Let be a formal power series over with radius of convergence . Let be the matrix norm introduced in this previous exercise.
Given a matrix over , we can construct a sequence of matrices by taking the th partial sum of . That is, . This gives us sequences where is the entry of . Suppose converges to for each , and let . In this situation, we say that converges to , and that . (In other words, converges precisely when each entrywise sequence converges.)
- Prove that if , then converges.
- Deduce that for all matrices , the following power series converge: , , and .
[Disclaimer: My facility with even simple analytical concepts is laughably bad, but I’m going to give this a shot. Please let me know what’s wrong with this solution.]
We begin with a lemma.
Lemma: For all , , where the subscripts denote taking the entry. Proof: By the definition of matrix multiplication, we have , where and . (I’m abusing the notation a bit here; is an element of , which we think of as a function on .) Now by the triangle inequality. Note that is the sum of all possible -fold products of (absolute values of) entries of , and that we have bounded above by a sum of some distinct -fold products of (absolute values of) entries of . In particular, since the missing terms are all positive. Thus .
Let us define the formal power series by . What we have shown (I think) is that , using the triangle inequality.
Now recall, by the Cauchy-Hadamard theorem, that the radius of convergence of satisfies . In particular, has the same radius of convergence as . So in fact the sequence converges. Now the sequence is bounded and monotone increasing, and so must also converge. That is, is absolutely convergent, and so is convergent. Thus converges.
Now we will address the convergence of the series . Recall that the radius of convergence of satisfies . Now if is even and if is odd. So if is even and if is odd.
Brief aside: Suppose . Certainly , so that . Now , so that . In particular, if , then .
By our brief aside, we have that if is odd, and if is even. So (skipping redundant terms) we have satisfies . We know from the Brief Aside that this sequence is monotone decreasing. Now let . Consider now the sequence , , , et cetera. only finitely many terms of this sequence are less than or equal to 1. So there must exist some such that the product is greater than 1. Now , so that . Thus the sequence converges to 0, and so the radius of convergence of is .
By a similar argument, the radii of convergence for and are also .