Inverse operator in the space of infinite bounded sequences $l_2$

Question

Define operator $A$ as $Ax=(\lambda x_1+x_3,\lambda x_2,\lambda x_3,...\lambda x_n,...)$. How do I find the inverse operator $A^{-1}$ for $\lambda ≠ 0$?

I'm trying to find the operator $A^{-1}$ such that $(A \circ A^{-1})x=x$. But I'm not sure what $A\circ A^{-1}$ means.

Answer 1

We compose operators as functions. So, if we want to find $A^{-1}$ so that $(A\circ A^{-1})(x)=x,$ this is equivalent to wanting $A^{-1}$ to satisfy $A(A^{-1}(x))=x.$ Spoiler below:

Define $A^{-1}$ by $$A^{-1} x=(\lambda^{-1} x_1-\lambda^{-2}x_3, \lambda^{-1} x_2,\lambda^{-1} x_3,\cdots,\lambda^{-1}x_n,\cdots ).$$ Then, if we apply $A$ to this, we get $$A(A^{-1} x)=(\lambda(\lambda^{-1} x_1-\lambda^{-2} x_3)-\lambda^{-1} x_3, \lambda (\lambda^{-1} x_2), \lambda(\lambda^{-1} x_3),\cdots, \lambda(\lambda^{-1}x_n),\cdots )=x.$$ You can see this by noting that the only "non-trivial" portion to invert is the 1st and 3rd component. This can be written as a matrix $$\begin{bmatrix}\lambda & 1\\ 0 & \lambda \end{bmatrix},$$ which has inverse $$\begin{bmatrix}\lambda^{-1} & -\lambda^{-2}\\ 0 & \lambda^{-1} \end{bmatrix}.$$