Jordan canonical form with more than 1 eigenvector

Question

I need to solve: $$ v'=\left(\begin{array}{ccc} 5 & -1 & -2\\ -4 & 5 & 4\\ 4 & -2 & -1 \end{array}\right)v $$ After finding the characteristic polynomial and finding the eigenvectors I found
$$ p(A)=(x-3)^3,ker(A-3I)=span\left\{ \left(\begin{array}{c} 1\\ 2\\ 0 \end{array}\right),\left(\begin{array}{c} 1\\ 0\\ 1 \end{array}\right)\right\} $$ Now I want to find the Jordan chains and I'm struggling a bit. $$ \left.\begin{array}{ccc} 2 & -1 & -2\\ -4 & 2 & 4\\ 4 & -2 & -4 \end{array}\right|\begin{array}{c} \alpha+\beta\\ 2\alpha\\ \beta \end{array}\xrightarrow[2R_{1}-R_{3}\to R_{3}]{\substack{2R_{1}+R_{2}\to R_{2}} }\left.\begin{array}{ccc} 2 & -1 & -2\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right|\begin{array}{c} \alpha+\beta\\ 4\alpha+2\beta\\ 2\alpha+\beta \end{array}\Rightarrow\beta=-2\alpha $$
After that I took $ \alpha =1,\beta =-2 $ and tried continuing the chain with
$$\left.\begin{array}{ccc} 2 & -1 & -2\\ -4 & 2 & 4\\ 4 & -2 & -4 \end{array}\right|\begin{array}{c} -1\\ 2\\ -2 \end{array}\Rightarrow2x-y-2z=-1\Rightarrow\left(\begin{array}{c} x\\ y\\ z \end{array}\right)=\left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right) $$ After doing this, I cant keep my chain as the 3rd vector I need has no solution: $$ \left.\begin{array}{ccc} 2 & -1 & -2\\ -4 & 2 & 4\\ 4 & -2 & -4 \end{array}\right|\begin{array}{c} 0\\ 1\\ 0 \end{array}\Rightarrow \text{no solution} $$ How do I fix it and reach the Jordan form? I know how to make it work when I have only 1 eigenvector but I can't find examples for this case.

Answer 1

you don't need a third vector, you have:
$$ker(A-3I)=span\left\{ \left(\begin{array}{c} 1\\ 2\\ 0 \end{array}\right),\left(\begin{array}{c} 1\\ 0\\ 1 \end{array}\right)\right\}=span\left\{ \left(\begin{array}{c} -1\\ 2\\ -2 \end{array}\right),\left(\begin{array}{c} 1\\ 0\\ 1 \end{array}\right)\right\}$$ so we can have $$A=\begin{pmatrix} 1&-1&0\\ 0&2&1\\ 1&-2&0 \end{pmatrix} \begin{pmatrix} 3&0&0\\ 0&3&1\\ 0&0&3 \end{pmatrix} \begin{pmatrix} 1&-1&0\\ 0&2&1\\ 1&-2&0 \end{pmatrix}^{-1}$$ so, for your differential equation $$e^{tA}=\begin{pmatrix} 1&-1&0\\ 0&2&1\\ 1&-2&0 \end{pmatrix} e^{t\begin{pmatrix} 3&0&0\\ 0&3&1\\ 0&0&3 \end{pmatrix}} \begin{pmatrix} 1&-1&0\\ 0&2&1\\ 1&-2&0 \end{pmatrix}^{-1} =\begin{pmatrix} 1&-1&0\\ 0&2&1\\ 1&-2&0 \end{pmatrix} {\begin{pmatrix} e^{3t}&0&0\\ 0&e^{3t}&e^{3t}t\\ 0&0&e^{3t} \end{pmatrix}} \begin{pmatrix} 1&-1&0\\ 0&2&1\\ 1&-2&0 \end{pmatrix}^{-1} = \begin{pmatrix} e^{3t}(2t+1)&-e^{3t}t&-2e^{3t}t\\ -4e^{3t}t&e^{3t}(2t+1)&4e^{3t}t\\ 4e^{3t}t&-2e^{3t}t&e^{3t}(1-4t) \end{pmatrix}$$