Rank of $A^n$ and $A^{n+1}$ Suppose $A$ is a $n\times n$ real matrix.
Then is it always true that Rank($A^n$) = Rank ($A^{n+1}$) for a matrix $A$?
This doubt came while solving the attached question:
If A is a 10×10 real matrix, then which of the following is true:

*

*rank($A^8$)=rank($A^9$)

*rank($A^9$)= rank($A^{10}$)

*rank($A^{10}$)=rank($A^{11}$)

*rank($A^8$)=rank($A^7$)

Attempt: I can take a nilpotent matrix of maximal index 10 for a matrix of order 10 and therefore option 1,2,4 are rejected but option 3 still is correct.
So I thought is there any generalization or I analyzed the question incorrectly?Please throw some light.
 A: Yes, it is always true.  One argument is to use Jordan form: a matrix $A$ is necessarily similar to a block diagonal matrix of the form
$$
\pmatrix{M & 0\\0& N},
$$
where $M$ is invertible and $N$ is nilpotent. Since $N^n = 0$, the ranks of $A^n,A^{n+1}$ must be equal to the ranks of
$$
\pmatrix{M & 0\\0& N}^n = \pmatrix{M^n & 0\\0& 0}, \quad 
\pmatrix{M & 0\\0& N}^{n+1} = \pmatrix{M^{n+1} & 0\\0& 0},
$$
which is to say that the rank of $A^n$ and of $A^{n+1}$ is simply the size of $M$.  So, the ranks are indeed equal.
A: We have
$$
n = \operatorname{rank}(A^0) \ge \operatorname{rank}(A^1) \ge \operatorname{rank}(A^2) \ge \cdots 
$$
This sequence cannot decrease beyond $0$ and each time it decreases it decreases by at least $1$. Therefore, the sequence is constant after $n$ terms. (It may happen earlier of course.)
A: If you take a nilpotent matrix, then if the $\mathrm{Rank}(A^i)=n$ for some $i\leq n$, then $\mathrm{Rank}(A^{i+1})=n$.
Now using the fact that for nilpotent matrix we have $A^i=0$ for some $i\leq n$ you can understand why $\mathrm{Rank}(A^i)=n$ for some $i\leq n$ holds, and then using linear independence of the basis prove that $\mathrm{Rank}(A^{i+1})=n$.
