How can I find the rank of $A^6$ if $A$ is a matrix of order $3$ and rank of $A^3$ is $2$? I have tried it in the following manner. We know that $$rank(XY)\ge rank(X)+rank(Y)-n$$ where $X$ and $Y$ are two matrices of order $n$ and also $$rank(XY) \le \min{\{rank(X),rank(Y)\}}$$ If we take $X=Y=A^3$ then using the above two inequalities we obtain $1\le rank(A^6)\le 2$ i.e. rank of $A^6=1 \text{ or } 2$. Is it correct at all? I have failed to obtain a definite rank of $A^6$ by the above process. Is it ok?
 A: Assume $A \in M_3(\mathbb{C})$. Let $a,b,c$ be the root of characteristic polynomial of $A$ or say eigenvalue of $A$. So $a^3,b^3,c^3$ will be eigenvalue of $A^3$. As kernel of $A^3$ is one dimensional one of the eigen value of $A^3$ will be $0$, say $c^3 = 0$. So $c=0$.
First note $a,b$ simultaneously can't be $0$, in that case characteristic polynomial of $A$ must be $x^3$ and $A^3=0$.
Now we will show neither $a$ nor $b$ is zero. Assume $b=0$ and $a\neq 0$. Using Jordan form, in that case matrix of $A$ w.r.t some basis will be one of the following form
\begin{align}
\begin{pmatrix}
a&0 &0\\
0&0&1\\
0&0&0
\end{pmatrix}, \begin{pmatrix}
a&0 &0\\
0&0&0\\
0&0&0
\end{pmatrix}
\end{align} 
In either case we will have $A^3$ has rank $1$, since in those case we will have
\begin{align}
A^3 &= \begin{pmatrix}
a^3&0 &0\\
0&0&0\\
0&0&0
\end{pmatrix}
\end{align}
So we have neither $a$ nor $b$ equal to $0$.
Now consider the case when $a=b$.  In that case matrix of $A$ w.r.t some basis will be one of the following form
\begin{align}
\begin{pmatrix}
a&0 &0\\
0&a&0\\
0&0&0
\end{pmatrix}, \begin{pmatrix}
a&1 &0\\
0&a&0\\
0&0&0
\end{pmatrix}
\end{align} 
As $a \neq 0$, In either case $A^3$ and $A^6$ both will have rank $2.
Now consider the last case $a,b$ are distinct non zero number. Then there exist a eigenbasis so that matrix of $A$  w.r.t that basis will be of the following form
\begin{align}
\begin{pmatrix}
a&0 &0\\
0&b&0\\
0&0&0
\end{pmatrix}
\end{align}
In this case also $A^3$ and $A^6$ both will have rank $2$.
If $A\in M_3(\mathbb{R})$, use the fact that $\{v_1,v_2,...v_k\} \subset \mathbb{R}^n$ is Linearly independent over $\mathbb{R}$ iff they are Linearly independent over $\mathbb{C}$. So that Real rank of $A$ and complex rank of $A$ remain same. 
