On the convergence of formal power series of matrices

Let G(x) = \sum_{k \in \mathbb{N}} \alpha_kx^k be a formal power series over \mathbb{C} with radius of convergence R. Let ||A|| = \sum_{i,j} |a_{i,j}| be the matrix norm introduced in this previous exercise.

Given a matrix A over \mathbb{C}, we can construct a sequence of matrices by taking the Nth partial sum of G(A). That is, G_N(A) = \sum_{k = 0}^N a_kA^k. This gives us n^2 sequences \{c_{i,j}^N\} where c_{i,j}^N is the (i,j) entry of G_N(A). Suppose c_{i,j}^N converges to c_{i,j} for each (i,j), and let C = [c_{i,j}]. In this situation, we say that G_N(A) converges to C, and that G(A) = C. (In other words, G_N(A) converges precisely when each entrywise sequence G_N(A)_{i,j} converges.)

  1. Prove that if ||A|| \leq R, then G_N(A) converges.
  2. Deduce that for all matrices A, the following power series converge: \mathsf{sin}(A) = \sum_{k \in \mathbb{N}} \dfrac{(-1)^k}{(2k+1)!}A^{2k+1}, \mathsf{cos}(A) = \sum_{k \in \mathbb{N}} \dfrac{(-1)^k}{(2k)!}A^{2k}, and \mathsf{exp}(A) = \sum_{k \in \mathbb{N}} \dfrac{1}{k!} A^k.

[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 (i,j), (A^k)_{i,j} \leq ||A||^k, where the subscripts denote taking the (i,j) entry. Proof: By the definition of matrix multiplication, we have A^k = [\sum_{t \in n^{k-1}} \prod_{i=0}^{k-1} a_{t(i), t(i+1)}], where t(0) = i and t(k+1) = j. (I’m abusing the notation a bit here; t is an element of \{1,\ldots,n\}^{k-1}, which we think of as a function on \{1,\ldots,k-1\}.) Now |A^k_{i,j}| \leq \sum_{t \in n^{k-1}} \prod_{i=0}^{k-1} |a_{t(i), t(i+1)}| by the triangle inequality. Note that ||A||^k is the sum of all possible k-fold products of (absolute values of) entries of A, and that we have |A^k_{i,j}| bounded above by a sum of some distinct k-fold products of (absolute values of) entries of A. In particular, |A^k_{i,j}| \leq ||A||^k since the missing terms are all positive. Thus |\alpha_k A^k_{i,j}| = |(\alpha_k A^k)_{i,j}| \leq |\alpha_k| ||A||^k.

Let us define the formal power series |G|(x) by |G|(x) = \sum |\alpha_k| x^k. What we have shown (I think) is that \sum_{k=0}^N |\alpha_k A^k_{i,j}| \leq |G|_N(||A||), using the triangle inequality.

Now recall, by the Cauchy-Hadamard theorem, that the radius of convergence of G(x) satisfies 1/R = \mathsf{lim\ sup}_{k \rightarrow \infty} |\alpha_k|^{1/k}. In particular, |G| has the same radius of convergence as G. So in fact the sequence |G|_N(||A||) converges. Now the sequence \sum_{k=0}^N |\alpha_k A^k_{i,j}| is bounded and monotone increasing, and so must also converge. That is, \sum_{k=0}^N (\alpha_k A^k)_{i,j} = G_N(A)_{i,j} is absolutely convergent, and so is convergent. Thus G(A) converges.

Now we will address the convergence of the series \mathsf{sin}. Recall that the radius of convergence R of \mathsf{sin} satisfies R^{-1} = \mathsf{lim\ sup}_{k \rightarrow \infty} |\alpha_k|^{1/k} \mathsf{lim}_{k \rightarrow \infty} \mathsf{sup}_{n \geq k} |\alpha_k|^{1/k}. Now \alpha_k = 0 if k is even and (-1)^t/k! if k = 2t+1 is odd. So |\alpha_k|^{1/k} = 0 if k is even and |1/k!|^{1/k} = 1/(k!)^{1/k} > 0 if k is odd.

Brief aside: Suppose k \geq 1. Certainly 1 \leq k! < (k+1)^k, so that (k!)^{1/k} < k+1. Now 1 \leq (k!)^{1 + 1/k} < (k+1)!, so that (k!)^{1/k} < ((k+1)!)^{1/(k+1)}. In particular, if n > k, then (k!)^{1/k} < (n!)^{1/n}.

By our brief aside, we have that \mathsf{sup}_{n \geq k} |\alpha_k|^{1/k} = 1/(k!)^{1/k} if k is odd, and 1/((k+1)!)^{1/(k+1)} if k is even. So (skipping redundant terms) we have \mathsf{sin}(x) satisfies R^{-1} = \mathsf{lim}_{k \rightarrow \infty} 1/(k!)^{1/k}. We know from the Brief Aside that this sequence is monotone decreasing. Now let \varepsilon > 0. Consider now the sequence \varepsilon, 2 \varepsilon, 3 \varepsilon, et cetera. only finitely many terms of this sequence are less than or equal to 1. So there must exist some k such that the product \prod_{i=1}^k \varepsilon i is greater than 1. Now \varepsilon^k k! > 1, so that \varepsilon > 1/(k!)^{1/k}. Thus the sequence 1/(k!)^{1/k} converges to 0, and so the radius of convergence of \mathsf{sin}(x) is \infty.

By a similar argument, the radii of convergence for \mathsf{cos}(X) and \mathsf{exp}(X) are also \infty.

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