Difficult question on Jordan normal form 
(a) Does there exist a $9 \times 9$ matrix $B$, for which the matrix $B^2$ has the Jordan Normal Form with blocks of sizes $4,3,2$ appearing once, each block with eigenvalue $0$?
(b)Answer the same query in the analogous situation with blocks of sizes $4,4,1$.
Justify your answer.

Any ideas?
 A: The trick is to figure out the following: if $J_n$ is the Jordan block with size zero of size $n$, then what is the Jordan structure of $A = J_n^2$?  When $n = 1$, the answer is trivial, so we take $n \geq 2$ below.
In fact, we can figure this out using the so-called "Weyr characteristic," which I explain here (references on this are sparse as far as I can tell, but I learned about this from Horn and Johnson's Matrix Analysis).  Let $r_k$ denote the rank of $(A - \lambda I)^k = A^k$ (and we'll take $r_0 = n$).  You can verify that we have
$$
r_k = \begin{cases}
n - 2k & k \leq n/2\\
0 & k > n/2
\end{cases}.
$$
Now, we note that $s_k = r_{k-1} - r_k$ will always give us the number of Jordan blocks that $A$ has with size at least $k$.  In our case, we find that when $k < n/2$, we have $s_k = 2$.  In the case where $n$ is even, we will also have $s_{n/2} = 2$, and in the case where $n$ is odd, we will have $s_{(n+1)/2} = 1$.  This accounts for all non-zero $s_k$.
Finally, $m_k$, the number of Jordan blocks with size $k$, will satisfy $m_k = s_k - s_{k-1}$.  In the even case, we find that $m_{n/2} = 2$, with all other $m_k$ zero.  In the odd case, we find that $m_{(n-1)/2} = m_{(n+1)/2} = 1$, with all other $m_k$ zero.
In conclusion, $J_n^2$ will have the Jordan structure $n/2,n/2$ in the case where $n$ is even and $(n+1)/2,(n-1)/2$ in the case where $n \geq 2$ is odd.
Now, apply this to your question.  What is required in order for $B^2$ to have an odd number of blocks in its Jordan form? With this you should find that the answer to (a) is no, and the answer to (b) is yes.
