Eigenvalues of a truncated matrix with binomial entries This question is closely related to Eigenvalues of a matrix with binomial entries, but is a bit more general.
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 $0\le a\le b$ (this is the difference to the previous question). Therefore the matrix $M_{ij}$ has dimensions $(b-a+1) \times (b-a+1)$.
I would content myself with an approximate expression (if there is no exact analytical result), valid for large $a,b$. 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.
 A: EDIT 2017-07-26.
We have
$$
\sum_{j=a}^b 4^{-j} \binom{2j}{i} x_j = \lambda x_i
$$
This is declared for $i=a, a+1, ..., b$. 
Suppose we have found $\lambda$ and the $\{x_i \}, i=a, a+1, \dots, b$ through these equations. Let us then use the same equation for obtaining further  $x_i, i=0, \dots, a-1$ and  $x_i, i=b+1, \dots, 2b$.  
Now  apply to both sides, for arbitrary $q$, 
$$
\sum_{i=0}^{2 b} q^i \; (\cdot)
$$
This gives for the LHS
$$
\sum_{j=a}^b 4^{-j} x_j \sum_{i=0}^{2 b} \binom{2j}{i} q^i =\\
\sum_{j=a}^b 4^{-j} x_j \sum_{i=0}^{2 j} \binom{2j}{i} q^i 1^{2 j - i} = \\
\sum_{j=a}^b 4^{-j} x_j (1+q)^{2j}
$$
On the RHS, we have that the $x_i$ outside the $[a,b]$-range are obtained by the additional equations as introduced above. So we have
 for all $q$ for the extended set of unknowns $x_i ;  i=0, 1, ..., 2b$ and $\lambda$ :
$$
\sum_{j=a}^b x_j (\frac{1+q}{2})^{2j} = \lambda \sum_{j=0}^{2b}  x_j q^j
$$
So, applying different $q$, we have a set of polynomial equations from which to deduce the  $\{x_i \}, i=0, 1, \dots, 2b$ as well as $\lambda$. I cannot directly see how to solve those. 
Some examples are:
$$
q\to 0: \quad \sum_{j=a}^b 4^{-j} x_j  = \lambda \; x_0
$$
$$
q = 1: \quad \sum_{j=a}^b  x_j  = \lambda \sum_{j=0}^{2b}  x_j 
$$
$$
q\to \infty: \quad  (\frac{q}{2})^{2b}  x_b  = \lambda  \, x_{2b} \, q^{2b} \quad {\mathrm{; i.e.}} \quad 4^{-b}  x_b  = \lambda \, x_{2b} 
$$
The last one can also be seen directly from the original equations. From them, 
one can deduce for example for $i = 2b-1$:
$$
\sum_{j=a}^b 4^{-j} \binom{2j}{2b-1} x_j = \lambda \, x_{2b - 1} \quad {\mathrm{; i.e.}} \quad  2 b \, 4^{-b}  x_b  = \lambda \, x_{2b -1} 
$$
and for $i = 2b-2$:
$$
\sum_{j=a}^b 4^{-j} \binom{2j}{2b-2} x_j = \lambda \, x_{2b - 2} \quad {\mathrm{; i.e.}} \quad   4^{-b} \binom{2b}{2b-2} x_b  +  4^{-b-1}  x_{b -1} = \lambda \, x_{2b -2} 
$$
This sets the path for iteratively determining the  additional $x_i, i=b+1, \dots, 2b$.  
Hope someone else can help for a complete solution. 
A: I will give only a bound for the eigenvalues based on the 1-norm $\|-\|_1$ of $M$
defined by
$$
\|M\|_1=\sup_{\|x\|>0}\frac{\|M(x)\|_1}{\|x\|_1}=\max_{1\le j\le n}\sum_{i=1}^n|M_{ij}|
$$ 
We have 
$$
\sum_{i=1}^n|M_{ij}|= \sum_{i=a}^{b}\frac{1}{2^{2j}}\binom{2j}{i}\le \sum_{i=0}^{2j}\frac{1}{2^{2j}}\binom{2j}{i}=1
$$
and if $a>0$, then the inequality is strict, and if $a=0$, then we have equality.
Moreover, the absolute values of eigenvalues are bounded by $|\lambda|\le \|M\|_1$, since
$$
|\lambda|\|x\|_1=\|Mx\|_1\le \|M\|_1\|x\|_1.
$$
So in general we have for an eigenvalue $\lambda$:
$$
|\lambda|\le \|M\|_1=\max_{a\le j\le b}\sum_{i=a}^{b}\frac{1}{2^{2j}}\binom{2j}{i}.
$$
For big $a,b$ the binomial distribution given by $\frac{1}{2^{2j}}\binom{2j}{i}$ can be approximated by the Gauss normal curve with expected value $j$ and variation $\sigma^2=j/2$, and so 
$$
\|M\|_1\approx \max_{t\in[a,b]}\int_{a}^b \frac{1}{\sqrt{\pi t}}e^{-\frac{(s-t)^2}{t}}ds.
$$
This explains (or at least does not contradict) the fact that 
for big $a$, $b$, the largest eigenvalue tends to 1, since for constant ratio
$a/b$, the value of the integral for $t=(a+b)/2$ is near 1, since the interval $[a,b]$ contains several lengths of the standard variation $\sigma=\sqrt{t/2}$.
