When defining a subspace by equations, the dimension of the subspace is $n-k$, where $n$ is the ambient space (in this case, $n=4$) and $k$ is the number of linearly independent equations; in this case, $k=2$.
Hence, you need only find two ($4-2$) linearly independent vectors on your subspace, that is, two linearly independet solutions yo your equation.
For instance, if you take $x_1=1,x_2=0$ then $x_4=-1$ and $x_3=0$, so one of them might be $(1,0,0,-1)$. Similarly, another option might be $(2,1,0,-1)$. You can check that they're linearly independent.
Your question is similar to Example 2 in Wikipedia article "Generalized eigenvectors," on which I base my answer.
For a given eigenvalue, the number of chains equals the number of linearly-independent eigenvectors for that eigenvalue. So for your matrix A, each eigenvalue has one chain.
For eigenvalue 2, because the algebraic multiplicity is one, the chain length is one and consists of the corresponding eigenvector x₁ = [–1, 1, 0]ᵀ.
For eigenvalue 3, the algebraic multiplicity is two, but there is only one corresponding eigenvector, so you need to find one more generalized eigenvector to make chain of length two. To do that, solve the matrix equation (A – 3 I)y₂ = y₁ for y₂ where y₁ is your eigenvector [1, 0, 0]ᵀ. You get y₂ = [0, 1, 1]ᵀ.
Hence, you have one chain for each eigenvector, and a chain basis is {x₁, y₁, y₂}.
With the chain basis in that order, the first Jordan block is a one-by-one block for eigenvalue 2, and the second block is a two-by-two block for eigenvalue 3.
Best Answer
The equation can be rewritten as $x_1=x_2-x_3$ and you can assign arbitrary values to $x_2$ and $x_3$, thus getting all solutions. In order to find two linearly independent solutions, choose first $x_2=1$ and $x_3=0$; then $x_2=0$ and $x_3$, getting the two vectors $$ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} $$ These are obviously linearly independent, by just looking at the second and third row.