Let $a_n=2^{-n}$. What are the eigenvalues and eigenvectors of the $\ell^2$ operator represented by the infinite matrix below? $$A=\begin{pmatrix} a_1 & a_2 & a_3 & \dots \\ a_2 & a_3 & \dots \\ a_3 & \dots \\ \dots \end{pmatrix}$$
The relation is $(Av)_i=\sum_j\frac{\alpha_j}{2^{i+j-1}}=\lambda \alpha_i$ where $\alpha_i= \left\langle v,e_i \right\rangle $ and $e_i$ are the standard basis, but it looks to be a horrible mess. Maybe some nice decomposition of $A$ resolves everything? Perhaps the fact each row is half the row above it?
Attempt. Since each row is half the row above it, $(Av)_j=\frac 1{2^{j-1}}(Av)_1=\frac \lambda{2^{j-1}}v_1$, so $Av\propto(1,2^{-1},2^{-2},\dots)$. If $\lambda$ is to be an eigenvector we must have $\sum_n \frac\lambda{2^n}<\infty$, which always holds... What...
Suppose $Av=\lambda v$. Then as you noted we have $(Av)_j=\frac 1{2^{j-1}}(Av)_1=\frac \lambda{2^{j-1}}v_1$ but also $(Av)_j=\lambda v_j$. Thus we have that $\lambda v_j=\frac \lambda{2^{j-1}}v_1$ so that $v_j=\frac{v_1}{2^{j-1}}$. For $v_1$, we have that $\lambda v_1=(Av)_1=\sum_{n=1}^\infty\frac{v_1}{2^{n-1}2^n}$ which implies that $\lambda=2\sum_{n=1}^\infty\frac{1}{4^n}=\frac{2}{3}$.