How to determine the Jordan form and give a Jordan base for a matrix?

Question

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:

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

2. Eigenvectors are for $l_1 = 3$: $(0,1,0,0,0)$ and $(-1,0,0,1,0)$

for $l_2 = -3$: $(0,0,0,0,1)$

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

4. How can I get then the jordanbase?

Answer 1

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 $$ S = \pmatrix{0&1&0&-1&0\\1&0&0&0&0\\0&0&-1&0&0\\0&0&0&1&0\\0&0&0&0&1} $$