Is it possible to have $\operatorname{rank}(A^2) \lt \operatorname{rank}(A^3)?$

Question

This is a question of GATE 2021.

Let $$ be a square matrix such that $\det( − ) = $

$x^4( − )^2( − )^3$ $\DeclareMathOperator{\rank}{rank}$ , where $\det()$ denotes the determinant of a square matrix . If $\rank(A^2) < \rank(A^3)=\rank(A^4),$

then I need to find the geometric multiplicity of the eigenvalue $$ of $$.

But firstly can we have $\rank(^2)$ strictly less than $\rank(^)?$

If yes then how to proceed with this question? Thanks in advance.

Answer 1

The rank is the dimension of the image. The image of $A^3$ is contained in the image of $A^2$, as $A^3=A^2\circ A$, so $\operatorname{rank}(A^2)\geq \operatorname{rank}(A^3)$

Answer 2

No. It is well-known tht the rank of an $m \times n$- matrix $ \mathbf B$ agrees with the dimension of the image of the associated linear map $\phi_{ \mathbf B} : \mathbb R^n \to \mathbb R^m, \phi_{ \mathbf B} (x) = \mathbf B \cdot x$.

Let $\phi = \phi_{\mathbf A} : \mathbb R^n \to \mathbb R^n$. Then the rank of $\mathbf A^r$ is the dimension of $\phi^r(\mathbb R^n)$. But each linear map has the property that $\dim V \ge \dim \phi(V)$ for all linear subspaces $V \subset \mathbb R^n$. Thus $\dim \phi^{r+1}(\mathbb R^n) \le \dim \phi^r(\mathbb R^n)$ for all $r$. In other words, $\operatorname{rank} \mathbf A^{r+1} \le \operatorname{rank} \mathbf A^r$ for all $r$.