I am trying to find Jordan's form of this matrix: \begin{pmatrix} 0 & 1 & 0\\ -4& 4& 0\\ -2 & 1& 2 \end{pmatrix} The only eigenvalue $r$ is 2 and therefore the simplest eigenvector $v_{1}$ is (0, 0, 1) To get the other two independent vectors (generalized eigenvectors) for the P matrix we need to compute $(A-rI)v_{2}=v_{1}$ and $(A-rI)v_{3}=v_{2}$ .
However, in the calculation for 2 I get the following system which has no solution:
$$ \left[ \begin{array}{ccc|c} -2 & 0& 0&0 \\ 0& 1& 0&0 \\ 0& 0& 0 & 1 \end{array} \right] $$
Any Idea on what I am doing wrong? If the eigenvalue is 2, when we substract it the lowest row will always contain all zeros...
I like this method for hand calculations: first, calling your matrix $A,$ let $$ B = A - 2 I $$ $$ B = \left( \begin{array}{ccc} -2&1&0 \\ -4&2&0 \\ -2&1&0 \end{array} \right) $$ A basis for the genuine eigenvectors is given by the convenient $$ E = \left( \begin{array}{cc} 1&0 \\ 2&0 \\ 0&1 \end{array} \right) $$ We may or may not be using these eigenvectors in the form shown. We know that the characteristic equation for $A$ is just showing $B^3 = 0.$ However, the minimal polynomial for $A$ is $B^2 = 0,$ which you can check easily enough.
We are going to make a matrix $R$ with columns $u,v,w;$ on the far right, we take $w$ as any vector with $B^2 w = 0$ (automatic) but $Bw \neq 0.$ Then $v = Bw$ will be a genuine eigenvector. Finally, we will choose an independent eigenvalue $u.$
I like ones and zeros, I choose $$ w = \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) $$ Then from $v = B w$ we get $$ v = \left( \begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right) $$
This $v$ is a genuine eigenvector, it is the sum of the two columns of my $E.$ At last, we get to choose some $u$ eigenvector that is not a multiple of $v,$ I choose $$ u = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) $$
$$ R = \left( \begin{array}{ccc} 0&1&0 \\ 0&2&1 \\ 1&1&0 \end{array} \right) $$ Next we find $R^{-1} $ and $J = R^{-1}A R,$ which will be the Jordan form if we did it correctly. A piece of luck , because of choosing ones and zeros, the determinant of $R$ is small, actually $1,$ and we calculate
$$ R^{-1} = \left( \begin{array}{ccc} -1&0&1 \\ 1&0&0 \\ -2&1&0 \end{array} \right) $$ and
$$ J = R^{-1} A R = \left( \begin{array}{ccc} 2&0&0 \\ 0&2&1 \\ 0&0&2 \end{array} \right) $$ $$ $$ $$ R^{-1} A R = J $$ $$ \left( \begin{array}{ccc} -1&0&1 \\ 1&0&0 \\ -2&1&0 \end{array} \right) \left( \begin{array}{ccc} 0&1&0 \\ -4&4&0 \\ -2&1&2 \end{array} \right) \left( \begin{array}{ccc} 0&1&0 \\ 0&2&1 \\ 1&1&0 \end{array} \right) = \left( \begin{array}{ccc} 2&0&0 \\ 0&2&1 \\ 0&0&2 \end{array} \right) $$ $$ $$ $$ R J R^{-1} = A $$ $$ \left( \begin{array}{ccc} 0&1&0 \\ 0&2&1 \\ 1&1&0 \end{array} \right) \left( \begin{array}{ccc} 2&0&0 \\ 0&2&1 \\ 0&0&2 \end{array} \right) \left( \begin{array}{ccc} -1&0&1 \\ 1&0&0 \\ -2&1&0 \end{array} \right) = \left( \begin{array}{ccc} 0&1&0 \\ -4&4&0 \\ -2&1&2 \end{array} \right) $$
COMMENT: the line above "as any vector with $B^2 w = 0$ (automatic)" may appear silly. If, however, you were given a 5 by 5 matrix $A$ with characteristic polynomial $(x - 5)^3 (x-7)^2$ and minimal polynomial $(x - 5)^2 (x-7),$ the demand for a vector $w$ with $(A - 5I)^2 w = 0$ but $(A - 5I) w \neq 0$ would make some sense, as the eigenvalue $7$ is ignored in this condition.