I want to find the Jordan Normal Form of the matrix $$M=\begin{pmatrix} 19 & -7 & -1 & 5 & -7 \\ 3 & 9 & -1 & 3 & -5 \\ -10 & 14 & 16 & -10 & 14 \\ 0 & 0 & 0 & 14 & 0 \\ 2 & -2 & 0 & 2 & 12 \end{pmatrix}.$$ I am given that it has the following pair of rank two generalised eigenvectors: $$ \mathbf x_2=\begin{pmatrix} 1\\-1\\0\\-2\\0 \end{pmatrix} \qquad \text{and} \qquad \mathbf y_2=\begin{pmatrix} 2\\0\\1\\0\\1 \end{pmatrix},$$ and that its characteristic polynomial is $(x-14)^5$.
The fact that they are rank two means they generate two others, namely $\mathbf x_1 = ( M-14I)\mathbf x_2$ and $\mathbf y_1 = (M-14I)\mathbf y_2$. So if I define the matrix $P = (\mathbf x_1 \mid \mathbf x_2\mid\mathbf y_1\mid\mathbf y_2\mid\mathbf z)$, then we have that $$P^{-1}AP = \begin{pmatrix} 14 & 1\\ & 14 \\ & & 14 & 1\\ & & & 14 \\ &&&&14 \end{pmatrix},$$ but I still need to find the vector $\mathbf z$ for this to work. Is there some efficient way I can compute it, given what I know already? or do I have to go through the whole usual computation of finding eigenvectors?
I appreciate any help with this.