$\lim_{k \to \infty} A^k$ where $A$ is diagonalizable 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?
 A: 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 $PD^kP^{-1}$.
