Given the following facts about a square complex matrix $C$:
- Characteristic polynomial of C is $x^5 (x-1)^4$
- $\operatorname{rank}(C)=7$
- $\operatorname{rank}(C-I)=6$
- $\operatorname{rank}(C^2)=5$
Find the Jordan Canonical Form of $C$.
My attempt so far to find the JCF of C:
From the characteristic polynomials I know that the eigenvalues are $0$ and $1$ with algebraic multiplicities $5$ and $4$ respectively.
For evalue $\lambda =0$:
From $\operatorname{rank}(C)=7$, I know that the geometric multilicity is $9-7=2$.
So J-Blocks are either: $J_1(0) \oplus J_4(0)$ or $J_2(0) \oplus J_3(0)$.
For evalue $\lambda =1$:
From $\operatorname{rank}(C-I)=6$, I know that the geometric multilicity is $9-6=3$.
So J-Blocks are: $J_2(1) \oplus J_1(1) \oplus J_1(1)$.
I'm confused on how to use the last fact that $\operatorname{rank}(C^2)=5$ to determine which of the to JNFs for $\lambda=0$ to choose.
I read the section on the Uniqueness of the Jordan Normal blocks in Wikipedia where it explained the significance of the $\operatorname{rank}(A-\lambda I)^j$ in determining the JNF form but I think I require to know $k_1$ in $(A-\lambda I)^{k_1}=0$ to do so.
Lets $C = MJM^{-1}$. Then $C^2 = MJ^2M^{-1}$, where $J$ is the Jordan form of $C$.
Now for the first case, $J_1(0) \oplus J_4(0)$
$$ J_4^2(0) = \left[\begin{array}\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0 \end{array}\right]^2 = \left[\begin{array}\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right] $$
For the second case $J_2(0) \oplus J_3(0)$ $$ J_3^2(0 ) = \left[\begin{array}\ 0&1&0\\ 0&0&1\\ 0&0&0 \end{array}\right]^2 = \left[\begin{array}\ 0&0&1\\ 0&0&0\\ 0&0&0 \end{array}\right] $$
$$ J_2^2(0) = \left[\begin{array}\ 0&1\\ 0&0 \end{array}\right]^2 = \left[\begin{array}\ 0&0\\ 0&0 \end{array}\right] $$
So if $J(0) = J_1(0) \oplus J_4(0)$, then rank($C^2$) = 6 and if $J(0) = J_2(0) \oplus J_3(0)$, rank($C^2$) = 5.