Let $C$ be a $10\times 10$ matrix whose characteristic polynomial is $(t+2)^5(t-3)^5$. Suppose that $rank((C+2I)^2)=6$ and $rank((C-3I))=8$. What are the possible Jordan forms of $C$?
This is the first problem in which I have ever encountered Jordan forms, so when it says to find all possible Jordan forms of $C$, I take that to mean finding every combination of possible Jordan blocks. Since $rank((C-3I))=8$, I know by rank nullity theorem that the dimension of the eigenspace of $\lambda=3$ is $2$. This tells me that there will be two Jordan blocks for $\lambda=3$, but I don't know how to determine the respective sizes because I don't know what the ranks of higher powers of $(C-3I)$ are. Additionally, I don't know how to glean any information from $rank(C+2I)^2=6$.
Best Answer
Throughout the post, I use $J(c;m)$ to denote the Jordan block with diagonal entries $c$ of size $m\times m$.
According to the char. polynomial, the Jordan form $J$ of $C$ has five $-2$ and five $3$ on the diagonal. For Jordan blocks with diagonal entries $3$, yes you have 2 blocks. For Jordan blocks with diag. entries $-2$, easy to see that $\DeclareMathOperator\rank{rank} \rank (J(-2;m)+2I) = m-1$ and $\rank ((J(-2;m)+2I)^2) = m-2$. Restricting to the blocks with diagonal entries $-2$, the total dimension is $5$ by the char. polynomial, so $\rank(J(-2)^2)=1$, then there exists at least $1$ block $J(-2;3)$, and the rest of the blocks could be 2 $J(-2;1)$ or 1 $J(-2;2)$.
Conclusion: up to permutation, the possible Jordan forms are $\DeclareMathOperator\diag{diag} \diag(J(-2;3), J(-2;2), J(3;1), J(3;4))$, $\diag (J(-2;3), J(-2;2), J(3;2), J(3;3))$, $\diag(J(-2;3), -2, -2, 3, J(3;4))$ and $\diag (J(-2;3), -2, -2, J(3;2), J(3;3))$.
UPDATE
You could prove that $$ \rank\begin{bmatrix} A & O \\O & B\end{bmatrix} = \rank A + \rank B $$ where the LHS is a block diagonal matrix.
Then, for example, $$ J = \begin{bmatrix} -2 & 1 &&&\\ &-2&&&\\ &&3 & * & * \\ &&&3 &* \\ &&&&3\end{bmatrix} = \mathrm {diag}(J(-2), J(3)), $$ then $$ (J+2I)^2= \begin{bmatrix} 0 & 0 &&&\\ &0&&&\\ &&25 & * & * \\ &&&25 &* \\ &&&&25\end{bmatrix} = \mathrm {diag}((J(-2)+2I)^2, (J(3)+2I)^2), $$ and $\rank((J+2I)^2) = 3$, so $$ \rank((J(-2)+2I)^2) = 3 - \rank((J(3)+2I)^2) = 3-3 = 0. $$