9
$\begingroup$

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.

$\endgroup$
5
  • 3
    $\begingroup$ check out arxiv.org/abs/math/9902004 and arxiv.org/abs/math/0503507 : most probably a more general problem is already solved there. $\endgroup$ – Dima Pasechnik Jul 20 '17 at 21:07
  • $\begingroup$ @DimaPasechnik Thanks. It looks like he computes a lot of determinants, but does not do much about eigenvalues ... $\endgroup$ – becko Jul 20 '17 at 22:30
  • $\begingroup$ also mathoverflow.net/questions/258448/… $\endgroup$ – Pietro Majer Jul 21 '17 at 21:30
  • $\begingroup$ @DimaPasechnik Great reference, though. Thanks for pointing it out. $\endgroup$ – becko Jul 24 '17 at 14:25
  • $\begingroup$ I meant that perhaps a useful transformation may be found there to simplify the problem at hand. $\endgroup$ – Dima Pasechnik Jul 24 '17 at 14:56
7
$\begingroup$

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 $S_5$.

$\endgroup$
3
  • $\begingroup$ Yes, the eigenvalues seem to be real for most numerical tests I have done. Even if they are not real, maybe a close approximation can be found for lage $a,b$. How did you get this polynomial? Do you have a tractable formula for the coefficients of the characteristic polynomial? $\endgroup$ – becko Jul 20 '17 at 22:32
  • $\begingroup$ I used the CharacteristicPolynomial function in Maple, and checked the number of real roots using the sturm function. $\endgroup$ – Robert Israel Jul 20 '17 at 22:51
  • $\begingroup$ Oh, in my previous comment, I meant to say: "Even if there is no exact analytic result, maybe a close approximation can be found for large $a,b$." I just had the "real" eigenvalues in my mind when I wrote that. $\endgroup$ – becko Jul 20 '17 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.