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


*

*The characteristic polynomial is $p(x)=-x^5(x-1)^5(x+1)^5$

*The dimension of the range of $A$ is $13$ and $\dim \ker A^2=4$.

*The dimension of the range of $A-I$ is $12$ and the minimal polynomial is $(x-1)^2q(x)$ and $q(1)\neq 0$.

*The dimension of the range of $(A+I)^i$ is $10$ for $i\geq 5$.


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

A: *

*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}
