Facts about power series of matrices

Let G(x) = \sum_{k \in \mathbb{N}} \alpha_k x^k be a power series over \mathbb{C} with radius of convergence R. Let A be an n \times n matrix over \mathbb{C}, and let P be a nonsingular matrix. Prove the following.

  1. If G(A) converges, then G(P^{-1}AP) converges, and G(P^{-1}AP) = P^{-1}G(A)P.
  2. If A = B \oplus C and G(A) converges, then G(B) and G(C) converge and G(B \oplus C) = G(B) \oplus G(C).
  3. If D is a diagonal matrix with diagonal entries d_i, then G(D) converges, G(d_i) converges for each d_i, and G(D) is diagonal with diagonal entries G(d_i).

Suppose G(A) converges. Then (by definition) the sequence of matrices G_N(A) converges entrywise. Let G_N(A) = [a_{i,j}^N], P = [p_{i,j}], and P^{-1} = [q_{i,j}]. Now G_N(P^{-1}AP) = P^{-1}G_N(A)P = [\sum_\ell \sum_k q_{i,k} a_{k,\ell}^Np_{\ell,j}]. That is, the (i,j) entry of G_N(P^{-1}AP) is \sum_\ell \sum_k q_{i,k} a_{k,\ell}^Np_{\ell,j}. Since each sequence a_{k,\ell}^N converges, this sum converges as well. In particular, G(P^{-1}AP) converges (again by definition). Now since G_N(P^{-1}AP) = P^{-1}G_N(A)P for each N, the corresponding sequences for each (i,j) are equal for each term, and so have the same limit. Thus G(P^{-1}AP) = P^{-1}G(A)P.

Now suppose A = B \oplus C. We have G_N(B \oplus C) = \sum_{k=0}^N \alpha_k (B \oplus C)^k = \sum_{k=0}^N \alpha_k B^k \oplus C^k = (\sum_{k = 0}^N \alpha_k B^k) \oplus (\sum_{k=0}^N \alpha_k C^k) = G_N(B) \oplus G_N(C). Since G_N(A) converges in each entry, each of G_N(B) and G_N(C) converge in each entry. So G(B) and G(C) converge. Again, because for each (i,j) the corresponding sequences G_N(A)_{i,j} and (G_N(B) \oplus G_N(C))_{i,j} are the same, they converge to the same limit, and thus G(B \oplus C) = G(B) \oplus G(C).

Finally, suppose D is diagonal. Then in fact we have D = \bigoplus_{t=1}^n [d_t], and so by the previous part, G(D) = \bigoplus_{t=1}^n G(d_t). In particular, G(d_t) converges, and G(D) is diagonal with diagonal entries G(d_t) as desired.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: