Find $\lim_{k\rightarrow \infty} A^k x$ Let $\displaystyle A=\left(\begin{matrix}.5&1\\0&.75\end{matrix}\right),x=\left(\begin{matrix}-2\\1\end{matrix}\right)$.

*

*Find $\displaystyle \lim_{k\rightarrow \infty} A^k x$


*Find $\displaystyle \lim_{k\rightarrow \infty} \frac 1 {\Vert A^k x\Vert}A^k x$
I found the eigenvalues and eigenvectors to be $\displaystyle \lambda_1=.5, v_1=\left(\begin{matrix}1\\0\end{matrix}\right),\lambda_2=.75,v_2=\left(\begin{matrix}4\\1\end{matrix}\right)$.
x can be expressed as $x=-6v_1+v_2$, then $\displaystyle Ax=-6(.5)\left(\begin{matrix}1\\0\end{matrix}\right)+.75\left(\begin{matrix}4\\1\end{matrix}\right)=\left(\begin{matrix}-6(.5)+.75(4)\\.75\end{matrix}\right)$
Then $\displaystyle \lim_{k\rightarrow \infty} A^k x=\lim_{k\rightarrow \infty} \left(\begin{matrix}-6(.5)^k+(.75)^k(4)\\(.75)^k\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$ ?
And what is $\displaystyle \lim_{k\rightarrow \infty} \frac 1 {\Vert A^k x\Vert}A^k x=\lim_{k\rightarrow \infty} \left(\begin{matrix}\frac{-6(.5)^k+(.75)^k(4)}{\sqrt{(-6(.5)^k+(.75)^k(4))^2+((.75)^k)^2}}\\\frac{(.75)^k}{\sqrt{(-6(.5)^k+(.75)^k(4))^2+((.75)^k)^2}}\end{matrix}\right)=\left(\begin{matrix}\infty\\\infty\end{matrix}\right)$?
 A: Since you already did the first part, I will rewrite the second part in terms of the eigenvector decomposition
$$\lim_{k\to\infty} \frac{\left(\frac{1}{2}\right)^k\cdot(-6v_1)+\left(\frac{3}{4}\right)^k\cdot(v_2)}{\left|\left(\frac{1}{2}\right)^k\cdot(-6v_1)+\left(\frac{3}{4}\right)^k\cdot(v_2)\right|} \equiv \lim_{k\to\infty}\frac{av+bw}{|av+bw|}$$
The two vectors are not orthogonal, and the magnitude of the sum of non orthogonal vectors is given by
$$|av+bw| = \sqrt{(av+bw)\cdot(av+bw)} = \sqrt{a^2|v|^2+2abv\cdot w + b^2|w|^2}$$
Plugging in the vectors we have, this simplifies to
$$\sqrt{a^2+8ab+17b^2} = |b|\sqrt{\left(\frac{a}{b}\right)^2+ 8\frac{a}{b}+17}$$
Now why did we pull out $b$ instead of $a$ when the same formula could have applied either way? It's because $\frac{a}{b}\to 0$, thus we can take the limit without a L'Hopital situatuon occurring. Meaning the limit simplifies to
$$\lim_{k\to\infty} \frac{\frac{a}{b}v +  w}{\sqrt{\left(\frac{a}{b}\right)^2+ 8\frac{a}{b}+17}} \to \frac{w}{\sqrt{17}} = \frac{1}{\sqrt{17}}\begin{pmatrix} 4 \\ 1 \end{pmatrix}$$
because $b>0$. In general you can see that the second limit will approach the unit eigenvector with the largest (by absolute value) eigenvalue.
