# convergent infinite matrix product

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 $$m,n\longrightarrow\infty$$?

• Why the down vote? Commented Feb 21, 2020 at 1:47

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 $$0\le \prod_{k=m}^n(1 + \|A_k\|) -1\le \exp(\sum_{k=m}^n \|A_k\|) -1$$

• I was able to proceed similarly assuming invertibility. Essentially my problem boils down to: how do we ensure that those terms are eventually invertible? I do know I could proceed easily applying Gershgorin if I knew that the $U_i$ are eventually invertible. However I could not prove that they are, solely on the given assumptions... so would you mind expanding on that please?
– xyz
Commented Feb 21, 2020 at 12:40
• I see, I guess my feeling that additional hypotheses are needed for convergent matrix infinite products to be Cauchy could be correct after all. Let us say that for now we know that invertibility of the $U_i$ is sufficient. As soon as I have some spare time I'll investigate if it is necessary as well. At this stage it is somehow clear that it is connected to the fact that for matrix norm $\Vert AB\Vert=0$ does not imply $\Vert A\Vert=0$ or $\Vert B\Vert=0$, a property that would enable the result without having to assume invertibility.
– xyz
Commented Feb 21, 2020 at 13:17
• @Giordano Giambartolomei: I also used that $\prod U_n$ converges to an invertible matrix ( otherwise the remaining products do not converge to $I$ apriori...). So it's good to know what exactly they mean by convergent product... whether is necessarily has an invertible limit... Commented Feb 21, 2020 at 13:21
• Yes of course I forgot to mention it: the limit need not be non-singular in general, so we need also that assumption, since it does not follow from the invertibility of all $U_i$. Usually a convergent infinite product is simply a well defined non zero matrix.
– xyz
Commented Feb 21, 2020 at 13:30