there is matrix $ B=\left(\begin{array}{ccc}1 & -1 \\
1 & 3\\
\end{array}\right)$ ,
for vector $ v_1=\left(\begin{array}{ccc}1 \\
-1\\
\end{array}\right), B v_1=2v_1$ holds.
using
$v_1$ and $v_2=\left(\begin{array}{ccc}a \\
b\\
\end{array}\right) $ which is a linear independent vector of magnitude 1, we can express $B.v_2$ as linear combination of $v_1$ & $v_2$ such as $B.v_2=\alpha v_1+2v_2$.
a.) find $v_2$ & $\alpha$
b.) $P=\left(\begin{array}{ccc}1 & a \\
-1 & b\\
\end{array}\right)$, then we can express $BP=PC$, find matrix C, and find $B^n$
first try to find $v_2$ & $\alpha$, $(B-2E)v_2=v_1$
$ B=[\left(\begin{array}{ccc}1 & -1 \\
1 & 3\\
\end{array}\right)$ - $ \left(\begin{array}{ccc}2 & 0 \\
0 & 2\\
\end{array}\right) ] .v_2 =\left(\begin{array}{ccc}1 \\
-1\\
\end{array}\right)$
got $v_2=\left(\begin{array}{ccc}-b-1 \\
b
\end{array}\right)$ = $\left(\begin{array}{ccc}-1 \\
1\\
\end{array}\right)s + \left(\begin{array}{ccc}-1 \\
0\\
\end{array}\right)t$,
since $v_2$ is vector with magnitude 1 , i chose $ \left(\begin{array}{ccc}-1 \\
0\\
\end{array}\right)$ with $\alpha =1$, is my answer right?
then find matrix $C$ , i know matrix $C$ is matrix diagonal, $P^{-1}BP=C$, so with $P=\left(\begin{array}{ccc}1 & -1\\
-1 & 0\\
\end{array}\right)$ , after some calculation $ C=\left(\begin{array}{ccc}2 & 0 \\
1 & 2\\
\end{array}\right)$
But i wasnt so sure to find $B^n$ ?