# Thread: [SOLVED] infinite products of matrices

1. ## [SOLVED] infinite products of matrices

hey all,

the basic fact about infinite products of complex numbers is that $\displaystyle \prod_{j=1}^\infty a_j$ is convergent if and only if $\displaystyle \sum_{j=1}^\infty |1-a_j|$ converges. does anyone know if something of this kind holds for matrix-valued case?

I hope for $\displaystyle \prod_{j=1}^\infty A_j$ converges if $\displaystyle \sum_{j=1}^\infty ||A_j-I||<\infty$.

an easier related statement, which looks very true (but I still don't know solution yet) is to prove $\displaystyle \prod_{j=1}^\infty e^{B_j}$ converges if $\displaystyle \sum_{j=1}^\infty ||B_j||<\infty$ (here $\displaystyle B_j$'s are not assumed to commute, of course)

if anyone saw this matter addressed in any book or science paper, I'd appreciate the reference

2. Originally Posted by choovuck
I hope for $\displaystyle \prod_{j=1}^\infty A_j$ converges if $\displaystyle \sum_{j=1}^\infty ||A_j-I||<\infty$.
I prefer to write I+A_j in place of A_j, and to assume that $\displaystyle M \mathrel{\mathop=^{\text{d}\text{ef}}} \sum_{j=1}^\infty ||A_j||<\infty$.

Let $\displaystyle T_n = \prod_{j=1}^n (I+A_j)$. Then $\displaystyle \|T_n\|\leqslant \prod_{j=1}^n \|I+A_j\| \leqslant \prod_{j=1}^\infty (1+\|A_j\|) \leqslant e^M,$ by the usual estimate for infinite products.

Also, if m < n then $\displaystyle \|T_m-T_n\| = \left\|T_m\left(I-\prod_{j=m+1}^n (I+A_j)\right)\right\| \leqslant e^M\left\|I-\prod_{j=m+1}^n (I+A_j)\right\|.$ But $\displaystyle \left\|I-\prod_{j=m+1}^n (I+A_j)\right\|$ should lie between $\displaystyle 1-e^{-K}$ and $\displaystyle e^K-1$, where $\displaystyle K = \sum_{j=m+1}^n\|A_j\|.$ That makes it look to me as though (T_n) should be a Cauchy sequence, which would therefore converge. Of course, you won't want the infinite product to diverge to 0, so you would also need to investigate whether $\displaystyle \|T_n\|$ is bounded away from 0 (presumably by $\displaystyle e^{-M}$).

I don't pretend to have thought through this carefully, so I can't be sure that there is any mileage in this approach. Maybe you have already tried it. But it looks reasonably promising to me.

3. this seems to be working, thanks!