Could you help me to solve this exercise?
Let $A$ be an $8\times8$ nilpotent matrix over $\mathbb{C}$ with $\mathrm{rank}(A)=5$ and $\mathrm{rank}(A^2)=2$. List all possible Jordan canonical forms for $A$ and show that knowledge of $\mathrm{rank}(A^3)$ would allow one to determine the Jordan canonical form.
This is what I have done: $A$ is nilpotent so the characteristic polynomial is $x^8$ and the minumum polynomial is $x^n$ with $3\leq n\leq8$. But now I don't know how to continue, any idea?
Best Answer
Since the matrix is nilpotent, as you note the characteristic polynomial is $x^8$, and the minimal polynomial is $x^i$ for some $i$, $1\leq i \leq 8$. It's not $x$, because $A$ is not zero, and it's not $x^2$, because $A^2$ is not zero either. So $3\leq i\leq 8$.
Let's go form there.
Because $A$ has rank $5$, the Rank-Nullity Theorem tells you the nullity is $3$; that means that the dimension of the eigenspace corresponding to $0$ is $3$; this tells you, by the basic theory of the Jordan Canonical Form, the number of Jordan blocks associated to zero. So the Jordan canonnical form of $A$ has exactly 3 blocks, all corresponding to zero.
The key observation to make is that if $J$ is $k\times k$ Jordan block associated to $0$, then it has rank $k-1$, and $J^s$ has rank $k-s$ ($1\leq s\leq k$); just notice that each time the $1$s "move up" one row until they disappear.
So: the original matrix consists of three blocks; say the sizes are $n_1$, $n_2$, and $n_3$, with $n_1\leq n_2\leq n_3$. We know the largest block has size at least $3$ (the degree of the minimal polynomial is the size of the largest block). And we know $n_1+n_2+n_3=8$.
We cannot have both $n_1$ and $n_2$ equal to $1$; if we did, the matrix would be a $6\times 6$ block and the rest would be zeros; after squaring, the matrix would have rank $4$ (by the observation above), which is not the case. So $n_2\geq 2$. We cannot have $n_1=1$ and $n_2=2$, because then $n_3=5$, and after squaring we would get a matrix of rank $3$, which is not the case.
In fact, if the blocks are of size $1+n_2+n_3$, with $n_3\geq n_2\geq 2$, then the rank of the square would be $n_2+n_3-4 = 3$, which is not the case. So $2\leq n_1$.
That will give the right rank, since then we will have that the rank of $A^2$ is $(n_1-2)+(n_2-2)+(n_3-2) = n_1+n_2+n_3-6 = 8-6 = 2$.
So we must have $2\leq n_1\leq n_2\leq n_3$, $n_1+n_2+n_3=8$; any of these combinations will be a possible Jordan canonical form.
After finding all possibilities, verify that in each case you get a different value for $\mathrm{rank}(A^3)$, so that will show the last clause of the problem is true.