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.