This is closely related to this question: [Eigenvalues of a matrix with binomial entries](https://mathoverflow.net/q/275911/16615).


We consider the matrix:


$$M\_{ij} = 4^{-j}\binom{2j}{i}$$


where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k<0$ or $k>m$. The indices $i,j$ traverse a discrete finite range, $i,j \in \{a, a+1, \dots, b\}$, from $a$ to $b$, where $a,b$ are non-negative integers with $0\le a\le b$. Therefore the matrix $M\_{ij}$ has dimensions $(b-a+1) \times (b-a+1)$.


Can we find the inverse matrix, $M^{-1}$? Numerical computations of the determinant suggest that this is a non-singular matrix (for all $0 \le a < b$).


A close-form expression for the determinant could be useful.


**Why this question is not a duplicate:** Although in principle the eigenvalues and eigenvectors of a matrix are enough to invert it, [the other question](https://mathoverflow.net/q/275911/16615) focuses on the *largest positive* eigenvalue alone. Moreover the eigenvalues/eigenvectors (not even the largest alone) have not been solved, so maybe finding the inverse of this matrix turns out to be easier. So, if the other question suddenly received a complete response and all the eigenvalues / eigenvectors were found, then yes, this question would be automatically solved. But that does not seem likely to happen.



Inverse of a matrix with binomial entries
Let's refer everything to square matrices indexed from $0$ to $h$, that I will denote as
$$
{\bf M}\_{\,h} = \left\| {\;f(n,m)\;} \right\|\_{\,h} 
$$
with $n$ being the row index and $m$ the column index.


I will then denote by
$$
\left( {f(n) \circ {\bf I}\_{\,h} } \right)
$$
the diagonal matrix whose entries are equal to $f(n)$.


So I write the matrix you proposed as
$$ \bbox[lightyellow] { 
{\bf M}\_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{
 2n + 2a \cr 
 m + a \cr} \right)\;} \right\|\_{\,h} = \left( {4^{ - n - a} \circ {\bf I}\_{\,h} } \right)\;\left\| {\;\left( \matrix{
 2n + 2a \cr 
 m + a \cr} \right)\;} \right\|\_{\,h} 
 }$$
where $h=b-a$.


That premised, consider that in general
$$
\eqalign{
 & \left( \matrix{
 r\,n + t \cr 
 m + q \cr} \right) = {{\left( {r\,n + t} \right)^{\,\underline {\,m + q} } } \over {\left( {m + q} \right)!}} = {{\left( {r\,n + t} \right)^{\,\underline {\,q} } \left( {r\,n + t - q} \right)^{\,\underline {\,m} } } \over {\left( {m + q} \right)^{\,\underline {\,q} } \;m^{\,\underline {\,m} } }} = \cr 
 & = \left( {r\,n + t} \right)^{\,\underline {\,q} } \left( \matrix{
 r\,n + t - q \cr 
 m \cr} \right){1 \over {\left( {m + q} \right)^{\,\underline {\,q} } }} \cr} 
$$
where $x^{\,\underline {\,a} } $ denotes the falling factorial ($x^{\overline {\,a\,} } $ the rising) and
where, for the present problem, we consider $q$ to be a non-negative integer,
while $r$ and $t$ could be real (or even complex).


Then we have that we can write the binomial as
$$
\begin{gathered}
 \left( \begin{gathered}
 r\,n + s \\ 
 m \\ 
\end{gathered} \right) = \frac{1}
{{m!}}\left( {r\,n + s} \right)^{\,\underline {\,m\,} } = \frac{1}
{{m!}}\sum\limits\_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right)} {\left( \begin{gathered}
 m \\ 
 k \\ 
\end{gathered} \right)s^{\,\underline {\,m - k\,} } \left( {r\,n} \right)^{\,\underline {\,k\,} } } = \hfill \\
 = \frac{1}
{{m!}}\sum\limits\_{\left\{ \begin{subarray}{l} 
 \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ 
 \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) 
\end{subarray} \right.} {\left( \begin{gathered}
 m \\ 
 k \\ 
\end{gathered} \right)s^{\,\underline {\,m - k\,} } \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered}
 k \\ 
 j \\ 
\end{gathered} \right]r^{\,j} n^{\,j} } = \hfill \\
 = \sum\limits\_{\left\{ \begin{subarray}{l} 
 \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ 
 \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) 
\end{subarray} \right.} {n^{\,j} r^{\,j} \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered}
 k \\ 
 j \\ 
\end{gathered} \right]\frac{1}
{{k!}}\left( \begin{gathered}
 s \\ 
 m - k \\ 
\end{gathered} \right)} \hfill \\ 
\end{gathered} 
$$


Then in the last line we can replace $n^m$ with
$$
n^{\,m} = \sum\limits\_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left\{ \matrix{
 m \cr 
 k \cr} \right\}n^{\,\underline {\,k\,} } } = \sum\limits\_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left( \matrix{
 n \cr 
 k \cr} \right)k!\left\{ \matrix{
 m \cr 
 k \cr} \right\}} 
$$


Thus we arrive finally to
$$ \bbox[lightyellow] { 
\eqalign{
 & {\bf M}\_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{
 2n + 2a \cr 
 m + a \cr} \right)\;} \right\|\_{\,h} = \left( {4^{ - n - a} \circ {\bf I}\_{\,h} } \right)\;\left\| {\;\left( \matrix{
 2n + 2a \cr 
 m + a \cr} \right)\;} \right\|\_{\,h} = \cr 
 & = \left( {4^{ - n - a} \circ {\bf I}\_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} } \circ {\bf I}\_{\,h} } \right)\;\left\| {\;\left( \matrix{
 2n + a \cr 
 m \cr} \right)\;} \right\|\_{\,h} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}\_{\,h} } \right) = \cr 
 & = \left( {4^{ - n - a} \circ {\bf I}\_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} } \circ {\bf I}\_{\,h} } \right)\;{\bf B}\_{\,h} \left( {n! \circ {\bf I}\_{\,h} } \right)\;\overline {{\bf St}\_{{\bf 2}\,h} } \left( {2^{\,n} \circ {\bf I}\_{\,h} } \right)\;\overline {{\bf St}\_{{\bf 2}\,h} } ^{\,{\bf - }\,{\bf 1}} \left( {n! \circ {\bf I}\_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} \left( {{\bf I}\_{\,h} + \overline {{\bf E}\_{\,h} } } \right)^{\,{\bf a}} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}\_{\,h} } \right) \cr} 
\tag{1} }$$
with
$$
\eqalign{
 & {\bf B}\_{\,h} = \;\left\| {\;\left( \matrix{
 n \cr 
 m \cr} \right)\;} \right\|\_{\,h} \quad {\bf St}\_{{\bf 2}\,h} = \;\left\| {\;\left\{ \matrix{
 n \cr 
 m \cr} \right\}\;} \right\|\_{\,h} \quad {\bf I}\_{\,h} + {\bf E}\_{\,h} = \;\left\| {\;\left( \matrix{
 1 \cr 
 n - m \cr} \right)\;} \right\|\_{\,h} \cr 
 & \overline {\bf X} = transpose({\bf X}) \cr} 
$$


After that the determinant follows easily, since the matrices other than the diagonal ones
have unitary determinant
$$ \bbox[lightyellow] { 
\left| {\,{\bf M}\_{\,h} (a)\,} \right| = \left( {\prod\limits\_{0\, \le \,n\, \le \;h} {{{\left( {2\,\left( {n + a} \right)} \right)^{\,\underline {\,a} } } \over {2^{\,n + 2a} \left( {n + a} \right)^{\,\underline {\,a} } }}} } \right) = \left( {\prod\limits\_{0\, \le \,n\, \le \;h} {{{\left( \matrix{
 2\,\left( {n + a} \right) \cr 
 a \cr} \right)} \over {2^{\,n + 2a} \left( \matrix{
 n + a \cr 
 a \cr} \right)}}} } \right)
 \tag{2}}$$


Some notes concerning the inversion of identity (1), and further analysis you might possibly want perform on that.
For the Binomial
$$
{\bf B}\_{\,h} ^{\,{\bf r}} = \;\left\| {\;r^{\,n - m} \left( \matrix{
 n \cr 
 m \cr} \right)\;} \right\|\_{\,h} = \left( {r^n \circ {\bf I}\_{\,h} } \right){\bf B}\_{\,h} \;\left( {r^n \circ {\bf I}\_{\,h} } \right)^{\, - \,{\bf 1}} \quad \;\left| {\;r \in R,C} \right.
$$
where the second expression for $r=0$ is understood to be taken in the limit.  

So
$$
{\bf B}\_{\,h} ^{\, - \,{\bf 1}} = \left( {\left( { - 1} \right)^n \circ {\bf I}\_{\,h} } \right){\bf B}\_{\,h} \;\left( {\left( { - 1} \right)^n \circ {\bf I}\_{\,h} } \right)^{\, - \,{\bf 1}} = \left( {\left( { - 1} \right)^n \circ {\bf I}\_{\,h} } \right){\bf B}\_{\,h} \;\left( {\left( { - 1} \right)^n \circ {\bf I}\_{\,h} } \right)
$$
For the Stirling Numbers, 1st and 2nd kind are related by
$$
{\bf St}\_{{\bf 2}\,h} ^{\, - \,{\bf 1}} = \;\left\| {\;\left( { - 1} \right)^{\,n - m} \left[ \matrix{ n \cr m \cr} \right]\;} \right\|\_{\,h} =
 \left( {\left( { - 1} \right)^n \circ {\bf I}\_{\,h} } \right)\;{\bf St}\_{{\bf 1}\,h} \;\left( {\left( { - 1} \right)^n \circ {\bf I}\_{\,h} } \right)
$$
${\bf E}$ is the "shift", "first off-diagonal", .. matrix, i.e:
$$
{\bf E}\_{\,h} = \left\| {\;\left[ {1 = n - m} \right]\;} \right\|\_{\,h} = \left\| {\;\left( \matrix{ 0 \cr 
 n - m - 1 \cr} \right)\;} \right\|\_{\,h} 
$$
(where $[P]$ is the [*Iverson bracket*](https://en.wikipedia.org/wiki/Iverson_bracket)) 


then
$$
\eqalign{
 & {\bf E}\_{\,h} ^{\,{\bf q}} = \left\| {\;\left[ {q = n - m} \right]\;} \right\|\_{\,h} = \left\| {\;\left( \matrix{
 0 \cr 
 n - m - q \cr} \right)\;} \right\|\_{\,h} \quad \;\left| {\;0 \le q \in Z} \right. \cr 
 & \left( {{\bf I}\_{\,h} + {\bf E}\_{\,h} } \right) = \left\| {\;\left[ {0 \le n - m \le 1} \right]\;} \right\|\_{\,h} = \left\| {\;\left( \matrix{
 1 \cr 
 n - m \cr} \right)\;} \right\|\_{\,h} \cr 
 & \left( {{\bf I}\_{\,h} + {\bf E}\_{\,h} } \right)^{\,{\bf r}} = \sum\limits\_{0\, \le \,k\,\left( { \le \;h} \right)} {\left( \matrix{
 r \cr 
 k \cr} \right){\bf E}\_{\,h} ^{\,{\bf k}} } = \left\| {\;\left( \matrix{
 r \cr 
 n - m \cr} \right)\;} \right\|\_{\,h} \quad \;\left| {\;r \in R,C} \right. \cr} 
$$


and finally that ${\bf B}$ and ${\bf {I+E}}$ are tied by the similarity
$$
{\bf B}\_{\,h} \; = \left( {{\bf St}\_{\,{\bf 2}\,h} \left( {n! \circ {\bf I}\_{\,h} } \right)} \right)\left( {{\bf I}\_{\,h} + {\bf E}\_{\,h} } \right)\left( {{\bf St}\_{\,{\bf 2}\,h} \left( {n! \circ {\bf I}\_{\,h} } \right)} \right)^{{\bf - 1}} 
$$
and by a bunch of other relations, among which
$$
\left( {{\bf I}\_{\,h} + {\bf E}\_{\,h} } \right)^{\,{\bf - q}} \quad \left| {\;0 \le {\rm integer }q} \right.\quad = \left( {{\bf B}\_{\,h} \left( {n^{\,\underline {\, - q\,} } \circ {\bf I}\_{\,h} } \right)} \right)^{\,{\bf - 1}} \;\left( {\left( {n^{\,\underline {\, - q\,} } \circ {\bf I}\_{\,h} } \right)\;\;{\bf B}\_{\,h} } \right)
$$


