So far, you know that the eigenvalue $1$ has multiplicity five and the eigenvalue $-2$ has multiplicity $3$. This tells you that the total size of the Jordan blocks corresponding to $1$ is five and the total size of the Jordan blocks corresponding to $-2$ is three.
Let us look at $-2$ first. Since the nullity of $A + 2I$ is three, we know that the geometric multiplicity of $-2$ is three. Recall that the geometric multiplicity is the number of Jordan blocks corresponding to that eigenvalue. There are three blocks for $-2$ with total size three, therefore the only option we have is for all three blocks to be trivial.
Now for eigenvalue $1$, we know that the geometric multiplicity is two. Therefore there will be two Jordan blocks. The blocks must add to size five, therefore the possibilities are
- A block of size one and a block of size four
- A block of size two and a block of size three
The point at which your generalized eigenspaces stabilize (no longer increase in size) is the length of your longest chain. Your matrix stabilizes at $k = 3$ and so your longest chain is length three. Correspondingly, the size of your largest block is three and so we must default to option 2. There is a single Jordan block of size three and a Jordan block of size two corresponding to eigenvalue $1$.