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 $v_2 = \begin{pmatrix}
1\\ 0
\end{pmatrix}$ $?$ 
 A: 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.
A: To start, a brief digression from the specific case at hand:  suppose $B$ is any $2 \times 2$ matrix with a single, repeated eigenvalue $\lambda$; then we know there exists at least one vector $v_1 \ne 0$ such that $Bv_1 = \lambda v_1$.  If in addition there existed $v_2 \ne 0$, linearly independent from $v_1$ with $Bv_2 = \lambda v_2$, then for any vector $v = av_1 + bv_2$ we would have $Bv = aBv_1 + bBv_2 = a\lambda v_1 + b\lambda v_2 = \lambda v$, which shows that $B = \lambda I$, where $I$ is the $2 \times 2$ identity matrix.  We thus conclude that if $B$ is not of this form, there is at most a one-dimensional subspace of vectors $\alpha v_1$ such that $Bv_1 = \lambda v_1$.  Furthermore, we have $(B - \lambda I)^2 = 0$, so that for any vector $v$, $(B - \lambda I)(B - \lambda I)v =(B - \lambda I)^2 v =  0$; if we choose $v_2$ linearly independent of $v_1$, then by what we have seen $(B - \lambda I)v_2 \ne 0$, but $(B - \lambda I)(B - \lambda I)v_2 = 0$; this implies that we must have $(B - \lambda I)v_2 = \alpha v_1$ for some $\alpha$, so by linearity we can in fact take $(B - \lambda I)v_2 = v_1$; $(B - \lambda I)v_2$ is in fact an eigenvector of $B$, with eigenvalue $\lambda$.  $v_2$ is called a generalized eigenvector corresponding to eigenvalue $\lambda$; note that $Bv_2 = \lambda v_2 + v_1$; this terminology is of course well-known.
Now in such a situation if we form the matrix $E$ such that
$E = \begin{bmatrix} v_1 & v_2 \end{bmatrix}, \tag{1}$
i.e., the columns of $E$ are $v_1, v_2$ then it is clear that
$BE = \begin{bmatrix} Bv_1 & Bv_2 \end{bmatrix} = \begin{bmatrix} \lambda v_1 & \lambda v_2 + v_1 \end{bmatrix}. \tag{2}$
Now $E^{-1}$ exists by the linear independence of $v_1, v_2$, hence we have
$\begin{bmatrix} E^{-1}v_1 & E^{-1}v_2 \end{bmatrix} = E^{-1} \begin{bmatrix} v_1 & v_2 \end{bmatrix} = E^{-1} E = I, \tag{3}$
which shows that
$E^{-1}v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \tag{4}$
and
$E^{-1}v_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}; \tag{5}$
therefore
$E^{-1}BE  = \begin{bmatrix} \lambda E^{-1} v_1 & \lambda  E^{-1}v_2 +  E^{-1} v_1 \end{bmatrix} = \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}, \tag{6}$
which is the Jordan canonical form of $B$.
We can use the conclusions reached in the preceding discussion to show how to correctly find the Jordan canonical from of the matrix
$A = \begin{bmatrix} 15 & -4 \\ 49 & -13 \end{bmatrix}, \tag{7}$
which as we know has a single eigenvalue $\lambda = 1$ of multiplicity $2$.  We observe that
$A - \lambda I = A - I = \begin{bmatrix} 14 & -4 \\ 49 & -14 \end{bmatrix} \ne 0, \tag{8}$
which, according to the above, implies that $A$ has a one-dimensional eigenspace for it's single eigenvalue $1$.  As has been shown, we can take a non-zero vector in this eigenspace to be $v_1 = (2, 7)^T$:
$\begin{bmatrix} 15 & -4 \\ 49 & -13 \end{bmatrix} \begin{pmatrix} 2 \\ 7 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \end{pmatrix}. \tag{9}$
At this point, instead of using $(A - I)^2 = 0$ and choosing $v_2 \in \ker (A - I)^2$
arbitrarily, we need to solve
$(A - I)v_2 = v_1 \tag{10}$
or
$\begin{bmatrix} 14 & -4 \\ 49 & -14 \end{bmatrix} v_2 = \begin{pmatrix} 2 \\ 7 \end{pmatrix}; \tag{11}$
a solution is
$v_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}, \tag{12}$
but it is worth noting that $v_2 + \alpha v_1$ is also a solution for any $\alpha$, 
since $v_1 \in \ker (A - I)$; this fact explains the apparent discrepancy between the_candyman's answer, which effectively gives
$\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} \frac{2k + 1}{7} \\ k \end{pmatrix} \tag{13}$
for the possible generalized eigenvectors, whereas the present analysis yields
$v_2 + \alpha v_1 = \begin{pmatrix} 2 \alpha + 1 \\ 7 \alpha + 3 \end{pmatrix}; \tag{14}$
taking $\alpha = \frac{1}{7} (k -3)$ shows these two sets are the same.  The vector
$(1, 0)^T$ is not of this form; there is no $\alpha$ such that $(1, 0)^T = v_2 + \alpha v_1$.  In any event, we may take for our matrix $E$
$E = \begin{bmatrix} 2 & 2 \alpha + 1 \\ 7 & 7\alpha + 3 \end{bmatrix}, \tag{15}$
and we easily see that $\det (E) = -1$, in accord with the_candyman's result.  The columns of $E$ are therefore linearly independent for all $\alpha$, though this was already apparent from the independence of $v_1$ and $v_2$; being non-singular, $E$ is invertible and we may take its inverse, thus:
$E^{-1} = -\begin{bmatrix} 7 \alpha + 3 & -2 \alpha - 1 \\ -7 & 2 \end{bmatrix}; \tag{16}$
taking $E^{-1}AE$ will then yield
$E^{-1}AE = \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}, \tag{17}$
in accord with equation (6).
The key thing in the above is that we need to find the generalized eigenvector corresponding to $\lambda$ in the event that the matrix in question is not a scalar multiple of the identity matrix $I$.
