Prove a linear transformation is one-to-one and onto? 
My solution: T, linear function, is totally determined by the image any basis, for example by: 
T(1,0) and T(0,1) 
and as : (1,0) = (1/2) [ (1,1) + (1, -1) ] we get: 
T(1,0) = (1/2) [ T(1,1) + T(1 , -1) ] 
T(1,0) = (1/2) [ 1 - 2t + t + 2 t^2 ] 
---------------------------------------... = 1/2 - t /2 + t^2 = p1(t) 
and 
(0,1) = (1/2) [ (1,1) - (1, -1) ] we get: 
T(0,1) = (1/2) [ T(1,1) - T(1 , -1) ] 
T(0,1) = (1/2) [ 1 - 2t - t - 2 t^2 ] 
---------------------------------------... T(0,1) = 1/2 - (3/2)t - t^2 = p2(t) 
thus : 
T is perfect defined for any element (a,b) of R^2 
T(R^2) is the subspace (of dimension = 2) generated by p1(t) and p2(t) 
And I don;t know how to proceed after this. Thanks!
 A: The matrix for $T$ with respect to the basis $(1,x,x^2)$ is $$A=\begin{bmatrix} 1/2 & 1\\ -1/2 & -3/2\\ 1&-1\end{bmatrix}$$
The kernel of this is clearly $\{\mathbf{0}\}$, so $T$ is one-one. (This is because $Ax=Ay$ would imply $A(x-y)=0\implies x-y\in\ker A\implies x-y=0\implies x=y$)
It cannot be onto because $\mathbb{R}^2$ is $2$-dimensional whereas the target space/codomain is $3$-dimensional, so by that one theorem $\dim \text{im} \,A=\text{rank}\,A\le 2$.
A: You are exactly right that a linear function is always determined by its values on any basis. However, we can note that 
$$\left( \begin{array}{c}  
1 \\
1 \\
\end{array} \right) , \left( \begin{array}{c}  
1 \\
-1 \\
\end{array} \right).$$
is a basis for $\mathbb{R}^2$. So the image of $T$ is the span in $P_2$ of $T(\left( \begin{array}{c}  
1 \\
1 \\
\end{array} \right))$ and $T(\left( \begin{array}{c}  
1 \\
-1 \\
\end{array} \right))$. That is, the image of $T$ is $span\{1-2t, t + 2t^2\}$. Try and express this span as some set of polynomials. This will help you to determine whether $T$ is onto. However, I can give you another hint to check whether this function is onto: consider the dimension of the two vectorspaces.
Now, to see if $T$ is one-to-one, it suffices to check that $T$ has a trivial kernel. (Please comment if you don't know why this is true, or what the kernal is). However, it is immediate that $T$ has a trivial kernel, since it is non-zero on a basis. 
