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 $I$.
Here is the way to go: consider the sequence of kernels:
$$\{\,0\,\}\varsubsetneq\ker(A-2I)\varsubsetneq\ker(A-2I)^2\subset\dots$$
The sequence stops after step $2$ since
$$A-2I=\begin{bmatrix}-2&1&0\\-4&2&0\\-2&1&0\end{bmatrix}\qquad (A-2I)^2=\begin{bmatrix}
0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
$A-2I$ has rank $1$, hence its kernel (the eigenspace) has codimension $1$, i.e. has dimension $2$.
$(A-2I)^2$ is the null matrix, hence its kernel has dimension $3$. Take any vector in $\ker(A-2I)^2\smallsetminus\ker(A-2I)$, i.e. any vector of $\mathbf R^3$ which is not an eigenvector. As the eigenspace is defined by the equation $\; y=2x$, we'll take, say
$$e_3=(0,1,0). $$
Note $e'_2=(A-2I)e'_3=(1,2,1),\;$ is an eigenvector by construction. We complete this set of two vectors to a basis, by choosing another eigenvector, linearly independent from $e'_2$, say
$$e'_1=(1,2,0).$$
The definition of $e'_2$ from $e'_3$ can be written as $\; Ae'_3=2e'_3+e'_2$, so the matrix of the linear map in basis $(e'_1,e'_2,e'_3)$ is the Jordan form:
$$J=\begin{bmatrix}2&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$
Best Answer
If the minimal polynomial is $(\lambda -5)^2,$ you may, indeed, pick any column vector $w$ you like that is not already an eigenvector, make it the right hand column of $P. \; $ The rule is that the middle column must be $v = (A-5I) w.$ The left column of $P$ is then $u,$ a different eigenvector from $v.$ Sometimes it takes a little ingenuity to take the pair of eigenvectors you first calculated and revise to get $u.$
Then, with matrix $P,$ you get $J = P^{-1}AP.$
Try it.