# Prove $\text{rank}(A^2) — \text{rank}(A^3) \leq \text{rank}(A) — \text{rank}(A^2)$

Let $A$ be an $n \times n$ matrix over a field $K$. Prove that $\text{rank}(A^2) — \text{rank}(A^3) \leq \text{rank}(A) — \text{rank}(A^2)$

In particular - I am not aware of any relationship that must hold in general between $\text{rank}(A)$ and $\text{rank}(A^2)$.

• $\text{rank}(A^2)\le\text{rank}(A)$. This can be seen if you think about what happens when applying a linear transformation twice – Shuri2060 Jul 23 '17 at 21:33
• @Shuri2060 that makes sense to me.... are you telling me its just that easy then??? – tastykakes Jul 23 '17 at 21:35
• No - I was pointing out a relationship that holds between the two ranks. I think you might need something else to solve the question. (But perhaps using a similar concept) – Shuri2060 Jul 23 '17 at 21:36
• I'm not completely sure how to write it out formally, but I think you can see where the inequality comes from if you look at $A_1:\Bbb R^n\rightarrow\Bbb R^n$ (applying transformation $A$ on $\Bbb R^n$), $A_2:\text{Im}(A_1)\rightarrow\Bbb R^n$ (applying transformation $A$ on $\text{Im}(A_1)$) and $A_3:\text{Im}(A_2)\rightarrow\Bbb R^n$ (applying transformation $A$ on $\text{Im}(A_2)$). Then the LHS of the inequality is equal to $\ker(A_3)$ and the RHS is equal to $\ker(A_2)$. – Shuri2060 Jul 23 '17 at 21:56

## 1 Answer

Hint: Consider epimorphism $\operatorname{im}A\to \operatorname{im}A^2/\operatorname{im}A^3$ given by $x\mapsto Ax + \operatorname{im}A^3$ which factors through $\operatorname{im}A/\operatorname{im}A^2$.