}= \begin{cases}i+r-j-1,&2\leq j<i\leq n+1,\\ 1,&1\leq i=j\leq n+1,\\ 0,&\text{elsewhere}.\end{cases}  \tag{30}\]  
Now, the following result deduces the pivots and multipliers of the Neville elimination of \(r\)-Stirling matrices of the second kind and their inverses. **Theorem 3**. : _For \(r,n\in\mathbb{N}\) with \(r\leq n\), the \((n+1)\times(n+1)\) second kind \(r\)-Stirling matrix \(L_{n}^{r}\) admits a bidiagonal factorization_  
\[\bar{L}_{n}^{r}=F_{n}\ldots F_{1}, \tag{31}\]  
_where \(F_{i}\), \(i=1,\ldots,n\), are lower triangular bidiagonal matrices of the form (6), whose off-diagonal entries \(m_{i,j}\), \(1\leq j<i\leq n+1\), are given by_  
\[m_{i,j}=r+j-1,\quad 1\leq j<i\leq n+1. \tag{32}\]  
_Moreover,_  
\[(L_{n}^{r})^{-1}=\bar{F}_{n}\ldots\bar{F}_{1}, \tag{33}\]  
_where \(\bar{F}_{i}\), \(i=1,\ldots,n\), are lower triangular bidiagonal matrices of the form (6), whose off-diagonal entries \(m_{i,j}\), \(1\leq j<i\leq n+1\), are given by_  
\[m_{i,j}=-(r+i-j-1).\quad 1\leq j<i\leq n+1. \tag{34}\]  
Proof. : Let us define \(L^{(1)}:=\bar{L}_{n}^{r}\) and \(L^{(k)}=(l_{ij}^{(k)})_{1\leq i,j\leq n+1}\), \(r=2,\ldots,n\), the matrix obtained after \(k-1\) steps of the Neville elimination of \(L_{n}^{r}\). By induction on \(k\in\{1,\ldots,n+1\}\), we shall deduce that  
\[l_{i,j}^{(k)}=\left\{ \begin{aligned} i+r-1\\ j+r-1\end{aligned} \right\}_{r+k-1}. \tag{35}\]  
Taking into account (23), identities (35) clearly hold for \(k=1\). If (35) holds for some \(k\in\{1,\ldots,n\}\), using (20), we have  
\[l_{i,k}^{(k)}/l_{i-1,k}^{(k)}=\left\{ \begin{aligned} i+r-1\\ k+r-1\end{aligned} \right\}_{r+k-1}/\left\{ \begin{aligned} i+r-2\\ k+r-1\end{aligned} \right\}_{r+k-1}=\frac{(k+r-1)^{i-k}}{(k+r-1)^{i-k-1 }}=k+r-1. \tag{36}\]  
Since \(l_{i,j}^{(k+1)}=l_{i,j}^{(k)}-\left(l_{i,k}^{(k)}/l_{i-1,k}^{(k)}\right)l_{i-1,j}^{(k)}\) (see (1)), and taking into account (21), (35), and (36), we derive  
\[l_{i,j}^{(k+1)}=\left\{ \begin{aligned} i+r-1\\ j+r-1\end{aligned} \right\}_{r+k-1}-(r+k-1)\left\{ \begin{aligned} i+r-2\\ j+r-1\end{aligned} \right\}_{r+k-1}=\left\{ \begin{aligned} i+r-1\\ j+r-1\end{aligned} \right\}_{r+k},\]  
corresponding to the identity (35) for \(k+1\). Now, by considering (2), (20), and (35), we can easily deduce that \[p_{i,j}=l_{i,j}^{(j)}=\left\{ \begin{aligned} & i+r-1\\ & j+r-1\end{aligned} \right\}_{r+j-1}=(r+j-1)^{i-j},\]  
and \(p_{i,i}=1\), for \(i=1,\ldots,n+1\). For the multipliers we have  
\[m_{i,j}=p_{i,j}/p_{i-1,j}=(r+j-1)^{i-j}/(r+j-1)^{i-j-1}.=r+j-1,\quad 1\leq j<i \leq n+1.\]  
Clearly, \(p_{i,i}>0\), for \(i=1,\ldots,n+1\), \(m_{i,j}>0\), for \(1\leq j<i\leq n+1\) and, using Theorem 1, we conclude that \(\bar{L}_{n}^{r}\) is strictly totally positive. Finally, taking into account the bidiagonal factorization (7) for the inverse of a nonsingular totally positive matrix, the factorization (33) for \((\bar{L}_{n}^{r})^{-1}\) can be deduced. The provided bidiagonal factorization (5) of the matrix \(\bar{L}_{n}^{r}\) can be stored in a compact form through \(BD(\bar{L}_{n}^{r})=\left(BD(\bar{L}_{n}^{r})_{i,j}\right)_{1\leq i,j\leq n+1}\) with  
\[BD(\bar{L}_{n}^{r})_{i,j}= \begin{cases}r+j-1,&2\leq j<i\leq n+1,\\ 1,&1\leq i=j\leq n+1,\\ 0,&\text{elsewhere}.\end{cases}  \tag{37}\]  
**Remark 1**.