find all possible vectors, with a matrix

Question

I don't know how to approach the next problem.

they give me

that are bases of a vector space V and let f: V → V be the linear transformation such that

All v∈V such that are

they gave me these 4 answers:

I did several accounts but none of them gave me similar, then I saw that it says f (v) = f (v), does it mean that the transform gives a transform? I'm very confused

Answer 1

$f(2v_1 + v_2 - v_3)$ is some vector. Let us call that $\mathbf w$. It turns out that $\mathbf w = 2v_1 + 3v_2 - v_3$ but we don't actually need to know it as anything more than "$\mathbf w$" for the solution.

Anyways, so we have an equation $f(\mathbf v) = \mathbf w$ and you will hopefully recall that the solution set of such an equation is a translate of the kernel (AKA null space) of $f$.

$$\{\mathbf v : f(\mathbf v) = \mathbf w\} = \mathbf v_0 + \ker f = \{\mathbf v_0 + \mathbf u : \mathbf u \in \ker f\} $$

where $\mathbf v_0$ is any particular solution to $f(\mathbf v) = \mathbf w$.

So for this problem we need two things:

1. we need $\mathbf v_0$
2. we need $\ker f$

For 1., we know already that $f(2v_1 + v_2 - v_3) = \mathbf w$ since that was the definition of $\mathbf w$. Therefore, we can take $\mathbf v_0 = 2v_1 + v_2 - v_3$ as a particular solution. This explains also why all but one of the proposed answers includes $2v_1 + v_2 - v_3$.

So now if $\ker f = \operatorname{Span}\{\mathbf u\}$ (which I'll leave up to you to solve), then the solutions to $f(\mathbf v) = \mathbf w$ is

$$\mathbf v = 2v_1 + v_2 - v_3 + \alpha \mathbf u$$

for some $\alpha \in \mathbb{R}$.

If you need a hint on solving for $\ker f$, notice that the first two columns of $M_{B',B}(f)$ are identical.