How to find the size of a Jordan Block In the solution below, for the individual Jordan blocks Ji, I am not sure how the size of each block was found. In part 2 it is obvious, as there are 3 blocks for a 4x4 space, therefore there must be 2 1x1 and 1 2x2, but for part 1, why is there a 3x3 block and a 2x2 block? I honestly have no idea where those numbers were found.
The question
 A: If you have an $m\times m$ Jordan block $J$ with eigenvalue $\lambda$, then $J-\lambda I$ is nilpotent of order $m$; that is, $(J-\lambda I)^m=0$, $(J-\lambda I)^{m-1}\ne0$.
Note also that when you take an $m\times m$ Jordan block $J$ and $\lambda$ is not the eigenvalue for $J$, then $J-\lambda I$ has rank $m$ and so do all its powers (because $J-\lambda I$ is invertible). If, on the other hand, $\lambda $ is an eigenvalue of $J$, then $J-\lambda I$ has rank $m-1$.
So they tell you first that for the eigenvalue $3i$, the matrix $A-3i I$ has rank 7, $(A-3iI)^2$ has rank 5, and further powers have rank 4. The fact that the rank is $7=9-2$ tells you that there are two blocks with eigenvalue $3i$. When you take powers of Jordan blocks with diagonal zero, the rank diminishes by 1. The decrease from 7 to 5 tells you that both blocks have size $\geq 2$. One of these blocks does not contribute in the next step, so its second power was zero: its size is $2\times 2$. The other is $3\times 3$, as the rank does not decrease afterwards.
Then we know that $A+I$ has rank 6, so there are three blocks with eigenvalue $-1$. Only one of them decreases rank when squaring, so two blocks are $1\times 1$. The remaining one is $2\times 2$, since it doesn't decrease rank starting with the power 3.
