I'm preparing for my exam and while studying an old exam I came across this exercise. In essence I'm supposed to find the Jordan normal form of the following matrix:
$$ A = \begin{bmatrix} 0 & 5 & 6 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ \end{bmatrix} $$
I found the characteristic polynomial with Laplace expansion along the first column (then 2nd then 3rd): $$ P(\lambda)=(1-\lambda)^5 $$
I got three Eigenvectors out of $\lambda = 1$, those are: $$ \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ \end{bmatrix} \begin{bmatrix} 0\\ -6/5\\ 1\\ 0\\ 0\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 1\\ \end{bmatrix} $$
So I'm looking for another 2 vectors since the algebraic multiplicity $m_a(\lambda) = 5$ and the geometric multiplicity $m_g(\lambda)=5$.
We learned that in order to find the remaining vectors if $m_a \neq m_g$, we have to take the power of the matrix we used for the initial Eigenvector(s) and solve it the same way. $$\Rightarrow Ker(A-\lambda*I_n)^2 $$
The problem ist, that this matrix is the Nullmatrix, so how am I supposed to find the remaining 2 vectors?