Given an unknown matrix $A \in \mathbb{R}^{n \times n}$ and assuming that for some eigenvalue $\lambda$ of $A$ I know the multiplitiy of the root corresponding to $\lambda$ in both the characteristic and the minimal polynomial, I want to find the sizes of the Jordan blocks corresponding to $\lambda$ in a Jordan normal form of $A$ (assuming one exists).
Specifically I am looking at the case $n = 6$, multiplicity in characteristic polynomial $m_c = 5$ and multiplicity in minimal polynomial $m_m = 3$.
Here's what I have come up with:
The size of the largest Jordan block corresponding to $\lambda$ must equal $m_m$, that leaves room for either one $2 \times 2$ or two $1 \times 1$ Jordan blocks. The number of Jordan blocks of size $k$ is equal to $2 \cdot \mathrm{dim}\ \mathrm{ker} (A – \lambda I)^k – \mathrm{dim}\ \mathrm{ker} (A – \lambda I)^{k + 1} – \mathrm{dim}\ \mathrm{ker} (A – \lambda I)^{k – 1}$. Since $m_c = \mathrm{dim}\ \mathrm{ker} (A – \lambda I)$ is also equal to the total number of Jordan blocks, we can either have the case $m_c = 2$ or $m_c$ = 3. Letting $m_k = \mathrm{dim}\ \mathrm{ker} (A – \lambda I)^k$, that results in the following systems of equations:
For $m_c = m_1 = 2$:
\begin{align*}
2 m_3 – m_3 – m_2 &= 1 \\
2 m_2 – m_3 – 2 &= 1 \\
4 – m_2 &= 0
\end{align*}
For $m_c = m_1 = 3$:
\begin{align*}
2 m_3 – m_3 – m_2 &= 1 \\
2 m_2 – m_3 – 3 &= 0 \\
6 – m_2 &= 2
\end{align*}
The first system of equations has no solution, but the second does, so I'm concluding that the Jordan normal form of $A$ has one Jordan block of size 3 and two of size 1 corresponding to $\lambda$. Is my reasoning correct? And if so, could I have arrived at this conclusion in a more simple manner?
EDIT: Well, this is embarrassing, both systems of equations do have a solution ($m_2 = 4$, $m_3 = 5$ in both cases) so clhpetersons answer does not contradict my solution, I've just made a very simple arithmetic mistake.
Best Answer
This is how I think about it: suppose I have a matrix consisting of just one Jordan block of size $k$ and eigenvalue $\lambda$. Then what is the minimal polynomial? An easy computation shows that it is $(x-\lambda)^k$. Now suppose I have a matrix which consists of any number of Jordan blocks all with eigenvalue $\lambda$ and the max size of any of the Jordan blocks is $k$. Then what is the minimal polynomial? Again it is $(x-\lambda)^k$ (since the matrix is block diagonal).
If I have an $n$ by $n$ matrix which consists of any number of Jordan blocks all with eigenvalue $\lambda$, then the characteristic polynomial is of course $(x-\lambda)^n$.
Note that it is not true that the minimal polynomial and characteristic polynomial completely determine the Jordan block structure. Suppose I have a 4 by 4 matrix $A$ with 2 Jordan blocks of size 2 and eigenvalue $\lambda$, and a 4 by 4 matrix $B$ with 3 Jordan blocks, one of size 2, and two of size 1 with eigenvalue $\lambda$. Then for both $A$ and $B$ the minimal polynomial will be $(x-\lambda)^2$ and characteristic polynomial will be $(x-\lambda)^4$.
So, if we know that the multiplicity of $\lambda$ in the characteristic polynomial is 5 and in the minimal polynomial it is 3, then all we know is that the largest Jordan block for $\lambda$ is size 3, and the sum of the sizes of all Jordan blocks is 5. One quickly sees that the only options are: