How to invert $\infty \times \infty$ matrix? I have an equation $AX=B$, where $A$ is $\infty \times \infty$ matrix, $X$ is $\infty \times 1$ vector and $B$ is $\infty \times 1$ vector.
$A$ and $B$ are known and I need to determine $X$.
For this, I think that I should calculate inverse of $A$ (if it exists) and obtain $X$ as $A^{-1}B$.
However, I know almost nothing about inverses of infinite matrices, and I do not know when they exist and how to calculate them.
I am an amateur even when it comes to finite matrices, but, the problem that I am trying to solve led me to this equation with infinite matrices and vectors, and, I do not know what to do?
I was thinking that I could find inverse for $A_n$, where $\lim_{n \to \infty}A_n=A$, and then pass to a limit, but I do not know is the limit of inverses an inverse of the limit.
What to do?
 A: I would first look whether a triangular decomposition of $A$ is helpful. For instance by the well known LDU-decomposition. This procedure finds
 $$ L \cdot D \cdot U = A$$ 
where $L$ is lower triangular ("row-finite"), $D$ is diagonal, $U$ is upper triangular ("column-finite") and moreover, $L$ and $U$ are normed to have diagonal-entries of $1$. 
Of course, to invert $D$ we need that no diagonal-entry is zero. Let us assume this for the example computation.
With this you can invert the triangular/diagonal matrices to, say, $K=L^{-1}$, $E=D^{-1}$ and $T=U^{-1}$ to as many rows/columns you want.
After that, you can easily compute
$$ \small \begin{array} {} A\cdot X &= (LDU)\cdot X &= B &&\text{the formal decomposition}\\
 K \cdot(LDU)X &= DU \cdot X &= K\cdot B && \text {computable because $K$ is rowfinite }\\
 E \cdot(DU)X &= U \cdot X &= EK\cdot B && \text {computable because $E$ is diagonal }\\
 T \cdot(U)X &=  X & \underset{\text{}problem!}= T \overset{\color{red}{???}}\cdot (EK B) && \text {in RHS is the problem because $T$ not rowfinite! }\\
\end{array}$$
In the last row in the RHS occurs that basical problem of matrix-inversion of infinite size in your example-constellation.      


*

*Sometimes it happens, that $(EKB)$ is actually column-finite, then you are fine and can compute $X=T\cdot(EKB)$ to $(LDU)^{-1}\cdot B = A^{-1}B = X$ as solution.       

*If this is not column-finite you must try for each row-column-dotproduct of $T \cdot (EKB)$ whether you have an analytically determinable expression evaluatable to a finite value.      

*Again with a bit of luck that dotproducts are either convergent or at least Cesaro-, Euler- or Borel-summable and you can have finite entries for $X$.       

*If nothing works, there is no computable solution.


For an example where it luckily came ot to be doable, see my essay on "is the inverse of this infinite matrix the Null-matrix?" A final solution for the still open problem in my essay see here an answer in MathOverflow.
