Finding the Jordan Form of a matrix Let $A$ be a $7 \times 7$ matrix with a single eigenvalue $q\in C$.
It is know that $\rho (A-qI) = 2$ and that $\rho (A-qI)^2 = 1$.
How can I find the Jordan form of $A$ (+ the minimal polynomial)?
 A: I tend to think about these things in terms of the kernel.  In particular: for any $k \geq 1$, the number of Jordan blocks that $A$ has of size at least $k$ is given by
$$
\dim \ker (A^k) - \dim \ker (A^{k-1})
$$
where $A^0 = I$ by definition. To recast this in terms of rank, we can apply the rank-nullity theorem to note that this difference can be rewritten as
$$
\rho(A^{k-1}) - \rho(A^k)
$$
Now, applying this to your question: we find that $A$ has:


*

*$7 - 2 = 5$ blocks of size at least $1$

*$2 - 1 = 1$ block of size at least $2$.


Since $A$'s only eigenvalue is $q$, this information is enough to determine the Jordan form of $A$.
A: One can think of the number of Jordan blocks of various sizes. We know that there are a total of $5$ $q$-blocks because the dimension of the nullspace of $A - qI$ is $5$. Now since the rank of the nullspace of $(A-qI)^2$ is $6$ we know that there is exactly one block of size greater than one (since we only gained one dimension by squaring). Therefore with five total blocks, and only one of size greater than one, whose sizes add to $7$, we must have one $3$ block, and $4$ $1$-blocks, as in 
$$\left[ \begin{matrix}q & 1\\ &q&1 \\ &&q \\&&&q \\&&&&q \\ &&&&&q \\ &&&&&&q \end{matrix} \right].$$
Since there is only one eigenvalue, the minimal polymomial is $(x-q)^k$ for some $k$, and we can see here that it is $(x-q)^3$.
