Determining the Jordan Canonical Form $18\times 18$ matrix 
Let A be an $18\times 18$ matrix over $\mathbb{C}$ with characteristic polynomial equal to
  $$
(x-1)^6(x-2)^6(x-3)^6
$$
  and a minimal polynomial equal to
  $$
(x-1)^4(x-2)^4(x-3)^3.
$$
  Assume $(A-I)$ has nullity $2$, $(A-2I)$ has nullity 3, and $(A-3I)^2$ has nullity 4. Find the Jordan canonical form of A.

For $\lambda=1$, I figure the largest block is a $4\times 4$ (since the multiplicity of the root is $4$ in the minimal polynomial), and since the nullity is $2$, that there are $2$ blocks in total, thus forcing the other block to be $2\times2$.
For $\lambda=2$, I figure the largest block is a $4\times4$, there are 3 blocks, thus making the others $1\times 1$ matrices.
For $\lambda=3$, The largest block is a $3\times 3$ matrix. However, I am not sure how the nullity of $(A-3I)^2$ determines the nullity of $(A-3I)$. 
Any insight into this problem would be extremely valuable as I am trying to teach myself this process.
 A: The followig auxiliary result is needed to settle this question. Let us consider an $n\times n$ Jordan block
$$
B=\left(\begin{array}{ccccc}
\lambda&1&0&\cdots&0\\
0&\lambda&1&\cdots&0\\
\vdots&\ddots&\ddots&\ddots&\vdots\\
0&\cdots&0&\lambda&1\\
0&\cdots&\cdots&0&\lambda
\end{array}\right).
$$
If $0\le k\le n$, then we see that $(B-\lambda I)^k$ has a diagonal of length $n-k$
filled with $1$s. Therefore its rank is $n-k$, and thus its nullity is equal to $k$. If $k>n$, then the nullity of this block is, of course $n$. We can summarize this by saying that the nullity of $(B-\lambda I)^k$ is $\min\{n,k\}$.
The OP can use this to handle the case $\lambda=3$. Assume that the Jordan blocks
associated to this eigenvalue have size $n_1\ge n_2\ge\cdots\ge n_m$. The characteristic polynomial tells us that $n_1+n_2+\cdots+n_m=6.$ For its part the minimal polynomial tells us that $n_1=3$. We are left with three possibilities:
A) $m=2,n_1=n_2=3$, B) $m=3, n_1=3,n_2=2,n_3=1$, C) $m=4, n_1=3, n_2=n_3=n_4=1$.
Applying the above observation to all those Jordan blocks we see that the nullity of $(A-3\lambda I)^2$ is 
$$
\sum_{j=1}^m\min\{n_j,2\},
$$
which adds up to $2+2=4$ in case A, $2+2+1=5$ in case B, and $2+1+1+1=5$ in case C. Thus we are left with case A as the only possibility, and can conclude that the matrix $A$ has two $3\times3$ Jordan blocks beloging to $\lambda=3$.
