T= 2x2 matrix with inputs [5/4 -1/4] and [3/4 1/4] across. Set xn=T^n (1,0) for n >=1. Let $T= \begin{bmatrix}\frac{5}{4} &  -\frac{1}{4} \\ \frac{3}{4} & \frac{1}{4}\end{bmatrix}$. Set $x_n=T^n \begin{pmatrix}1 \\ 0\end{pmatrix}$ for $n\geq 1$. Need to prove that $(x_n)$ converges and find the limit $y$. Also have to find an explicit $N$ so that $\left\|x_n-y\right\| < (1/2)10^{-100}$ for all $n \geq N$. I am not sure how to solve this question at all. I know that to find an explicit $N$, we can use induction but I don't know how to go about doing that.
 A: This diagonalizes to
$$ T = SJS^{-1} = \pmatrix{\frac{1}{3} & 1 \\ 1 & 1}\pmatrix{\frac{1}{2} & 0 \\ 0 & 1}\pmatrix{-\frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & -\frac{1}{2}} .$$
Noting that $T^n = SJ^nS^{-1}$ we then have 
$$x_n = \pmatrix{\frac{1}{3} & 1 \\ 1 & 1}\pmatrix{(\frac{1}{2})^n & 0 \\ 0 & 1}\pmatrix{-\frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & -\frac{1}{2}}\pmatrix{1 \\ 0}.$$
Multiplying $x_n$ out we have 
$$ x_n = \pmatrix{-2^{-(n + 1)} + \frac{3}{2} \\ -3\cdot2^{-(n+1)} + \frac{3}{2}}. $$
Thus, noting that $\lim_{n \to \infty} x_n = \pmatrix{\frac{3}{2} \\ \frac{3}{2}}$, we know
$$|| x_n - p || = || \pmatrix{-2^{-(n + 1)} & -3\cdot2^{-(n+1)}}^T || = 2^{-(n - 1)} < \frac{1}{2}10^{-100},$$
assuming that the norm is just the max of the sum of the absolute values of each column. If you are using a different definition of the norm, then change as required. 
Using this we now have 
$$2^{-(n - 1)} < \frac{1}{2}10^{-100} \iff \ln{2^{-n + 2}} < \ln{10^{-100}} \iff -n < \frac{\ln{10^{-100}}}{\ln 2} - 2 \iff n > -\frac{\ln{10^{-100}}}{\ln 2} + 2.$$
