Finding the limit of a matrix and showing if it exists For the matrix
\[
A=\frac{1}{10}\cdot \begin{pmatrix} 
1 & 7 \\
7 & 1 
\end{pmatrix}\]
I try to find the limit of $A^n$ when $n$ goes to infinity.
 A: I would suspect that you're meant to take the limit of $$\begin{pmatrix}.1 & .7\\.7 & .1\end{pmatrix}^n$$ as $n\to\infty.$ One way to do this is to diagonalize the matrix $$A=\begin{pmatrix}.1 & .7\\.7 & .1\end{pmatrix},$$ that is, to rewrite as $A=PDP^{-1},$ where $P$ is some invertible $2\times 2$ matrix and $D$ has the form $$D=\begin{pmatrix}d_1 & 0\\0 & d_2\end{pmatrix}.$$ It can be shown by induction that $$A^n=PD^nP^{-1}=P\begin{pmatrix}d_1^n & 0\\0 & d_2^n\end{pmatrix}P^{-1}$$ for all $n\ge 1.$ Since matrix multiplication is continuous, then, $$\lim_{n\to\infty}A^n=P\left[\lim_{n\to\infty}D^n\right]P^{-1}.$$
A: Assuming you are good with guessing eigenvalues there is a a similar way which in this case is kind of more easy.
Because of
\[ \frac{1}{10}\begin{pmatrix} 1 & 7 \\ 7 & 1 \end{pmatrix} \cdot 
\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{4}{5} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix}\]
and 
\[ \frac{1}{10}\begin{pmatrix} 1 & 7 \\ 7 & 1 \end{pmatrix} \cdot
\begin{pmatrix} 1 \\ -1 \end{pmatrix} = -\frac{3}{5} \cdot \begin{pmatrix} 1 \\ -1\end{pmatrix}\] 
So instead of calculating the limit of $A^n$, I look at the limit of $A^n \cdot v$ for any $v$. 
As the geometric sequence $q^n$ converges to zero iff $|q|<1$ we see that here $A^n\cdot v$ converges
to zero for any $v$. Hence $A^n$ must converge to zero. 
