How to order the basis vectors that put a matrix in jordan canonical form

Question

I have the following matrix A \begin{pmatrix} 1 & -3 & -1\\ 1 & 5 & 1 \\ -2 & -6 & 0 \end{pmatrix} I want to find a basis that puts the matrix in Jordan canonical form. The matrix A has eigenvalue 2 with an algebraic multiplicity of 3 and a geometric multiplicity of 2. I choose the first two vectors in my basis to be two eigenvectors, \begin{pmatrix} 1 \\ 0\\-1\end{pmatrix} \begin{pmatrix} 0 \\ 1\\-3\end{pmatrix} and find the last basis vector by solving the equation $(A-2I)^{2}\textbf{x}=0$, and take $\textbf{x}$ to be the follwing vector \begin{pmatrix} 0 \\ 0\\1\end{pmatrix} My problem is that when I construct the Matrix P as \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ -1 & -3 & 1 \end{pmatrix} and calculate the Jordan Canonical form using J=$P^{-1}AP$ I get that J is \begin{pmatrix} 2 & 0 & -1\\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}. Why is there a -1 in the first row? And does the order I put my basis vectors in the matrix P matter? Thank you!

Answer 1

no. when everything is integers, I do this: the minimal polynomial is $(x-2)^2.$ The genuine eigenvectors make up a subspace. The, I choose a column vector $w$ for which $(A-2I)^2 w = 0,$ that one is automatic here, but $(A-2I) w \neq 0.$ Then name $v = (A-2I) w.$ This $v$ is a genuine eigenvector. Finally, pick a genuine eigenvector $u$ that is independent from $v.$ The change of basis matrix $P$ will have columns $u,v,w.$ Our $P$ will have integer elements, but its determinant may not be $\pm 1,$ so $P^{-1} = \frac{1}{\det P} \tilde{P} $ where $ \tilde{P} $ is another matrix of integers. Finally $$ \frac{1}{\det P} \tilde{P} A P = J $$ is in Jordan form

When possible, I like $w = (0,0,1)^T$ (a column) and that is possible here.

Answer 2

To have 1 at the right place in J, we must find U, V, W such that

AU=2U, AV=2V and AW=V+2W

i.e. (A-2I)U=0, (A-2I)V=0 and (A-2I)W=V

Without taking care of U, we must find W non-zero in $N((A-2I)^2)$ such that $V := (A-2I)W$ is non-zero (V will automatically be in $N(A-2I)$)

Example: I choose $W=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ such that $(A-2I)^2 W=0$

and I define

$V:=(A-2I)W=\begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}$