Rank of some power of a given matrix $A$ is a $3\times 3$ real matrix such that rank of $A^3$ is $2$. What is rank of $A^6$? Since $A^3$ has rank $2$, $A$ also has rank $2$ because the determinant is zero. Can I decisively conclude that $A^6$ has always rank $2$? Please help. 
 A: Let me prove a more general result as follows:

Let $A$ be any $n \times n$ matrix on a field $F$, then
$$ \text{rank}(A^{n}) = \text{rank}(A^{n + 1}) = \text{rank}(A^{n + 2}) = \cdots \tag{1}$$

In view of this result, $\text{rank}(A^6) = \text{rank}(A^3) = 2$.
To prove the result, let $\text{Ker}(A^j), j = 1, 2, \ldots$ denote the kernel spaces of the linear transformations $A^j, j = 1, 2, \ldots$. Since $n = \dim(\text{Ker}(A^j)) + \text{rank}(A^j)$, it suffices to show
$$\dim(\text{Ker}(A^{n})) = \dim(\text{Ker}(A^{n + 1})) = \dim(\text{Ker}(A^{n + 2})) = \cdots. \tag{2}$$
By kernel's definition, it is easily seen
\begin{equation}
\text{Ker}(A) \subseteq \text{Ker}(A^2) \subseteq \cdots \subseteq \text{Ker}(A^k) \subseteq \cdots \subseteq F^n,
\end{equation}
which implies
$$\dim(\text{Ker}(A)) \leq \dim(\text{Ker}(A^2)) \leq \cdots \leq \dim(\text{Ker}(A^{k})) \leq \cdots \leq n.$$
Therefore there must exist some positive integer $k$, such that $\dim(\text{Ker}(A^k)) = \dim(\text{Ker}(A^{k + 1}))$, which necessarily entails $\text{Ker}(A^k) = \text{Ker}(A^{k + 1})$. We now show this in turn leads to $\text{Ker}(A^{k + 1}) = \text{Ker}(A^{k + 2})$.
Let $\alpha \in \text{Ker}(A^{k + 2})$, then $A^{k + 2}(\alpha) = A^{k + 1}(A(\alpha)) = 0$, this means $A(\alpha) \in \text{Ker}(A^{k + 1}) = \text{Ker}(A^{k})$, thus $A^k(A(\alpha)) = A^{k + 1}(\alpha) = 0$, i.e., $\alpha \in \text{Ker}(A^{k + 1})$. This shows $\text{Ker}(A^{k + 2}) \subseteq \text{Ker}(A^{k + 1})$. Together with $\text{Ker}(A^{k + 1}) \subseteq \text{Ker}(A^{k + 2})$, we have $\text{Ker}(A^{k + 1}) = \text{Ker}(A^{k + 2})$. Similar argument generates
$$\text{Ker}(A^k) = \text{Ker}(A^{k + 1}) = \text{Ker}(A^{k + 2}) = \cdots. $$
Let $k_0$ be the smallest positive integer such that the above expression holds so that
$$\text{Ker}(A) \subsetneq \text{Ker}(A^2) \subsetneq \cdots \subsetneq \text{Ker}(A^{k_0 - 1}) \subsetneq \text{Ker}(A^{k_0}) = \text{Ker}(A^{k_0 + 1}) = \cdots \subseteq F^n,$$
which implies
\begin{align}
& \dim(\text{Ker}(A)) < \dim(\text{Ker}(A^2)) < \cdots < \dim(\text{Ker}(A^{k_0 - 1})) < \dim(\text{Ker}(A^{k_0})) \\
= & \dim(\text{Ker}(A^{k_0 + 1})) = \cdots \leq n. \tag{3}
\end{align}
If $\dim(\text{Ker}(A)) = 0$, then $\text{rank}(A) = n$, in other words, $\det(A) \neq 0$, thus $\det(A^j) = (\det(A))^j \neq 0, j \geq n$, $(1)$ holds. Otherwise,
$\dim(\text{Ker}(A)) \geq 1$, under this condition, $(3)$ then implies $k_0 \leq n$ as there are $k_0 - 1$ strict inequalities in $(3)$. Consequently
\begin{equation}
\text{Ker}(A^n) =\text{Ker}(A^{n + 1}) = \text{Ker}(A^{n + 2}) =\cdots,
\end{equation}
hence $(2)$. This completes the proof of $(1)$.

If you are allowed to use Jordan form theory, then the proof can be made shorter. Here is the discussion.
Since $A^3$ is $3 \times 3$ and $\textrm{rank}(A^3) = 2$, the Jordan form $J$ of $A^3$ must be one of the following two cases:
\begin{equation*}
\begin{pmatrix}
0 & 0 & 0 \\
0 & \lambda_1 & 0 \\
0 & 0 & \lambda_2
\end{pmatrix}, \quad
\begin{pmatrix}
0 & 0 & 0 \\
0 & \lambda_0 & 1 \\
0 & 0 & \lambda_0
\end{pmatrix},
\end{equation*}
where $\lambda_1, \lambda_2, \lambda_0 \neq 0$. Therefore, $A^6 = (A^3)^2$ is similar to $J^2$, where $J^2$ is one of
\begin{equation*}
\begin{pmatrix}
0 & 0 & 0 \\
0 & \lambda_1^2 & 0 \\
0 & 0 & \lambda_2^2
\end{pmatrix}, \quad
\begin{pmatrix}
0 & 0 & 0 \\
0 & \lambda_0^2 & 2 \\
0 & 0 & \lambda_0^2
\end{pmatrix},
\end{equation*}
both of which are of rank $2$.
