Does there exist a linear operator $A:\Bbb F^5\to\Bbb F^3$ s.t. $$\operatorname{Im} A=\left\{(x_1,x_2,x_3)\in\Bbb F^3,|x_1|\leqslant 1\right\}\quad?$$
Depending on the answer, give an example or explain why such an operator doesn't exist.
Let $y=(y_1,y_2,y_3)\in\operatorname{Im} A$. Then $|y_1|\leqslant 1$
Checking whether or not $Im$ is closed under scalar multiplication:
Let $\alpha\in\Bbb F$ be arbitrary.
$$\alpha x=(\alpha x_1,\alpha x_2,\alpha x_3)$$ $$\alpha x\in \operatorname{Im} A\implies|\alpha x_1|=|\alpha|\cdot|x_1|\leqslant 1$$ $$x_1\ne 0\implies|\alpha|\leqslant 1$$ so, $\operatorname{Im} A$ is closed under scalar multiplication only for $x_1=0$. The same holds for addition. $\implies \dim (\operatorname{Im})\le 2\implies\operatorname{Ker} A\geqslant 3$, $$\operatorname{Im} A=\left\{(0,x_2,x_3)\in\Bbb F^3\right\}$$ We know that $L(\Bbb F^5,\Bbb F^3)\ne\emptyset$ and the trivial $0$-operator definitely satisfies the constraints. I took canonic bases $e$ for $\Bbb R^5$ and $f$ for $\Bbb R^3$ and thought of writing a composition of projectors $B,C,D$: $$B:\Bbb R^5\to\Bbb R^4$$ $$C:\Bbb R^4\to\Bbb R^3$$ $$D:\Bbb R^3\to\Bbb R^2$$ $$A=DCBx$$ So as to get: $yz$-plane $\equiv\; \operatorname{Im} A$. but it didn't work due to the size of $[A]_e^f\in M_{3\times 5}$
How can I construct, if possible, such a nontrivial operator?
You showed that $\{ (x_1, x_2, x_3) \in \mathbb F^3: |x_1| \le 1\}$ is not closed under scalar multiplication, and therefore is not a vector space. Since the image of a linear operator is a vector space, that proves there is no such operator.