Consider the operator $T:\ell^\infty({\mathbb N})\to\ell^\infty({\mathbb N})$ defined by $$ (Tx)_m=\sum_{k=m+1}^\infty p_{k,m} \ \ x_k, $$ where $$ p_{k,m}=\frac k{(k-1)(k-m)(k-m+1)}. $$ Then $T$ is a bounded operator of norm $\zeta(2)=\frac{\pi^2}6$ as an easy calculation shows. I need to know the dimension of the eigenspace to the eigenvalue $1$, i.e. $$ E=\{x\in \ell^\infty: Tx=x\}. $$ Ideally, I would like to have $\dim(E)=1$, however, I already don't know whether this space is finite-dimensional. So my question is, what is the dimension of the space $E$?