$A$ is nilpotent matrix and satisfies the equation $A^5=3A^3-2A.$ Prove $A=0.$

Question

I'm stuck at this question , I had an idea how to solve it but I think it's not good tho.

The Question :

$A$ is nilpotent and $A^5=3A^3-2A,$ prove $A=0.$
( $A^k = 0 :$ for certain $k\in \mathbb{N}$)

My try : I tried changing the equation to the form $\Rightarrow A=\frac{A^3(3I-A^2)}{2}$ and then multiplying both sides by $A^{k-3}$ and proceed with $\Rightarrow A^{k-2} = 0$.

But I'm not sure how to proceed with it now.

Answer 1

Hint If $A^n=0$ and $n \geq 2$ then, multiply your relation by $A^{n-2}$ to get $$A^{n+3}=3A^{n+1}-2A^{n-1}$$

Show that $A^{n-1}=0$. Repeat.

P.S. If you learned about the minimal polynomial, then the following is a shorter alternate solution: The minimal polynomial of $A$ must divide some $X^n$ and $X^5-3X^3+2X$, and hence it must divide their GCD which is $X$.

Answer 2

Hint If $A$ is nilpotent but $A \neq 0$, then there is some $v \in \Bbb F^n$ such that $A v \neq 0$ but $A^2 v = 0$. Now, multiply both sides of the given equation, $$A^5 = 3 A^3 - 2 A ,$$ by $v$.

Answer 3

Here is another approach:

$$A(\underbrace{A^4 - 3A^2}_{B:=}+2I)=O$$

Note that $B$ is nilpotent because $A$ is nilpotent. Hence, the only eigenvalue of $B$ is $ \lambda = 0$. Hence, $B+2I$ is invertible (otherwise $\lambda = -2$ would be an eigenvalue of a nilpotent matrix which is a contradiction).

It follows by multiplying by $(B+2I)^{-1}$ from the right that $A =O$.