I want to determine the Jordan normal form of a complex matrix $A$ with characteristic polynomial $\chi_A(x)=(x+1)^4(x+2)^2$, minimal polynomial $m_A(x)=(x+1)^2(x+2)^2$ and that has the property that the matrix $A+I_6$ has order $3$.
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I have done the following:
The eigenvalues of $A$ are the roots of $\chi_A$. So the eigenvalues of $A$ are $-1$ and $-2$, where the multiplicity of the eigenvalue $-1$ is $4$ and the multiplicity of the eigenvalue $-2$ is $2$.
From the minimal polynomial we get the information that there is a Jordan block for the eigenvalue $-1$ that is $2\times 2$ (since we have the power $2$). We also have a Jordan block for the eigenvalue $-2$ that is $2\times 2$ (since we have the power $2$).
Is everything correct so far?
What information do we get from the fact that the matrix $A+I_6$ has order $3$ ?
Best Answer
The information you have so far is correct. This narrows it down to two forms:
$$ \begin{bmatrix} -1 & 1 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 1\\ 0 & 0 & 0 & 0 & 0 & -2\\ \end{bmatrix} \quad\text{ and } \quad \begin{bmatrix} -1 & 1 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 1 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 1\\ 0 & 0 & 0 & 0 & 0 & -2\\ \end{bmatrix}$$
Now the first matrix has $\operatorname{rank} (A + I) = 3$ and the second one has $\operatorname{rank} (A + I) = 4$.
Therefore the first one is the one you need.