An infinite matrix $[a_{ij}]_{i,j\in\mathbb{N}}$ is called invertible, if for any convergent sequence $(y_m)$ there exists exactly one sequence $(x_m)$ such that $y_m=\sum_{n\ge 1}a_{mn}x_n$ for all $m\in\mathbb{N}$.
Find two infinite invertible matrices $A,B$ such that $AB$ is not invertible.
What about these two?
$A=\left[\begin{array}{cccccc} 1&1&1&1&1&\cdots\\ 0&1&0&0&0&\cdots\\ 0&0&1&0&0&\cdots\\ \vdots & \vdots & \vdots &\vdots & \vdots & \ddots \end{array}\right]$
and
$B=\left[\begin{array}{cccccc} 1&0&0&0&0&\cdots\\ -1&1&0&0&0&\cdots\\ 0&-1&1&0&0&\cdots\\ 0&0&-1&1&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array}\right]$
Then the entire first row of $AB$ is zeros so it cannot be invertible. But I believe both $A$ and $B$ are invertible. If you write down the equations from the first one we have $x_n=y_n$ for all $n>1$ and $x_1=y_1-\sum_{n\geq2}y_n$. And from the second one $x_1=y_1$, $x_2=y_2+y_1$, $x_3=y_3+y_2+y_1$, etc. So both $A$ and $B$ are invertible.