Let $\textbf{A}=\left(\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 0\\ -60 & 0 & 26 & 0 \end{array}\right)$ be, find a chain of generalized eigenvectors.
Eigenvalues : $\lambda_{1}=0 ;\quad \lambda_{2}=0 ;\quad \lambda_{3}=2 ;\quad \lambda_{4}=0$.
For $\lambda_{1}=0$, $v_{1}= \left(\begin{array}{c} v_{11} \\ v_{12} \\ v_{13} \\ v_{14} \end{array}\right)$
$$(\textbf{A}-\lambda_{1} \textbf{I})v_{1}=0 ; \quad \left(\begin{array}{cccc} -\lambda_{1} & 0 & 0 & 0\\ 0 & -\lambda_{1} & 0 & 0\\ 0 & 0 & 2-\lambda_{1} & 0\\ -60 & 0 & 26 & -\lambda_{1} \end{array}\right)\left(\begin{array}{c} v_{11} \\ v_{12} \\ v_{13} \\ v_{14} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\right)$$
$$ \left \{ \begin{array}{c} -\lambda_{1}v_{11}=0 \\ -\lambda_{1}v_{12}=0 \\ (2-\lambda_{1})v_{13}=0 \\ -60v_{11}+26v_{13}-\lambda_{1}v_{14}=0 \end{array}\right.;\quad v_{11}=v_{13}=0;\quad v_{1}=\left(\begin{array}{c} 0 \\ v_{12} \\ 0 \\ v_{14} \end{array}\right)$$
For $\lambda_{2}=0$, $v_{2}= \left(\begin{array}{c} v_{21} \\ v_{22} \\ v_{23} \\ v_{24} \end{array}\right)$
$$(\textbf{A}-\lambda_{2} \textbf{I})v_{2}=v_{1} ; \quad \left(\begin{array}{cccc} -\lambda_{2} & 0 & 0 & 0\\ 0 & -\lambda_{2} & 0 & 0\\ 0 & 0 & 2-\lambda_{2} & 0\\ -60 & 0 & 26 & -\lambda_{2} \end{array}\right)\left(\begin{array}{c} v_{21} \\ v_{22} \\ v_{23} \\ v_{24} \end{array}\right)=\left(\begin{array}{c} 0 \\ v_{12} \\ 0 \\ v_{14} \end{array}\right)$$
$$ \left \{ \begin{array}{c} -\lambda_{2}v_{21}=0 \\ -\lambda_{2}v_{22}=v_{12} \\ (2-\lambda_{2})v_{23}=0 \\ -60v_{21}+26v_{23}-\lambda_{2}v_{24}=v_{14} \end{array}\right.;\quad v_{23}=v_{12}=0;\quad v_{1}=\left(\begin{array}{c} 0 \\ 0 \\ 0 \\ v_{14} \end{array}\right);\quad v_{2}=\left(\begin{array}{c} -\dfrac{v_{14}}{60} \\ v_{22} \\ 0 \\ v_{24} \end{array}\right)$$
For $\lambda_{3}=2$, $v_{3}= \left(\begin{array}{c} v_{31} \\ v_{32} \\ v_{33} \\ v_{34} \end{array}\right)$
$$(\textbf{A}-\lambda_{3} \textbf{I})v_{3}=0 ; \quad \left(\begin{array}{cccc} -\lambda_{3} & 0 & 0 & 0\\ 0 & -\lambda_{3} & 0 & 0\\ 0 & 0 & 2-\lambda_{3} & 0\\ -60 & 0 & 26 & -\lambda_{3} \end{array}\right)\left(\begin{array}{c} v_{31} \\ v_{32} \\ v_{33} \\ v_{34} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\right)$$
$$ \left \{ \begin{array}{c} -\lambda_{3}v_{31}=0 \\ -\lambda_{3}v_{32}=0 \\ (2-\lambda_{3})v_{33}=0 \\ -60v_{31}+26v_{33}-\lambda_{3}v_{34}=0 \end{array}\right.;\quad v_{31}=v_{32}=0; v_{34}=13v_{33}\quad v_{3}=\left(\begin{array}{c} 0 \\ 0 \\ v_{33} \\ 13v_{33} \end{array}\right)$$
how can I find the last generalized eigenvector?
The eigenvalues are $0$ and $2$.
Calculating $\ker A$, we get:
$$Ax = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 0\\ -60 & 0 & 26 & 0}\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4} = 0 \implies \left \{ \begin{array}{c} 2x_3=0 \\ -60x_1+26x_3 =0 \\ \end{array}\right. \implies x_1 = x_3 = 0$$
so $\ker A = \operatorname{span}\{e_2, e_4\}$. We have $\dim\ker A = 2$, which is less than the algebraic multiplicity of $0$. Therefore, we continue onto $\ker A^2$:
$$A^2x = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 52 & 0}\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4} = 0 \implies 52x_3 = 0 \implies x_3 = 0$$
so $\ker A^2 = \operatorname{span}\{e_1, e_2, e_4\}$. Now $\dim\ker A^2 = 3$ is equal to the algebraic multiplicity of $0$, so there are no more generalized eigenvectors for $0$.
For the eigenvalue $2$, we have
$$(A - 2I)x = \pmatrix{-2 & 0 & 0 & 0\\ 0 & -2 & 0 & 0\\ 0 & 0 & 0 & 0\\ -60 & 0 & 26 & -2}\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4} = \left \{ \begin{array}{c} -2x_1 = 0\\ -2x_2 = 0\\ -60x_1+26x_3 -2x_4 =0 \\ \end{array}\right. \implies \left \{ \begin{array}{c} x_1 = x_2 = 0 \\ x_4 = 13x_3 \\ \end{array}\right. $$
so $\ker (A - 2I) = \operatorname{span}\{e_3 + 13e_4\}$.
Therefore, one set of generalized eigenvectors is given by
$$\left\{\pmatrix{1 \\ 0 \\ 0 \\ 0},\pmatrix{0 \\ 1 \\ 0 \\ 0},\pmatrix{0 \\ 0 \\ 0 \\ 1},\pmatrix{0 \\ 0 \\ 1 \\ 13}\right\}$$