Sequence of matrices and convergence There is question in Matrix analysis by Horn (p. 300) that:
Let $f(x) = \sum_{k=0}^{\infty} a_{k} x^{k}$ with radius of convergence $R>0$. Then, $f(A)$ converges for all $A \in M_{n}$ such that $\rho(A)<R$, where $\rho(A)$ is the spectral radius of $A$.
I would like to prove it. Would you give me a hint?
 A: There exists $R_1 \in (0,R)$ such that $\rho (A) <R_1$. By spectral radius formula $||A^{n}|| <R_1^{n}$ for $n$ sufficiently large. Hence $\sum ||a_n A^{n}||$ is dominated by the series $\sum a_n R_1^{n}$ which is convergent since $R_1$ is less than the radius of convergence.
A: A useful result is that for any $\epsilon>0$ there is a submultiplicative matrix norm $\|\cdot\|_*$ such that $\|A\|_* < \rho(A)+\epsilon$.
All norms are equivalent on finite dimensional normed spaces, so it is sufficient to prove convergence with $\|\cdot\|_*$.
Let $\epsilon>0$ be such that $\rho(A)+\epsilon < R$, and $\|\cdot\|_*$  the
corresponding norm.
Then $\sum_k |a_k| \|A\|_*^k \le \sum_k |a_k| (\rho(A)+\epsilon)^k < \infty$.
In particular, the series $\sum_k a_k A^k$ is Cauchy and hence converges.
Aside:
To prove the result stated above, suppose $J= V^{-1} AV$ is the Jordan form of $A$ and let $R = \operatorname{diag} (r_1,...,r_n)$. Then
$[RJR^{-1}]_{ij} = {r_i \over r_j} [J]_{ij}$. Note that the diagonal elements of $RJR^{-1}$ are the same as those of $J$ and the off diagonal elements are of the form ${r_{j-1} \over r_j}$, for $j=2,...,n$. 
In particular, we can choose $r_n=1$ and the subsequent $r_k$ so that the
numbers ${r_{n-1} \over r_n}, ..., {r_1 \over r_2}$are arbitrarily close to zero.
Now note that $\|J\|_2 = \rho(A)$ and that the norm is a continuous function.
In particular, for any $\epsilon>0$ we can choose $R$ such that
$\|A\|_* = \| R V^{-1} A V R^{-1}\|_2 < \rho(A)+\epsilon$.
Now check that $\|B\|_* = \| R V^{-1} B V R^{-1}\|_2$ is indeed a submultiplicative norm.
