Problem:

If $T:R^2 \to R^3$ is a linear transformation such that $T(1,1)=(1,1,1); T(1,0)=(2,0,1)$, is $T$ one to one? Is $T$ onto? What is $T(2,4)$? 

My attempt:

let an arbitrary $x$ belonging to range$(T), x=(a,b,c)$

let an arbitrary $y$ belonging to domain$(T), y=(e,f)$

then since T is linear, we have T(y) = (C1\*e+C2\*f, C3\*e+C4\*f, C5\*e+C6\*f)

and from our case we get the system of equations

1) C1+C2=1

2) C3+C4=1

3) C5+C6=1

4) C1=2

5) C3=0

6) C5=1

Solve and we get T(y)=(2e-f,f,e)

T(2,4)=(2\*2-4,4,2)=(0,4,2)

Is it one-to-one? 
yes since N(T) = 0 because the zero vector of the codomain (0,0,0) requires "e" and "f" to be 0, and therefore requires the zero vector of the domain (0,0).

Ok I'm confused as how to express that, first idk if my wording above to conclude it's one-to-one sounds mathematical or convincing... I hope someone can tell me how to word it better. And how to word it in a generalized case or how to prove in a generalized case? 

Is it onto? 
no because dim(R3)>dim(R2), and the map is one-to-one, therefore onto is impossible. 

Also for the above I kinda feel like it's gonna be impossible to unto as long as the dim of codomain > dim of domain when the map is linear, no need to one-to-one. But I'm not sure on that can someone help me out with a theorem or something? 

Yea so in general I'm always very confused doing this sort of problems. As I feel it's always very very difficult to express what I have in mind using mathematical notations. And expressiong in words are often pretty imprecise, so that's where my main difficulties in linear algebra lies. I can't express my ideas clearly! Anyone have some tips thanks!