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I need your help for understanding how to compute jordan base for this matrix:

$\begin{pmatrix} 2 &1 &2 \\ -1 &0 &2 \\ 0&0 & 1 \end{pmatrix}$


This is what I tried to do:

I found the Minimal polynomial: $(x-1)^{3}$,

so I know that the Jordan normal form is: $\begin{pmatrix} 1 &0 &0 \\ 1& 1 & 0\\ 0&1 & 1 \end{pmatrix}$

Now I remember that I need to find: $\ker (A-I), \ker (A-I)^{2}, \ker(A-I)^{3}$.

I found that $\ker (A-I)$ is: $\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix}$ and $\ker (A-I)^{2}$ is: $\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}$ , $\begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}$

and obivously $\ker(A-I)^{3}$ is all $\mathbb{R}^{3}$, What should I do from here? I can't remember the whole algorithem, and I can't find it anywhere.

Thank you, Have a great week-end.

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You can choose a vector from $\ker(A-I)^{3}$ which is not in $\ker(A-I)^{2}$.

$\vec a=(0,0,1)^T$

$\vec b=(A-I)\vec a=A\vec a-\vec a=(2,2,0)^T$ belongs to $\ker(A-I)^{2}$ but not to $\ker(A-I)$.

$\vec c=(A-I)\vec b=(4,-4,0)$

The vectors $\vec c$, $\vec b$, $\vec a$ form a base in which the linear transformation $A$ has form $J$.

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Two side notes:

  • The minimal polynomial (EDIT: I mean characteristic polynomial here - as was pointed out in comments) being $(x-1)^3$ does not determine the Jordan form uniquely. Only after you find out that the eigenspace has dimension 1 you know, that there is only one Jordan block.

  • I have more often seen different form with 1's above the diagonal.

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  • $\begingroup$ @Martin: Thank you for the answer. If I found that the minimal polynomial is $(x-1)^3$, I know for sure that I must have a jordan block of size 3, so it leaves me with no other choices, right? $\endgroup$ – user6163 Apr 16 '11 at 12:36
  • $\begingroup$ @Nir: you are right. @Martin: you might have mixed up the minimal polynomial with the characteristic polynomial. $\endgroup$ – Fabian Apr 16 '11 at 12:38
  • $\begingroup$ Why should I even need to compute $\ker (A-I)$? We don't use it anyway. $\endgroup$ – user6163 Apr 16 '11 at 12:43
  • $\begingroup$ @Nir: you don't need to compute it. But knowing that it has dimension 1 is equivalent to showing that $(x-1)^3$ is the minimal polynomial (how did you show that?). $\endgroup$ – Fabian Apr 16 '11 at 12:45
  • $\begingroup$ I simply showed that $(x-1)^{2}$ or $(x-1)$ is not 0. $\endgroup$ – user6163 Apr 16 '11 at 12:49

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