Jordan form of $15 \times 15$ matrix

Question

Let $A \in M(15,15,\mathbb{R})$ be a matrix that satisfies:

1. The characteristic polynomial is $p(x)=-x^5(x-1)^5(x+1)^5$
2. The dimension of the range of $A$ is $13$ and $\dim \ker A^2=4$.
3. The dimension of the range of $A-I$ is $12$ and the minimal polynomial is $(x-1)^2q(x)$ and $q(1)\neq 0$.
4. The dimension of the range of $(A+I)^i$ is $10$ for $i\geq 5$.

How I can determinate the Jordan form of $A$?

Answer 1

Hint:

• The characteristic polynomial gives you the eigenvalues, with multiplicity.
• From 2, we have $\dim \ker A$ and $\dim \ker A^2$. $\dim \ker A$ is the number of Jordan blocks associated with $0$, and $\dim \ker A^2 - \dim \ker A$ is the number of Jordan blocks associated with $0$ that have size at least $2$.
• Similarly, $\dim \ker(A - I)$ is the number of Jordan blocks associated with $1$. The minimal polynomial tells you that the largest Jordan block associated with $1$ has size $2$.
• I'm not sure how to interpret 4. If the 4th point is supposed to imply that the dimension of the range of $(A + I)^4$ is strictly less than $10$, then we know that there is exactly one Jordan block associated with $-1$. Otherwise, it gives us no new information; the fact that $(A + I)^i$ has range of dimension $10$ for all $i\geq 5$ follows from the characteristic polynomial.

Answer 2

• The eigenvalue $0$:

$\dim\ker A = 2$ tells you there are two Jordan blocks associated with $0$. Furthermore, $\dim\ker A^2 = 4$ tells you that both of them are of size $\ge 2$. Since the sum of their sizes has to be $5$, we conclude that there is one block of size $3$ and one of size $2$.

• The eigenvalue $1$:

$\dim\ker A = 3$ tells you there are three Jordan blocks associated with $1$. The fact $(x-1)^2 \mid\mid m_A$ implies that the size of the largest block is $2$. Since the sum of blocks' sizes has to be $5$, the only option for the block sizes is $2, 2, 1$.

• The eigevalue $-1$:

$\dim\ker(A+I)^i = 5, \forall i \ge 5$ tells you nothing, since you already know that the kernels of $(A+I)^i$ have to stablize to the maximum dimenision of $5$ when $i \ge 5$ or sooner. Therefore, you only know that the sum of the blocks' sizes is $5$ so the possibilities are the seven partitions of $5$:

\begin{align} &5\\ &4,1\\ &3,2\\ &3,1,1\\ &2,2,1\\ &2,1,1,1\\ &1,1,1,1,1 \end{align}