Show whether this linear transformation is one-to-one and onto. $T:P_2$ $\rightarrow$ $R^3$ is a linear transformation defined by $$T(a+bx+cx^2) =  \left[ \begin{array}{ccc}
2a-b \\
a+b-3c \\
c-a \end{array} \right]$$
This linear transformation is neither one-to-one nor onto, but I don't really get why. I have read the definition of what a one-to-one linear transformation is and what an onto linear transformation is, but I still don't fully grasp the concept. If you could show me why it is neither one-to-one nor onto that would be a big help.
 A: Hint:First note that as $dim(P_2(R))=3=dim(R^3)$ so $T$ is one one iff T is onto{A linear transformation $T:V\to V$ is one-to-one if and only if it is onto} Now show that $T$ is not one to one by showing that ker(T) is nonzero
A: Let
\begin{align*}                                                                                 
  f_0(t) &= 1 & f_1(t) &= t & f_2(t) &= t^2                                                    
\end{align*}
and note that $\beta=\{f_0,f_1,f_2\}$ is a basis for $P_2$. Let
$\alpha=\{e_1,e_2,e_3\}$ be the standard basis for $\Bbb R^3$. Then the
computations
\begin{array}{lcrcrcr}
T(f_0) & = & \color{red}2\,e_1    & + & \color{blue}1\,e_2 & + & \color{green}{-1}\,e_3 \\
T(f_1) & = & \color{red}{-1}\,e_1 & + & \color{blue}1\,e_2 & + & \color{green}0\,e_3 \\
T(f_2) & = & \color{red}0\,e_1   & + & \color{blue}{-3}\,e_2 & + & \color{green}1\,e_3
\end{array}
imply that the matrix of $T$ relative to the bases $\alpha$ and    $\beta$    is
$$
[T]_\alpha^\beta=
\begin{bmatrix}
\color{red}2 &\color{red}{-1} & \color{red}0 \\
\color{blue}1 & \color{blue}1 & \color{blue}{-3} \\
\color{green}{-1} & \color{green}0 & \color{green}1
\end{bmatrix}
$$
But $\DeclareMathOperator{rank}{rank}\rank[T]_\alpha^\beta=2$ so $T$ is neither injective nor surjective.
A: In the case of a linear transformation (as opposed to a more general function) it is relatively simple to tell whether or not it is one-to-one.

Theorem.  A linear transformation $T$ is one-to-one if and only if the equation $T(v)=0$ has the unique solution $v=0$.

In this case you can see by trial and error (or more systematically if you need to) that $c=a$, $b=2a$ gives a zero vector for your RHS, that is,
$$T(1+2x+x^2)=\pmatrix{0\cr0\cr0\cr}\ .$$
So $T$ is not one-to-one.
