Is the following linear operator surjective? Define for $\lambda \neq 0$ the linear operator
$$T_\lambda: l^2(\mathbb{N}) \to l^2(\mathbb{N}): (a_0,a_1, \dots) \mapsto (\lambda a_0-a_1/2, \lambda a_1 - a_2/4, \lambda a_2- a_3/8,\dots )$$
In my notes, it is claimed that this is a bijective linear map.
I was able to show injectivity and that it is a linear map, but I can't show surjectivity.
Given $(b_0, b_1, \dots) \in l ^2$, I want to construct $(a_0, a_1, \dots) \in l^2$ with $T_\lambda (a_0, a_1, \dots) = (b_0,b_1, \dots)$. I tried to solve this "system" but could not succeed.
Any help is appreciated!
 A: Suppose $a_0=c$. Then we can express each next $a_n$ from an infinite recurrence system sequentially. We obtain regularity $a_n=2^{\frac{n^2+n}{2}}\lambda^nc-\displaystyle\sum\limits_{k=1}^{n}b_{n-k}\lambda^{k-1}2^{\frac{k}{2}(2n-k+1)}$. Let's make a replacement $l=n-k$ in the sum: $$a_n=2^{\frac{n^2+n}{2}}\lambda^nc-\displaystyle\sum\limits_{l=0}^{n-1}b_{l}\lambda^{n-l-1}2^{\frac{n-l}{2}(n+l+1)}=2^{\frac{n^2+n}{2}}\lambda^nc-\lambda^{n-1}2^{\frac{n^2+n}{2}}\displaystyle\sum\limits_{l=0}^{n-1}b_{l}\lambda^{-l}2^{\frac{-l^2-l}{2}}.$$ From here $\dfrac{a_n}{2^{\frac{n^2+n}{2}}\lambda^{n-1}}=c\lambda-\displaystyle\sum\limits_{l=0}^{n-1}b_{l}\lambda^{-l}2^{\frac{-l^2-l}{2}}$. Left hand side tends to $0$, so the right hand side are the same. Therefore $c\lambda=\displaystyle\sum\limits_{l=0}^{\infty}b_{l}\lambda^{-l}2^{\frac{-l^2-l}{2}}$, and $\dfrac{a_n}{2^{\frac{n^2+n}{2}}\lambda^{n-1}}=\displaystyle\sum\limits_{l=n}^{\infty}b_{l}\lambda^{-l}2^{\frac{-l^2-l}{2}}$. From here we obtain $a_n$. It remains only to show that $(a_n)\in l_2$. Can you do it yourself?
