How to express yourself in linear algebra 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! 
 A: Your computation of $T(2,4)$ is correct. You would have worked less if you had used the fact that $(2,4)=4\times(1,1)-2\times(1,0)$.
And, yes, $\ker T=\{0\}$ and therefoe $T$ is one-to-one.
Finally, no linear map from $\mathbb{R}^2$ into $\mathbb{R}^3$ can possibly be onto, because the dimension of the image is $2$, at most, and $\dim\mathbb{R}^3=3$.
A: 
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?

Your wording is a bit confusing. You use N(T), but you don't define N. From context, it appears to mean "null space". If so, then saying that the zero of the codomain is reached only by the zero of the domain is simply restating "null space is just the zero vector". There are several conditions that are logically equivalent to being one-to-one (for linear operators), and trivial null space is indeed one of them. Another one is that every linearly independent set of vectors get sent to a linearly independent set of vectors, and this condition can be checked by finding a basis of the domain, and verifying that it is sent to a linearly independent set. You are given a basis, and the outputs are linearly independent, so that shows that T is one-to-one. If you weren't given a set of vectors that span the domain, then you would not have been able to verify that T is one-to-one; at most you could have said that the information is consistent with T being one-to-one.

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?

Suppose $d$ is the dimension of the domain, $c$ is the dimension of the codomain, $r$ is the dimension of the range, and $n$ is the dimension of the null space. "T is 1-1" is equivalent to "$n=0$" and "T is onto" is equivalent to "$c=r$". We have that $d=r+n$ An example of a proof of that is http://mathonline.wikidot.com/the-dimension-of-the-null-space-and-range . So substituting in $n=0$, we get that 1-1 is equivalent to $d=r$. Since $c \geq r$, if $d > c$ then $d>r$, so we can't have $d=r$, thus T is not 1-1. On the other hand,  $d=r+n$ means that $r \leq d$, which means that if $d < c$, then $r < c$ and so it's not 1-1. If $d=c$, then being 1-1, onto, and having $n=0$ are all equivalent.
A: HINT
Your way is correct, more simply not that
$$(2,4)=4\cdot (1,1)-2\cdot (1,0)\implies T(2,4)=4\cdot T(1,1)-2\cdot T(1,0)$$
Note that the image of $T$ is a $2$ dimensional subspace of $R^3$ therefore it can't be onto.
A: It is 1-1 as (1,1,1) and (2,0,1) are linearly independent.
It is not onto, as you say, the dimension of domain is $\mathbb R^2$ and the codomain is $\mathbb R^3$
$T(2,4) = 4T(1,1) - 2T(1,0) = 4(1,1,1)-2(2,0,1) = (0,4,2)$
A: Solving a system of linear equations as you have done, to find the image of a vector, or even a matrix representation of your linear transformation, is time consuming and not particularly illuminating. 
If you look at a linear algebra text/book, you will see that theorems are given, and procedures that make these calculations efficient and straightfirward (of course one needs to put time in to learn the material).
Similarly it is impirtant to know definitions exactly to avoid confusion.
There is for example a theorem that states that the dimension of the domain equals the dimension of the nul/kernel space plus the dimension of the column space/range (also rank of matrix representation of the lin. transf.). The proof should be easy to understand.
From this the dimension considerations become pretty trivial. 
