convergent infinite matrix product

Question

For an infinite matrix product $\prod_{i=1}^\infty (U_i+A_i)$ a result states that if for some consistent norm (basically sub-multiplicative) $\Vert U_i \Vert =1$, $\prod_{i=r}^\infty U_i$ coverges for every $r$ and $\sum_i \Vert A_i \Vert <\infty$ then the product converges. The question is: would it also apply a matrix analogue of the Cauchy property that usually holds for convergent infinite products, that is $\prod_{i=m}^n (U_i+A_i)\longrightarrow I$ as $m,n\longrightarrow\infty$?

Answer 1

Yes, basically you can show the convergence of sequence in a complete topological group by showing it is Cauchy.

Details:

You may assume that all the terms $(U_i + A_i)$ are invertible (they certainly are for $i$ large enough).

Now consider $X_n=\prod_{i=1}^n(U_i + A_i)$. You can show that $$\|X_m^{-1} X_{n}-I\|\to 0$$ as $n < m$ approach $\infty$ (check below) Now the group of invertible matrices is locally compact and so complete ( but we can show the completeness directly). Therefore $X_n$ is convergent to an invertible matrix.

Check: $$\|\prod_{i=m}^n (U_k+A_k)-I\|\le \|\prod_{i=m}^n U_k-I\| + \prod_{k=m}^n(1 + \|A_k\|) -1$$ and $$0\le \prod_{k=m}^n(1 + \|A_k\|) -1\le \exp(\sum_{k=m}^n \|A_k\|) -1$$