Find the Jordan normal form of a nilpotent matrix $N$ given the dimensions of the kernels of $N, N^2, N^3$ 
Let $N\in \text{Mat}(10 \times 10,\mathbb{C})$ be nilpotent. Furthermore let $\text{dim} \ker N =3 $, $\text{dim} \ker N^2=6$ and $\text{dim} \ker N^3=7$. What is the Jordan Normal Form?

The only thing I know is that there have to be three blocks, since $\text{dim} \ker N = 3$.
Thank you very much in advance for your help.

Since $N$ is nilpotent, it has $0$ as unique eigenvalue. Since $\dim\ker N=3$, by the rank nullity theorem, the rank of $N$ is $7$. Hence the JNF of $N$ has seven 1's. Assume that there are $k$ blocks, and denote by $n_i$ the size of the $i$-th block. Of course, $n_1+\cdots+n_k=10$. Now, the rank of $N$ is $(n_1-1)+(n_2-1)+\cdots+(n_k-1)=10-k$, and we know that this rank is 7 (by rank nullity theorem). Hence k=3. You already got to that point.
Now, let $p$ be the number of blocks of size $1$. Then, the rank of $N^2$ is $(n_1-2)+\cdots+(n_k-2)+p=10-2k+p$. Since $\dim\ker N^2=6$, we can conclude (by the rank–nullity theorem again) that the rank of $N^2$ is $4$. Hence (since $k=3$) we conclude that $p=0$. So all blocks have a size at least 2.
Finally, let $q$ be the number of blocks of size $2$. Then, the rank of $N^3$ is $(n_1-3)+\cdots+(n_k-3)+q=10-3k+q$. Since $\dim\ker N^3=7$, we conclude that the rank of $N^3$ is $3$. Since $k=3$, we conclude that $q=2$. Hence $N$ has 2 blocks of size $2$ and one block of size $6$: we hence conclude that the JNF is (up to reordering of the blocks):
$$\left(\begin{array}{cc|cc|cccccc}0&1&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\\hline0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\\hline0&0&0&0&0&1&0&0&0&0\\0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&1&0&0\\0&0&0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&0&0&1\\0&0&0&0&0&0&0&0&0&0\end{array}\right).$$
