I'm reviewing diagonalization and am wondering if the following makes sense. Let $A \in \mathcal{M}_{n \times n}(\mathbb{R})$ be a diagonalizable matrix. That is, there exist matrices $D$ and $P$ such that
$$ A = PDP^{-1} $$
where the columns of $P$ are linearly independent eigenvectors of $A$ and the $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $A$ (repeated based on their respective multiplicities). Does it follow that
$$ \lim_{k \to \infty} A^k = 0 $$
if each eigenvalue of $A$ is in the range $(-1, 1)$? My reasoning is that (from elementary linear algebra) we can show that
$$ A = PD^kP^{-1} $$
if $A$ is diagonalizable. Since $P$ and $P^{-1}$ are finite, the product should approach zero since each (diagonal) entry of $D$ will approach zero since
$$ \lim_{k \to \infty} \lambda^k = 0 $$
if $-1 < \lambda < 1$. Does this make sense or is my reasoning flawed?
Taking into account that all the operations (sums and products of real numbers) involved in the product $PD^kP^{-1}$ are continuous, you get
$$ \mathrm{lim}_{k\rightarrow \infty}\ A^k = P\cdot \mathrm{lim}_{k\rightarrow \infty}\ D^k\cdot P^{-1} \ . $$
Then, for
$$ D = \begin{pmatrix} \lambda_1 & 0 & \dots & 0 \\ 0 & \lambda_2 & \dots & 0 \\ \dots \\ 0 & 0 & \dots & \lambda_n \end{pmatrix} $$
you obviously also have
$$ \mathrm{lim}_{k\rightarrow \infty}\ D^k = \mathrm{lim}_{k\rightarrow \infty}\ \begin{pmatrix} \lambda_1^k & 0 & \dots & 0 \\ 0 & \lambda_2^k & \dots & 0 \\ \dots \\ 0 & 0 & \dots & \lambda_n^k \end{pmatrix} = \begin{pmatrix} \mathrm{lim}_{k\rightarrow \infty}\ \lambda_1^k & 0 & \dots & 0 \\ 0 & \mathrm{lim}_{k\rightarrow \infty}\ \lambda_2^k & \dots & 0 \\ \dots \\ 0 & 0 & \dots & \mathrm{lim}_{k\rightarrow \infty}\ \lambda_n^k \end{pmatrix} $$
from which your conclusion follows: if for all $\lambda_i$ you have $|\lambda_i | < 1$, then $\mathrm{lim}_{k\rightarrow \infty}\ \lambda_i^k = 0$, hence $\mathrm{lim}_{k\rightarrow \infty}\ D^k = 0$, hence $\mathrm{lim}_{k\rightarrow \infty}\ A^k = P \cdot 0 \cdot P^{-1} = 0$.
Remark. To be picky, that "obviously" over there means something like you are considering the product topology in the space of matrices ${\cal M}_{m\times n}(\mathbb{R}) = \mathbb{R}^{m\times n}$ and then the continuity of a function of matrices is checked component-wise. So the limit of a function of matrices is the limit of everyone of its components. A similar remark applies for the first statement about the continuity of $PD^kP^{-1}$.