How do I calculate generalized eigenvectors? I have the matrix 
$$A=\begin{pmatrix} 5 & 1 & 0\\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$
and I should determine generalised eigenvectors, if they exist. 
I found one eigenvalue with algebraic multiplicity $3$.
$$\lambda=5$$
I calculated two eigenvectors:
$$\vec{v_{1}}  =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \qquad{}   
 \vec{v_{2}}  =\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$
Also, I know this formula for generalized vector
$$\left(A-\lambda I\right)\vec{x} =\vec{v}$$
Finally, my question is:

How do I know how many  generalised eigenvectors I should calculate? 

For every eigenvector one generalised eigenvector or?
My university book is really confusing, and I saw there that they calculated generalised eigenvector only for some eigenvectors, and for some not. But I don't understand how to know that.
 A: Your  matrix is in Jordan normal form. You can read on the matrix that $e_1$ and $e_3$ are eigenvectors for the eigenvalue $2$, and $e_2$ is a generalised eigenvector.
A: The generalised eigenspace of $A$ for an eigenvalue $\lambda$ is the kernel of $(A-\lambda I)^k$ for sufficiently large $k$ (meaning that the kernel won't get bigger by further increasing $k$). The multiplicity of $\lambda$ as root of the characteristic polynomial is always sufficiently large. In the example $(A-\lambda)^2=0$ so $k=2$ suffices and the generalised eigenspace is the whole space.
It is common to find a basis for the kernel with exponent $1$ first (the ordinary eigenspace) then extend to a basis for exponent$~2$, and so forth until$~k$. This basis is somewhat better than just any basis for the generalised eigenspace, but it remains non unique in general. Though there are infinitely many generalised eigenvectors, it is not useful to list linearly dependent ones among them, so one stops having found a basis for the generalised eigenspace (here after $3$ independent vectors).
A: The dimension of the nullspace of A minus lamda*I will give you the number of 'generalizable' eigenvectors for any particular eigenvalue. The sum of this for all different eigenvalues is the dimension of the eigenspace. Your matrix does not have 3 generalizable eigenvectors so it is not diagonizable. 
Sorry for the lack of formulas. Doing this on my phone. Will come back later to edit. 
