# solution to infinite matrix equation

• Feb 10th 2013, 08:40 AM
hatsoff
solution to infinite matrix equation
Let $\displaystyle b_{i,j}\in\mathbb{C}$, and suppose that for each $\displaystyle i$ we have

$\displaystyle \sum_{j=1}^\infty|b_{i,j}|<\infty$ and $\displaystyle \sum_{j=1}^\infty|b_{i,j}|\leq\sum_{j=1}^\infty|b_ {i+1,j}|$.

I seek to determine whether a solution $\displaystyle X$ to the equation $\displaystyle AX=B$ exists, where $\displaystyle A,B,X$ are infinite matrices and $\displaystyle a_{i,j}=b_{i+1,j}$. In other words, I seek to show that there exists some $\displaystyle X=(x_{i,j})$ with complex entries satisfying the following equation:

$\displaystyle \begin{bmatrix}b_{2,1}&b_{2,2}&b_{2,3}&\cdots\\b_{ 3,1}&b_{3,2}&b_{3,3}&\cdots\\b_{4,1}&b_{4,2}&b_{4, 3}&\cdots\\\vdots&\vdots&\vdots&\end{bmatrix}$$\displaystyle \begin{bmatrix}x_{1,1}&x_{1,2}&x_{1,3}&\cdots\\x_{ 2,1}&x_{2,2}&x_{2,3}&\cdots\\x_{3,1}&x_{3,2}&x_{3, 3}&\cdots\\\vdots&\vdots&\vdots&\end{bmatrix}$$\displaystyle =\begin{bmatrix}b_{1,1}&b_{1,2}&b_{1,3}&\cdots\\b_ {2,1}&b_{2,2}&b_{2,3}&\cdots\\b_{3,1}&b_{3,2}&b_{3 ,3}&\cdots\\\vdots&\vdots&\vdots&\end{bmatrix}$

What tools do I use to deal with a problem like this? Clearly, if $\displaystyle A$ is invertible then $\displaystyle X=A^{-1}B$ exists. But how do I show that an infinite matrix is invertible? If $\displaystyle A$ is not invertible, do I have other options?

Any help would be much appreciated, thanks!