How to find Jordan basis of a matrix

Question

Assume matrix

$$A= \begin{bmatrix} -1&0&0&0&0\\ -1&1&-2&0&1\\ -1&0&-1&0&1\\ 0&1&-1&1&0\\ 0&0&0&0&-1 \end{bmatrix} $$

Its Jordan Canonical Form is $$J= \begin{bmatrix} -1&1&0&0&0\\ 0&-1&0&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&1\\ 0&0&0&0&1 \end{bmatrix} $$

I am trying to find a nonsingular $P$, let $P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}$ s.t. $J=P^{-1}AP\Leftrightarrow AP=PJ$. I came up with the Wikipedia article on JCF and I think I need to find the generalized eigenvectors so that $AP=PJ=\begin{bmatrix}-\mathbf{p_1}&\mathbf{p_1}-\mathbf{p_2}&-\mathbf{p_3}&\mathbf{p_4}&\mathbf{p_4}+\mathbf{p_5}\end{bmatrix}$ yielding the systems $$(A+I)\mathbf{p_1}=\mathbf{0}$$ $$(A+I)^2\mathbf{p_2}=\mathbf{0}$$ $$(A+I)\mathbf{p_3}=\mathbf{0}$$ $$(A-I)\mathbf{p_4}=\mathbf{0}$$ $$(A-I)^2\mathbf{p_5}=\mathbf{0}$$

I solved each of these systems making sure that the vectors $\mathbf{p_i}$ I chose are linearly independent. So I chose $$P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}=\begin{bmatrix}1&2&-2&0&0\\1&1&2&0&1\\1&1&2&0&0\\0&0&0&1&1\\1&1&-2&0&0\end{bmatrix}$$ which even though is nonsingular I am not getting $AP=PJ$.

What am I doing wrong?

Answer 1

Note that the condition $AP=PJ$ is equivalent to

• $Ap_1=-p_1 \to p_1$
• $Ap_2=p_1-p_2\to p_2$
• $Ap_3=-p_3\to p_3$
• $Ap_4=p_4 \to p_4$
• $Ap_5=p_4+p_5 \to p_5$

Since the set up is equivalent, from you results seems that there is something wrong in the calculation indeed $Ap_1\neq p_1-p_2$.

Notably from

• $Ap_1=-p_1 \implies (A+I)p_1=0$
• $Ap_2=p_1-p_2\implies (A+I)p_2=p_1$

we obtain

• $p_1=(0,1,1,0,0)$
• $p_2=(-1,0,0,0,0)$

from

• $Ap_3=-p_3 \implies (A+I)p_3=0$

excluding $p_1$ we obtain

• $p_3=(1,0,0,0,1)$

and from

• $Ap_4=p_4 \implies (A-I)p_4=0$

• $Ap_5=p_4+p_5 \implies (A-I)p_5=p_4$

we obtain

• $p_4=(0,0,0,-1,0)$
• $p_5=(0,-1,0,0,0)$

Answer 2

I got the Jordan blocks in slightly different order.

What you seem to be missing is the consistency part: in my

$$ P = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \end{array} \right) $$

we have a place where we do have $(A+I)^2 p_3 = 0,$ but we have consistency in that $p_2 =(A+I)p_3 .$ It follows automatically that $(A+I)p_2 = (A+I)^2 p_3 = 0.$

We also have $(A-I)^2 p_5 = 0,$ then $p_4 =(A-I)p_5.$ As a result $(A-I) p_4 = 0.$

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

To get the Jordan form in the order they report, use my columns but permuted, $p_2p_3p_1p_4p_5$ and then correct $P^{-1}.$ We can correct $P^{-1}$ by permuting the rows to 23145.