# What is the inverse of a triangular matrix whose nonzero elements are binomial coefficients? What is the closed-form solution to a recursive relation?

Let $$\begin{equation*} \begin{split} M_m &=\begin{pmatrix} -\binom{1}{0} & \binom{2}{0} &-\binom{3}{0} &\dotsm & (-1)^{m-1}\binom{m-1}{0} & (-1)^m\binom{m}{0}\\ 0 & \binom{2}{1} &-\binom{3}{1} &\dotsm & (-1)^{m-1}\binom{m-1}{1} & (-1)^m\binom{m}{1}\\ 0 & 0 &-\binom{3}{2} &\dotsm & (-1)^{m-1}\binom{m-1}{2} & (-1)^m\binom{m}{2}\\ \vdots & \vdots &\vdots &\ddots & \vdots & \vdots\\ 0 & 0 & 0 &\dotsm & (-1)^{m-1}\binom{m-1}{m-2} & (-1)^m\binom{m}{m-2}\\ 0 & 0 & 0 &\dotsm & 0 & (-1)^m\binom{m}{m-1} \end{pmatrix}_{m\times m}\\ &=(M_{i,j})_{m\times m}, \end{split} \end{equation*}$$ where $$\begin{equation*} M_{i,j}= \begin{cases} (-1)^{j}\dbinom{j}{i-1}, & 1\le i\le j\le m;\\ 0, & 1\le j For $$m=5$$, by the famous software Mathematica, we obtain $$\begin{pmatrix} -1 & 1 & -1 & 1 & -1 \\ 0 & 2 & -3 & 4 & -5 \\ 0 & 0 & -3 & 6 & -10 \\ 0 & 0 & 0 & 4 & -10 \\ 0 & 0 & 0 & 0 & -5 \\ \end{pmatrix}^{-1} = \begin{pmatrix} -1 & \frac{1}{2} & -\frac{1}{6} & 0 & \frac{1}{30} \\ 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} & 0 \\ 0 & 0 & -\frac{1}{3} & \frac{1}{2} & -\frac{1}{3} \\ 0 & 0 & 0 & \frac{1}{4} & -\frac{1}{2} \\ 0 & 0 & 0 & 0 & -\frac{1}{5} \\ \end{pmatrix}.$$ What is the inverse of the triangular matrix $$M_m$$ for $$m\in\mathbb{N}=\{1,2,\dotsc\}$$?

The matrix $$M_m$$ comes from the recursive relation $$$$\label{beta(m+1minus1)} \sum_{k=j+1}^{m}(-1)^{k}\binom{k}{j}\beta_{m+1,k} =(-1)^{j+1} \beta_{m,j}, \quad 0\le j\le m-1,$$$$ where the first few $$\beta_{m,j}$$ are \begin{align*} \beta_{1,0}&=1, & & & & & &\\ \beta_{2,0}&=\frac{5}{3}, & \beta_{2,1}&=1, & & & &\\ \beta_{3,0}&=\frac{11}{5}, & \beta_{3,1}&=\frac{13}{6}, & \beta_{3,2}&=\frac{1}{2}, & &\\ \beta_{4,0}&=\frac{93}{35}, & \beta_{4,1}&=\frac{101}{30}, & \beta_{4,2}&=\frac{4}{3}, & \beta_{4,3}&=\frac{1}{6}. \end{align*} We can also derive \begin{align*}%\label{beta(m+1)m-form} \beta_{m,m-1}&=\frac{1}{(m-1)!}, \quad m\ge1,\\ \beta_{m,m-2}&=\frac{3m+4}{6(m-2)!}, \quad m\ge2,\\ \beta_{m,m-3}&=\frac{15 m^2+35 m+24}{120(m-3)!}, \quad m\ge3,\\ \beta_{m,m-4}&=\frac{105 m^3+315 m^2+364 m+176}{5040(m-4)!}, \quad m\ge4. \end{align*} We guess that $$\begin{equation*} \beta_{m,m-k}=\frac{1}{(2k-1)!(m-k)!}\sum_{\ell=0}^{k-1}\theta_{k,\ell} m^\ell, \quad m\ge k, \end{equation*}$$ where $$\theta_{k,\ell}$$ is a sequence of positive integers.

What is the explicit or closed-form expression of the sequence $$\theta_{k,\ell}$$ for $$0\le\ell\le n-1$$? What is the explicit or closed-form expression of the sequence $$\beta_{m,j}$$ for $$0\le j\le m-1$$?

• Numerics suggest that that this inverse is Faulhaber's triangle of fractions or some straightforward modification of it -- I'd look through the references of that linked OEIS entry. Dec 17, 2022 at 1:28
• Let's forget $(-1)^j$ because it's easily handled. Then your matrix represents the linear transformation $f \mapsto (x+1)f(x+1)-xf(x)$ on the space of polynomials of degree $<m$. To find the inverse, you have to solve the deifference equation $g(x+1)-g(x)=x^j$. Dec 17, 2022 at 4:15
• Sorry, but I still don't see how you get the $\beta_{m,j}$ from the recursion relation without specifying, e.g., $\beta_{m,0}$. Jan 3 at 11:24
• Yes, I understood that. However, with your definition the matrix $M_m$ has to be applied to the vector $(\beta_{m+1,1},\ldots,\beta_{m+1,m})$, and therefore $\beta_{m+1,0}$ is not used. Consequently, the inverse $M_m^{-1}$ cannot reconstruct $\beta_{m+1,0}$. See also my comment below. Jan 3 at 13:15
• So, will you clarify your question? Jan 9 at 10:52

This may answer the question, the sequence is the only implicit thing. I consider $$Q=\begin{pmatrix}Q_{i,j}\end{pmatrix}_{n\times n}$$ for $$n\ge 3$$, where $$Q_{i,j}= \begin{cases} \dbinom{j}{i-1}, & 1\le i\le j\le n;\\ 0, & 1\le j This is the same as the one defined in the question up to multiplying it by a $$\pm 1$$ diagonal matrix. Define sequence $$u_k$$ by $$u_1=1$$ and $$u_k=-\sum_{i=0}^{k-2}\dfrac{u_{i+1}}{(k-i)!}$$ for every $$k>1$$. The $$n\times n$$ matrix $$Q^{-1}=\begin{pmatrix}q_{i,j}\end{pmatrix}_{n\times n}$$ is given by $$\begin{cases} q_{i,j}=0,&1\le j The proof is formal and should be direct, also the sequence $$u_k$$ has some zero entries.
• The inverse of $𝑀_𝑚$ should be $𝑀^{−1}_𝑚=\begin{pmatrix}(−1)^iq_{𝑖,𝑗}\end{pmatrix}_{𝑚\times 𝑚}$. Is there an explicit expression of the sequence $𝑢_𝑘$? Dec 31, 2022 at 1:02
• Hi, ok for the edit and the expression of $M_m^{-1}$, i just kept the 'multiplying of a diagonal matrix $D$' as $(QD)^{-1}=D^{-1}Q^{-1}$. For the sequence i don't think so (i am not an expert in that). Dec 31, 2022 at 9:06
• I just computed the equations and got the identites, we compute the first diagonal of the upper triangular $Q^{-1}$ (easy) then the second, then the third (off) diagonal etc. the induction would appear and could be verified. Jan 2 at 11:06
Using the answer of @Toni, the sequence $$u_k$$ can be related to the Bernoulli numbers $$B_k$$ for $$k\geq 0$$, \begin{align}\tag{1} u_{k+1}= \frac{B_{k}}{k!}. \end{align} After some algebra, the inverse of $$M_m$$ is given by \begin{align}\tag{2} M_m^{-1} = \left[ \, (-1)^i \binom{j}{i} \frac{B_{j-i}}{j} \, \right]_{i,j=1}^{m} \end{align} for $$m>0$$.
• Dear Fred, you are right, I derived the same result from the identity $$\sum_{k=0}^n\binom{n+1}{k}B_k=\delta_{n,0}.$$ What is the explicit expression of the sequence $\beta_{m,j}$? Jan 2 at 13:45
• I don't see the precise relation between $M_m$ and the recursion, please recheck. Either, in the definition of $M_m$ index $j$ must be shifted by one to run from $0,\ldots,m-1$ and it should read (-1)^{i-j}, or the recursion for $\beta$ is not correct. Jan 2 at 14:22