Eigenspaces and Jordan normal from of a matrix Let $A\in \operatorname{End}(V)$ be a linear transformation of $V=\mathbf{C}^3$ which is represented by $$[A]=\begin{pmatrix}
1 &0  &0 \\ 
x_1 &2  &0 \\ 
x_2 &x_3  &1 
\end{pmatrix}$$
w.r.t the standard basis, where $x_1,x_2,x_3\in \mathbf{C}$. The problem I am struggling with is, to determine the generalized eigenspaces and a Jordan normal form. 
I found out that the eigenvectors of $[A]$ are $(0,1/x_3,1)$ and $(0,0,1)$ for eigenvalues $2$ and $1$, respectively. But I do not know how to determine the generalized eigenspaces. I checked it in Wikipedia, and it kind of confuses me as I am beginner to it.
 A: The eigenspaces of $A$ are the kernels of $A-1$ and $A-2$. Since the multiplicity of eigenvalue 2 is one, it has no generalized eigenvalues. 
For the eigenvectors with eigenvalue $1$, we find the kernel of $A-1$. 
We want to solve
$$
\left(\begin{matrix}
0 & 0 & 0\\
x_1 & 2 & 0\\
x_2 & x_3 & 1
\end{matrix}
\right)
\left(\begin{matrix}
v_1\\
v_2\\
v_3
\end{matrix}
\right)=0
$$
So we get $x_1v_1+v_2=0$ and $x_2v_1+x_3v_2=0,$ and substitution gives $(x_2-x_1x_3)v_1=0.$
So I agree that $(0,0,1)$ is the only independent eigenvector in this kernel, but only if we assume $x_2-x_1x_3\neq 0$, so that we can solve the system with $v_1=v_2=0$. 
On the other hand, if $x_2-x_1x_3=0$, then $(1,-x_1,0)$ and $(0,0,1)$ are both eigenvectors. In this case the matrix is diagonalizable, which also tells you its Jordan form.
So assuming that $x_2-x_1x_3\neq 0$, to find the generalized eigenvectors for eigenvalue 1, find the kernel of $(A-1)^2$. There is one independent vector there that is not also in the kernel of $A-1$ (so not an eigenvector), which is $(1, -x_1, 0)$.
In general to find the generalized eigenvectors for an eigenvalue $c$, you compute the kernel $(A-c)^k$, where $k$ is the power with which the $(x-c)$ factor occurs in the minimal polynomial.
Finally once you have the generalized eigenvector, you can write the Jordan canonical form by expressing the matrix in the ordered basis $(0,0,x_2-x_1x_3), (1,-x_1,0),(0,1,x_3)$. You get
$$
\left(\begin{matrix}
1 & 1 & 0\\
0 & 1 & 0\\
0 & 0 & 2
\end{matrix}
\right)
$$
A: For the eigenvalue $2$, since its algebraic multiplicity is $1$, so will be its geometric multiplicity. We have 
$$
\begin{bmatrix}2a\\ 2b\\ 2c\end{bmatrix}
=\begin{bmatrix}
1 &0  &0 \\ 
x_1 &2  &0 \\ 
x_2 &x_3  &1 
\end{bmatrix}
\begin{bmatrix}a\\ b\\ c\end{bmatrix}
=\begin{bmatrix}
a\\ ax_1+2b\\ ax_2+bx_3+c
\end{bmatrix}
$$
From $2a=a$ we get $a=0$. Then the third equation becomes $bx_3+c=2c$, or $c=bx_3$. Thus the eigenvectors are of the form
$$
\begin{bmatrix}
0\\ b\\ bx_3
\end{bmatrix}
=b\,\begin{bmatrix}
0\\ 1\\ x_3
\end{bmatrix}
$$
(note that I am not assuming that $x_3\ne0$, as you did). 
For the eigenvalue $1$, the equations become
$$
a=a, \ \ \ ax_1+2b=b, \ \  ax_2+bx_3+c=c.
$$
It follows that $b=-ax_1$ and that $ax_2-ax_1x_3=0$. This last equation can be rewritten as $$\tag{*}a(x_2-x_1x_3)=0.$$
So we have two cases:


*

*If $x_2=x_1x_3$, we get independent eigenvectors for the cases $a=0$ and $a\ne0$:
$$
\begin{bmatrix}
0\\ 0\\1
\end{bmatrix},\ \ \ \text{ and }\begin{bmatrix}1\\-x_1\\ 0\end{bmatrix}.
$$ 
In this case $A$ is diagonalizable and its Jordan form is $$\begin{bmatrix}1&0&0\\0&1&0\\ 0&0&2\end{bmatrix}.$$

*When $x_2\ne x_1x_3$, the equation $(*)$ forces $a=0$, and we only get the eigenvectors 
$$
\begin{bmatrix}
0\\ 0\\c
\end{bmatrix}
=c\,\begin{bmatrix}
0\\ 0\\1
\end{bmatrix}.
$$
In this case the Jordan form of $A$ will be $$\begin{bmatrix}1&1&0\\0&1&0\\ 0&0&2\end{bmatrix}. $$ For a generalized eigenvalue, we look at $(A-I)^2v=0$, i.e. 
$$
\begin{bmatrix}
0\\ 0\\0
\end{bmatrix}=\begin{bmatrix}0&0&0\\ x_1&1&0\\ x_1x_3&x_3&0\end{bmatrix}\begin{bmatrix}
a\\ b\\c
\end{bmatrix}
=\begin{bmatrix}
0\\ ax_1+b\\ ax_1x_3+bx_3
\end{bmatrix}
$$
As $ax_1x_3+bx_3=(ax_1+b)x_3=0$, the third entry gives us nothing. So $b=-ax_1$, and the generalized eigenvectors are of the form
$$
\begin{bmatrix}
a\\ -ax_1\\0
\end{bmatrix}.
$$
