# 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?

## 1 Answer

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.

• The approach given here is also applicable here and is a faster way to find the Jordan form of $J_n^2$. – Ben Grossmann Jan 22 '20 at 17:32