Finding possible jordan canonical forms when knowing the eigenvalues and specific ranks of matrices For a linear transformation $A: \mathbb{R}^{10} \rightarrow \mathbb{R}^{10}$ we know that $\lambda = 1$ and $ \lambda = 2$ are the only eigenvalues. Eigenspace for $\lambda = 1$ is 3-dimensional, and the eigenspace for $ \lambda = 2$ is 2-dimensional. It is also true that: $rank(A - I)^5 = rank (A - I)^4 = 3$.
What is the dimension of $Ker (A - I)^2$? Write all possible Jordan forms for transformation A.
Attempted solution: I know that because of the dimension of eigenspaces, in the Jordan form there will be 3 Jordan blocks for $\lambda = 1$, and 2 Jordan blocks for $\lambda = 2$. Similarly, because for $\lambda = 1$ the eigenspace is 3-dimensional, the dimension of $Ker (A - I) = 3$. Even though I know that from the above equation, $dimKer (A - I)^5 = dim ker(A - I)^4 = 10 - 3 = 7$, I dont know how this could help me to find $dimKer (A - I)^2$ or the possible Jordan forms.
 A: For a Jordan block $J$ with eigenvalue $0$, the rank of its powers decreases by $1$ until it vanishes. This means that if three Jordan blocks for $\lambda=1$ have dimension $m_1,m_2,m_3$, none of them can have $m_j\geq5$, because we would have $\operatorname{rank}(A-I)^4\ne\operatorname{rank}(A-I)^5$. So $m_1,m_2,m_3\leq4$. If the Jordan blocks for $\lambda=2$ have dimensions $n_1$ and $n_2$, we know that $n_1+n_2=3$, so $m_1+m_2+m_3=7$.
This means that $n_1,n_2$ are $2,1$, and $m_1,m_2,m_3$ are either $3,3,1$ or $3,2,2$ (up to reordering).
If we had $3,3,1$ then the rank of $(A-I)^2$ would be $(2-1)+(2-1)+0+3=5$ (the dimension $1$ block does not lose rank with powers, and the $3=1+1$ is contributed by the $\lambda=2$ blocks which are invertible in $A-I$ since they have nonzero diagonal), and then $\dim\ker (A-I)^2=5$. If we had $3,2,2$ the rank of $(A-I)^2$ would be $(2-1)+(1-1)+(1-1)+3=4$ and now $\dim\ker (A-I)^2=6$.
In summary, the Jordan blocks are either
$$
J_3(1)\oplus J_3(1)\oplus J_1(1)\oplus J_2(2)\oplus J_1(2), \qquad\dim\ker(A-I)^2=5
$$
or
$$
J_3(1)\oplus J_2(1)\oplus J_2(1)\oplus J_2(2)\oplus J_1(2), \qquad\dim\ker(A-I)^2=6.
$$
