How to determine if a Linear Transformation is 1-1, onto from $P_2$ to $\mathbb R^3$ I had a hard time picking a title for this post, so I apologize if it does not encompass everything i'm trying to ask here. 
I've been studying linear transformations of vector spaces for my Linear Algebra class and I really, really struggle to understand them. I've asked my professor and tutors to help explain them in a way I can understand I still don't understand them in a lot of cases. Often people on this site do a great job explaining these things in a way I can understand, so i'm bringing the question here in hopes that someone can help me figure it out.  
The following question is from my textbook, and is one i've struggled to understand more than most. The question is:
"Determine whether this Linear Transformation is 1-1, onto:"
$T:\mathscr P_2 \mapsto \mathbb R^3$ defined by:
$T(a + bx +cx^2)$ = $ \left[ 
                      \begin{matrix}
                    2a  -  b \\
                    a  +  b  -  3c \\
                    c  -  a \\
                    \end{matrix}
                    \right] $
Everyone I ask always starts by finding the $ker(T)$, so I guess i'll start there as well:
$\mathbf 1.)$ 
$ker(T) = T(a + bx + cx^2)$ = 
                      $ \left[ 
                  \begin{matrix}
                2a  -  b \\
                a  +  b  -  3c \\
                c  -  a \\
                \end{matrix}
                \right] $ = $\left[ \begin{matrix} 0\\ 0\\ 0\\ \end{matrix}               \right]$ $\implies$
$ \mathbf 2.)$ 
$2a-b = 0 \implies b=2a$
$a+b-3c = 0 \implies$ ...? I'm not entirely sure here
$c-a=0 \implies a=c$ 
$\implies$
$ \mathbf 3.)$ 
$\{a+bx+cx^2 : a=c, b=2c\}$ $\implies$
$\{ax + bx + cx^2: cx^2\}$
$= c(1 + x +x^2)$
$\implies ker(T) \neq \{0\}$
$\implies$ T is not 1-1
$\mathbf4.)$ As for onto, the following explanation was given on Chegg, but does not make any sense to me:
"Since the dimensions of the domain space and co-domain space are the same, T is not onto."
The book gives the answer as "neither 1-1 or onto". I can understand why it's not 1-1, but I do not understand why its also not onto. I attempted to make sense of this using the rank-nullity theorem, although I don't have confidence in my work:
$\mathbf 5.)$ 
$range(T)$ =  $\begin{bmatrix}
           2a  -  b \\
           a  +  b  -  3c \\
           c  -  a \\
           \end{bmatrix}$
= a$\begin{bmatrix} 2 \\ 1 \\-1 \\ \end{bmatrix}$ + b$\begin{bmatrix} -1 \\ 1 \\0 \\ \end{bmatrix}$ + c$\begin{bmatrix} 0 \\ -3 \\1 \\ \end{bmatrix}$
= span  { $\begin{bmatrix} 2 \\ 1 \\-1 \\ \end{bmatrix}$ , $\begin{bmatrix} -1 \\ 1 \\0 \\ \end{bmatrix}$ , $\begin{bmatrix} 0 \\ -3 \\1 \\ \end{bmatrix}$ }
But these 3 vectors are not linearly independent( the first vector can be written as a linear combo of the second 2 vectors) and thus not a basis for $range(T)$.  
If i'm not mistaken, 
$rank(T) = dim(range(T))$ and 
$nullity(T) = dim(ker(T))$ and
$rank(T) + nullity(T) = dim(V)$
Since the $dim(\mathscr P_2) = 3$, but the $rank(T) = 2$, this is why the transformation is not onto?
I'm confusing myself the more I type, so here are my major questions:
$\mathbf a.)$ How was the kernel calculated in steps 2 and 3? I can see the answer, but no matter how many times it's been explained to me I still don't understand it. I'm hoping someone can clarify in a way I understand. If it seems trivial I apologize, but i'm just not "getting it". 
$\mathbf b.)$ Why is this linear transformation not onto? Is my work in part 5 correct? In general, is there a easier method for verifying if something is onto? 
$\mathbf c.)$ What would a basis for the kernel look like in this case? This is mostly to satisfy my own curiosity/understanding of the topic. 
$\mathbf d.)$ This is more "in general", but if anyone has tips for how to work linear transformations between 2 different vector spaces, I would love to hear them. I REALLY struggle with these, but perhaps hearing generalized advice on how to approach these problems could help me in the future. 
It took me about an hour to type this whole thing out with proper syntax, so please don't tell me to "look it up". I've exhausted every resource available to me trying to understand problems like this one, and these types of problems in general. I'm seeking help from this site because i'm looking for a different perspective. 
I sincerely thank anyone who takes the time to try and help me understand this question, and this concept in general. Thank you so much in advance. 
 A: Finding the kernel first is a good idea. (though actually what we need is just to find the nullity)
Construct the augmented matrix (note that the last column is actually not needed):
$$\left[\begin{array}{ccc|c}2 & -1 & 0 & 0\\ 1 & 1 & -3 & 0 \\ -1 & 0  & 1 & 0\end{array}\right]$$
And find its corresponding RREF:
$$\left[\begin{array}{ccc|c}1 & 0 & -1 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0  & 0 & 0\end{array}\right]$$
See that the third column is not a pivot column, let $c=t$. from the second equation, $b-2c=0$, hence $b=2t$, from the first equation, $a-c=0$, hence $a=t$. 
Hence solution to the system $Tx=0$ is $(a,b,c)=t(1,2,1)$.
A basis of the kernel would be $\{ 1+2x+x^2\}$.
As mentioned at the start, the goal is to find the nullity. From the RREF, we can see that $rank(T)=2$ and the the number of columns is $3$, hence $nullity(T)=1 > 0$, hence it is not injective.
Your codomain is $\mathbb{R}^3$, however as you can see from the RREF, $rank(T)=2 < 3$, it does not have enough vectors to span $\mathbb{R}^3$.
A: From what you have in $(2)$:
$b=2a$ , $c=a$ , and $a+b-3c=0$.  Substitute the third equation using the first two, and get that $0=0$.  So, you have some degree of freedom.  Choose $a=1$, say, and compute $b=2$, and $c=1$, to get that $1+2x+x^2\in \ker T$.  Go ahead, apply $T$ to this polynomial and verify that it is $0$.  (There is a mistake in $(3)$, not sure if you made the mistake typing or if where you're getting your information made a mistake.)  Indeed, it's a mistake since $T(1+x+x^2)=\begin{pmatrix}-1\\-1\\0\end{pmatrix}$.  Since $1+2x+x^2\in\ker T$, and this is a non-zero polynomial, $\dim(\ker T)\ge 1$.  Using your rank nullity formula, what is the highest that $\dim(\text{Range} T)$ can be?  So, can this linear transformation be onto?  If it is onto, then the dimension of the range better be $3$ since this is the dimension of $\mathbb{R}^3$.
To answer your question (c), we investigate our degree of freedom from earlier a little bit more in depth.  We showed that any vector $v\in\ker T$ has the form $\begin{pmatrix}a\\2a\\a\end{pmatrix}$ for any $a$.  Factor an $a$ out of this and get that anything in the kernel has the form $a+2ax+ax^2=a(1+2x+x^2)$.  So, it appears that $\{1+2x+x^2\}$ is a basis for $\ker T$, the nullspace of $T$ since it spans the nullspace and is linearly independent.
To attempt to answer your question (d):  I'll just say that any two finite dimensional vector spaces are really the same vector space but look different.  Take $\mathcal{P_2}$ that you were working with here.  We can see that $\{1,x,x^2\}$ is a basis for this space.  Associating $(1,0,0)$ with $1$, $(0,1,0)$ with $x$, and $(0,0,1)$ for $x^2$ allows us to see $\mathcal{P_2}$ as $\mathbb{R}^3$.  In this light you can read the basis for the kernel above as the vector $\begin{pmatrix}1\\2\\1\end{pmatrix}$.  Really, just keep working problems involving two different vector spaces and you'll get better at it.  Ask questions like you're doing when you get stuck, but remember to push until you just can't get anywhere.
Hope this helps a bit.  Best of luck with your studies!
A: I take it you have already seen the algebraic formalization for linear transformations. Let me provide a "backward" definition that I often find useful, but requires me to first specify a basis.
One can think of a linear transformation as a transformation $T:V \to W$ so that  $T(0)=0$ but also has a very "rigid" geometric structure. In particular, if $\{v_1,\dots,v_n\}=S$ is a basis for $V$, then the linear transformation $T$ is a function that extends uniquely from $S$ to all of $V$. By this, I mean that once you specify the values of $T$ on $S$, this determines the linear transformation entirely. 
One can think of "injective" linear transformations as those that are good enough so that $T(v_1),\dots,T(v_n)$ forms a basis for the image of  $T$. This means exactly that the image of each basis vector is again linearly independent. Another way of saying this is that $a_1T(v_1)+ \dots a_nT(v_n)=0 \implies a_1=\dots=a_n=0$. But, using linearity, we see that this is finding exactly the kernel of $T$:
$a_1T(v_1)+ \dots a_nT(v_n)=0=T(a_1v_1+ \dots a_nv_n)=0$.
So, it is enough to find the kernel (here is the formal thing).
In your problem, note that you are studying the space of polynomials, and so we can take a basis $\{1,x,x^2\}$. To understand $T$, it is enough to understand where these guys are sent. $$1 \mapsto (2,1,-1)$$ $$x \mapsto (-1,1,0)$$ $$x^2 \mapsto (0,-3,1)$$
just by following the rules prescribed. Hence, we can form the matrix
$$T=\begin{pmatrix} 2 &-1 & 0 \\1 &1 &-3 \\-1 & 0 & 1
\end{pmatrix}
$$
Notice that this is actually our linear transformation since, for example, $T(x)=(-1,1,0)$.
Row-reducing, we see that the kernel is nontrivial.
Your first method was okay as well though, you just had to note that you found a nontrivial solution to $T(x)=0$.
