Showing that $(\phi^2(\vec x),\phi(\vec x),\vec x)$ is a free family, $\phi^2(\vec x)\neq \vec 0, \phi^3(\vec x)= \vec 0$

Question

Let be $A\in M_4(\mathbb C)$ such that with $O_4$ representing the null matrix of rank $4$,

$$A^2\neq O_4 \mbox{ and } A^3=O_4$$

Let be $\phi$ the endomorphism of $\mathbb C^4$ canonically associated to $A$.

1. How to justify the existence of $\vec x\in\mathbb C^4$ such that $\phi(\vec x)\neq\vec 0$
2. How to show that $(\phi^2(\vec x),\phi(\vec x),\vec x)$ is a free family?

I had no idea, I'm very bad to be creative with these kind of questions.

1. As far as $A^2\neq O$ then $A\neq O$ therefore $\phi\vec x\neq \vec0

2. this one is not easy for me: $\phi^2(\vec x)$ is free in the basis $(\phi^2(\vec x),\phi(\vec x),\vec x)$ iif $¬\exists\lambda,\mu, \phi^2(\vec x)=\lambda\phi(x)+\mu\vec x$ I supposed that it existed to find a contradiction but didn't found it.

   $$Ax^2-\lambda A\vec x -\mu \vec x=\vec 0$$

$$\Leftrightarrow (A^2-\lambda A-\mu I)\vec x=\vec0$$

....

Answer 1

reductio ad absurdum for both question.

1. What if $\phi(\vec x) = \vec0$ for all $\vec x$ ?
2. What if there is some $\lambda_i$ such $\lambda_2\phi^2(\vec x)+\lambda_1\phi(\vec x)+\lambda_0\vec x = \vec0$

The first one is also relevant to show there is a $\vec x$ such $\phi^2(\vec x) \neq \vec0$.

For the second one, you may want to make $\phi^3(\vec x)$ appear.

By applying $\phi$ to the equality.

You know something about it but don't use it yet.

One of the criticisms that I have about school system, when you are stuck, just look what they give to you and see what you don't have used, the solution is probably here.

[Solution]

First part:

If $\phi^2(\vec x) = \vec0$ for all $x$, $A^2 = O_4$ that is a contradiction.
So there is a $\vec x$ such $\phi^2(\vec x) \neq \vec0$.
Obviously $\phi(\vec x) \neq \vec0$ and $\vec x \neq \vec0$.

Second part:

If there is $\lambda_i$ such:
$\lambda_2\phi^2(\vec x)+\lambda_1\phi(\vec x)+\lambda_0\vec x = \vec0$
By applying $\phi$:
$\lambda_2\phi^3(\vec x)+\lambda_1\phi^2(\vec x)+\lambda_0\phi(\vec x) = \vec0$
But $\phi^3(\vec x) = \vec0$ so:
$\lambda_1\phi^2(\vec x)+\lambda_0\phi(\vec x) = \vec0$
By applying $\phi$:
$\lambda_1\phi^3(\vec x)+\lambda_0\phi^2(\vec x) = \vec0$
But $\phi^3(\vec x) = \vec0$ so:
$\lambda_0\phi^2(\vec x) = \vec0$
That is a contradiction. There is no such $\lambda_i$ and you got a free family.