<s>[INST] How to determine the Jordan form and give a Jordan base for a matrix? given is
$\begin{pmatrix} 3&0&-1&0&0 \\ 1&3&0&1&0 \\ 0&0&3&0&0 \\ 0&0&0&3&0 \\ 0&0&0&0&-3 \end{pmatrix}$
I have to determine the jordan form and also give a jordanbase.
I got this so far:


*

*Eigenvalues are
$l_1 = 3, l_2 = -3$ whereas $P_A(x) = (x-3)^4 * (x+3)$.

*Eigenvectors are
for $l_1 = 3$:
$(0,1,0,0,0)$ and $(-1,0,0,1,0)$
for $l_2 = -3$:
$(0,0,0,0,1)$


*This means, that we still need two vectors. How can I get them?

*How can I get then the jordanbase?
 [/INST] To get the full base, you need the generalized eigenvectors. In this case, we do so as follows:
I define
$$
A = 
\pmatrix{
3&0&-1&0&0 \\ 
1&3&0&1&0 \\ 
0&0&3&0&0 \\ 
0&0&0&3&0 \\ 
0&0&0&0&-3
}
$$
We find non-zero solutions to the following:
$$
(A - 3I)x = \pmatrix{0\\1\\0\\0\\0} \implies
x = \pmatrix{1\\0\\0\\0\\0}\\
(A - 3I)x = \pmatrix{-1\\0\\0\\1\\0} \implies
\text{no solution}\\
(A - 3I)x = \pmatrix{1\\0\\0\\0\\0} \implies
x = \pmatrix{0\\0\\-1\\0\\0}\\
$$
And we now have a base.  Our Jordan form $J = S^{-1}AS$ must be
$$
J = \pmatrix{3&1\\&3&1\\&&3\\&&&3\\&&&&-3}
$$
with base
$$
S = \pmatrix{0&1&0&-1&0\\1&0&0&0&0\\0&0&-1&0&0\\0&0&0&1&0\\0&0&0&0&1}
$$
</s>