# Finding Jordan basis of a matrix (4x4)

I'm facing a problem finding a Jordan basis for this ($4 \times 4$) matrix: $$\left(\begin{matrix}3&-1&1&7\\9&-3&-7&-1\\0&0&4&-8\\0&0&2&-4\end{matrix}\right)$$

I know that the characteristic polynomial is $\lambda^4=0$ and it's minimal polynomial is: $x^2$. This means that $(A-0I)^2=0$. Also the Jordan form for this matrix is $$\left(\begin{matrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{matrix}\right)$$ But I'm facing problems in the general procedure to compute the basis for such a matrix?

• You can find one linearly independent eigenvector and three generalized eigenvectors. Look up "chaining" + eigenvectors. For example, ms.uky.edu/~lee/amspekulin/jordan_canonical_form.pdf – Moo Jan 5 '16 at 19:07
• Very useful and clear document. Was very helpful! – Misha Jan 5 '16 at 20:48
• @Misha take also a look here math.stackexchange.com/questions/2557595/… – user Dec 9 '17 at 9:44
• @Misha Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Jan 24 '18 at 21:51

Since:

$J$ is obtained by $A$ from a similarity transfomation $$P^{-1}AP=J$$

$P$ can be found solving the system $$AP=PJ$$

First, compute the eigenvectors. Row reducing yields $$\pmatrix{ 3&-1&1&7\\ 9&-3&-7&-1\\ 0&0&4&-8\\ 0&0&2&-4} \leadsto \pmatrix{ 3&-1&0&9\\ 0&0&1&-2\\ 0&0&0&0\\ 0&0&0&0}$$ So, we have eigenvectors $$v_1= (1,3,0,0)^T, \quad v_2 = (-3,0,2,1)^T$$ Now, we need to solve $Aw_1 = v_1$ and $Aw_2 = v_2$. Once we have suitable solutions $w_1,w_2,$ the basis of the Jordan-form matrix can be given by $\{v_1,w_1,v_2,w_2\}$.

Thank you everyone for the tips, i manage to solve it. The basis is

$$\left(\begin{matrix}-7 & 0 & 3 & 1\\-1 & 0 & 9 & 0\\-8 & 0 & 0 & 0\\-4 & 1 & 0 & 0 \end{matrix}\right)$$ And $P^{-1}AP=J$

• Could you give a hint on how to solve it? I am facing a similar problem. Just cannot solve it. – user522841 Feb 8 '18 at 20:03