Suppose $A$ is a square matrix, and $\operatorname{Rank}(A^{2})=3$, then can we establish any relationship to determine the rank of $A$? If it helps, $A$ has three distinct eigenvalues and it is a $4\times 4$ matrix.
Also, I want to know if there is a theorem or general statement about figuring out ranks of matrices of higher orders than the original one.
Best Answer
$A^2$ has Eigenvalue $0$ and an Eigenspace of $0$ of dimension $1$. This means $A$ has also Eigenvalue $0$
(there are several ways to see that for example $\det(A^2)=0=\det(A)\det(A)$)
This means $\text{rank(A)}<4$. Now we need to verify that $\text{rank}(A^2)\le \text{rank}(A)$
Its sufficient to show $\ker(A)\subseteq \ker(A^2)$:
Let $v\in \ker(A) \Longleftrightarrow Av=0 \Longrightarrow A^2v=A(Av)=A0=0\Longrightarrow v\in\ker(A^2)$
$\ker(A)\subseteq \ker(A^2)$
concluding $\text{rank}(A)=3$
In general its not that easy, consider a nilpotent endomorphism.
$\text{rank}(A)\ge\text{rank}(A^2)\ge\text{rank}(A^3)...$
and you usually cannot say anything about the ranks without doing some calculations
A information which can help alot is knowing the algebraic and geometric multiplicity. It can help you to understand the structure and get informations about the ranks.
For your question in the comment:
Okey first of all, the case that the rank of $A^n$ is full, which means, $A^n$ is invertable, gives you (take for example the determinant argument I stated above) that the rank of $A$ is also full. So in general only endomorphisms with no full rank are interesting. And btw if $\lambda$ is an eigenvalue of $A^n$, so is $\lambda^{1/n}$ for $A$.
So $A^n$ has an eigenvalue $0$ (the other case is already clear from above). There are two cases which are important to understand:
Is the geometric multiplicity $\ne$ algebraic multiplicity for eigenvalue $0$ of $A$, then and only then $\text{rank}(A)\ne\text{rank}(A^n)$, $n>1$. It helps to look over the theory of the jordan canonical form.