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 $E$?