Let $V$ be a $3$-dimensional vector space over $\mathbb R$ with basis $\{v_1,v_2,v_3\}$. Check in each case whether the given map is a monomorphism/epimorphism/isomorphism. (Recall that it is enough to define a linear transformation on a basis, so in each case the linear transformation $\phi$ is well-defined.)
a) $\phi\colon V\to\mathbb R^2$ is the unique linear transformation with $\phi(v_1) = (1, 0)$, $\phi(v_2) = (2, 1)$, $\phi(v_3) = (2, 2)$
b) $\phi\colon V\to V$ is the unique linear transformation with $\phi(v_1) = v_1+v_2$, $\phi(v_2) = v_2$ and $\phi(v_3) = v_3$
The question also has a c) and a d), however I'm hoping if I am able to get help with how to do a) and b) then I should be able to attempt c) and d) completely by myself. I have the definitions for monomorphism, epimorphism, and isomorphism however I don't know how to go about answering the question using those definitions. Any help would be much appreciated, thank you.
$a)$ is not a monomorphism since $\phi(2v_2-v_1)=\phi(v_3)$ and clearly $2v_2-v_1 \neq v_3$ (they form a basis). So in particular it's not an isomorphism. We can check that it's an epimorphism if the canonical basis of $\mathbb{R}^2$ is in the image. Now $(1,0)=\phi(v_1)$ and $(0,1)=\phi(v_3-v_2)$. In particular, we can write every $(a,b) \in \mathbb{R}^2$ as $(a,b)=\phi(av_1)+\phi(b(v_3-v_2))$.
$b)$ we prove that $\phi$ is an isomorphism (and then both a monomorphism and an epimorphism) by finding its inverse. Now $\phi^{-1}:V \rightarrow V$ is such that $\phi^{-1}(v_1)=v_1-v_2$, $\phi^{-1}(v_2)=v_2$ and $\phi^{-1}(v_3)=v_3.$