Non-linearly Independent Result in Calculating Generalized Eigenvectors

Question

I am trying to find a basis for the generalized eigenvectors of the operator whose matrix is $T=\begin{bmatrix}2&1&0&4\\0&2&-1&0\\0&0&1&1\\0&0&0&3\end{bmatrix}$, following the method on Wikipedia for generalized eigenvectors. From what I've gathered so far, the eigenvalues are 2 multiplicity 2, 1 multiplicity 1, and 3 multiplicity 1. The eigenvectors for these eigenvalues are spanned by the respective vectors: $v_1=\begin{pmatrix}1\\0\\0\\0\end{pmatrix},v_2\begin{pmatrix}1\\-1\\-1\\0\end{pmatrix},v_3=\begin{pmatrix}7\\-1\\1\\2\end{pmatrix}$. Since 2 has multiplicity 2, it should have generalized eigenvectors spanned by some $v_4$, found by solving $Tv_4=\begin{bmatrix}2&1&0&4\\0&2&-1&0\\0&0&1&1\\0&0&0&3\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=v_1$ where $v_4=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$. When I do this, I seem to get $v_4=\begin{bmatrix}\frac{1}{2}\\0\\0\\0\end{bmatrix}$, since $3d=0$ so $d=0$, $c+d=c=0$, $2b-c=2b=0$ so $b=0$ which isn't linearly independent from $v_1$. Where is my mistake in using this method?

Answer 1

when everything involved is integers, I like reversing and finding the generalized eigenvector(s) first; here the minimal polynomial is equal to the characteristic. An actual eigenvector comes from $(T-2I)v = 0.$ To make things consistent for the Jordan form, We choose column vector $w$ with $(T-2I)w \neq 0$ but $(T-2I)^2w = 0.$ We might as well take $w = (0,1,0,0)^t.$ The corresponding eigenvector is $ u =(T-2I)w \neq 0,$ we get $u = (1,0,0,0)^t.$

$$ R = \left( \begin{array}{rrrr} 1&0&-1&7\\ 0&1&1&-1\\ 0&0&1&1\\ 0&0&0&2\\ \end{array} \right) $$

gives $R^{-1}TR = J $ in Jordan form.

Answer 2

I see what I did, I was solving $Tv_4=v_1$ instead of $(T-2I)v_4=v_1$. Solving $(T-2I)v_4=v_1$ yields $\begin{pmatrix}a\\1\\0\\0\end{pmatrix}$, or $\begin{pmatrix}0\\1\\0\\0\end{pmatrix}$ for $a=0$.