Eigenvalues of a matrix with binomial entries

Question

I am trying to determine the eigenvalues and eigenvectors of the following 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 $(b-a+1) \times (b-a+1)$.

I would content myself with an approximate asymptotic expression (if there is no exact analytical result), valid for large $a,b$ (say for each fixed ratio $a/b$). I am mostly interested in the largest eigenvalue (not in absolute value, but the largest positive eigenvalue) and the corresponding eigenvector.

Also a recurrence relation would be useful. Anything that helps...

To be explicit, we want to solve the following eigenquation:

$$\sum_{j=a}^b 4^{-j} \binom{2j}{i} x_j = \lambda x_i,\quad i=a, a+1, ..., b$$

for the eigenvalues $\lambda$ and eigenvectors $x_i$.

Update: Numerical experiments suggest that if $a,b\rightarrow \infty$ with a fixed ratio $a/b$, the largest eigenvalue $\rightarrow 1$ always, irrespective of the value of the ratio $a/b$. There is a leading order correction proportional to $1/\sqrt{b}$, and the value of the proportionality constant depends on the value of the ratio $a/b$. I do not know how to prove any of these statements, and I cannot be sure they are correct. It would be nice if we could compute the value of leading order coefficient.

Answer 1

I don't have an answer, but it appears that the eigenvalues are always real. I don't have a proof, but have checked this using Sturm sequences for $1 \le a \le b \le 30$.

You're unlikely to get "an exact analytic result", as the characteristic polynomial seems to be irreducible over the rationals unless $a=0$ (in which case $\lambda - 1$ is a factor). Thus for $a=1, b=5$, the characteristic polynomial is $${\lambda}^{5}-{\frac {437\,{\lambda}^{4}}{256}}+{\frac {29823\,{ \lambda}^{3}}{32768}}-{\frac {85687\,{\lambda}^{2}}{524288}}+{\frac { 15115\,\lambda}{2097152}}-{\frac{1}{32768}} $$ which has Galois group $S_5$.