for which $k$ the matrix $A^k=0$ 
Let $A\in\mathbb{C}^{10\times 10}$ with one eigenvalue $\lambda$ such that $\operatorname{Rank}(A-\lambda I)=4$ and $\operatorname{Rank}(A-\lambda I)^3=3$
Calculate $\operatorname{Rank}(A-\lambda I)^k$ for every $k>3$

It is true to say that if rising a matrix power in one cause it to lose a rank because $\operatorname{Ker}(A)\subseteq \operatorname{Ker}(AA)$?
 A: Here Rank surely means the dimension of the span of the columns of the matrix. Let me rephrase this question in (equivalent) terms of linear operators and the dimension of the range, replacing $A - \lambda I$ with a linear operator $T$:
Proposition: Suppose $T$ is a linear operator from a vector space $V$ to $V$ and $\dim \text{range } T = 4$ and $\dim \text{range } T^3 = 3$. Then $\dim \text{range } T^k = 3$ for all $k > 3$.
Proof: Clearly
$$
\text{range } T \supset \text{range } T^2 \supset \text{range } T^3 \supset \cdots
$$
If an inclusion in the chain above is ever an equality, then it is an equality for all further terms (proof: Suppose $\text{range } T^n = \text{range }T^{n+1}$. Suppose $v \in \text{range } T^{n+1}$. Then $v = T^{n+1} u = T(T^n u)$ for some $u \in V$. But $T^n u = T^{n+1}w$ for some $w \in V$, and thus $v = T^{n+2} w \in \text{range }T^{n+2}$. Hence $\text{range } T^{n+1} = \text{range } T^{n+2}$.)
Now the dimension has decreased by only one (from 4 to 3) in going from the first term to the last term in the following chain of length 2: $\text{range } T \supset \text{range } T^2 \supset \text{range } T^3$. Thus one of these inclusions must be an equality. Thus we have an equality for all further inclusions in the chain displayed above. In other words, $\dim \text{range } T^k = 3$ for all $k > 3$.
