Finding Jordan Basis of a matrix Having trouble finding the Jordan base (and hence $P$) for this matrix 
$A = \begin{pmatrix}
15&-4\\ 49&-13
\end{pmatrix}$
I know that the eigenvalue is $1$, this gives an eigenvector $\begin{pmatrix}
2\\ 7
\end{pmatrix} $
Now to create the Jordan basis and find $P$ (of which its columns will consist of the two basis vectors) I'm aware that I need to find $v_1$ s.t $(A-I)v_1 = 0$, so $v_1 = \begin{pmatrix}
2\\ 7
\end{pmatrix}$. Now to find $v_2$ I need to  do $ker(A-I)^2$ but $(A-I)^2 = 0$ so any non zero vector is in the kernel right$?$ So why doesn't it work if I choose $v_2 = \begin{pmatrix}
1\\ 0
\end{pmatrix}$ $?$ 

Let
$$P=\left[ \begin{array}{cc}2 & a\\ 7 & b\end{array} \right]$$
You are meant to solve the equation $AP = PJ$, with 
$$J=\left[ \begin{array}{cc}1 & 1\\ 0 & 1\end{array} \right].$$
$J$ has the previous form since its only eigenvalues $\lambda=1$ has an eigenspace of dimension $1$.
Expanding $AP = PJ$, you get the following non-trivial equations:
$$\left\{ \begin{array}{rcl}15a - 4b & = & a+2\\49a-13b & = & b+7\end{array} \right.$$
$$\left\{ \begin{array}{rcl}14a - 4b & = & 2\\49a-14b & = & 7\end{array} \right.$$
These equations are linearly dependent, so you can choose to solve the first one:
$$\left\{ \begin{array}{rcl}a & = & \frac{1+2k}{7}\\b & = & k\end{array} \right.$$
If you fix $k$, say $k=0$, then
$$P=\left[ \begin{array}{cc}2 & \frac{1}{7}\\ 7 & 0\end{array} \right]$$
Note that $$det(P(k))=\left[ \begin{array}{cc}2 & \frac{1+2k}{7}\\ 7 & k\end{array} \right] = -1 \neq 0 ~\forall k,$$ so the choice of $k$ is arbitrary.
