Show that a linear transformation $T$ is one-to-one Problem:
Consider the transformation $T : P_1 -> \Bbb R^2$, where $T(p(x)) = (p(0), p(1))$ for every polynomial $p(x) $
in $P_1$. Where $P_1$ is the vector space of all polynomials with degree less than or equal to 1.
Show that T is one-to-one.
My thoughts:
I'm a little stuck on this one. I've already proven that T is a linear transformation.
I know that to prove that T is one to one, for any $p(x)$ and $q(x)$ in $P_1$, 
 (1) $T(p(x) + q(x)) = T(p(x)) + T(q(x))$
(2) $T(cp(x)) = cT(p(x))$
I'm going about it in a proof oriented way:
Let p(x), q(x) be a polynomial in P1 such that T(p(x)) = T(q(x)). Let $c$ be a real number.
(1) $T(p(x)) = T(q(x))$ -> $(p(0), p(1)) = (q(0), q(1))$
This means $p(0) = q(0)$ and $p(1) = q(1)$. I can't figure out how to prove that $p(x) = q(x)$ from knowing $p(0) = q(0)$ and $p(1) = q(1)$.
Am I going about this in the wrong way? I know that another way to prove that $T$ is one to one is if you can find that the standard matrix $A$ for $T$ is invertible, but I can't figure out how to get a standard matrix for the linear transformation $T$ in this case.
Any help or hints would be appreciated.
 A: First note that if $p(x)$ and $q(x)$ have degree at most $1$, then $d(x)=p(x)-q(x)$ has degree at most $1$ (or is the zero polynomial, depending on how you define the degree of the zero polynomial). Thus if $p(0)=q(0)$ and $p(1)=q(1)$, then $d(0)=d(1)=0$. Thus $d(x)$ has two roots, but has degree at most $1$. Thus $d(x)$ is identically zero, and $p(x)=q(x)$.
A: There are lots of ways to do this. To finish your proof, note that 
\begin{align*}
f(t)&=a_0+a_1t&g(t)=b_0+b_1t
\end{align*}
satisfy $T(f)=T(g)$ if and only if $f(0)=g(0)$ and $f(1)=g(1)$. But $f(0)=g(0)$ and $f(1)=g(1)$ if and only if 
\begin{align*}
a_0 &= b_0 & a_0+a_1&=b_0+b_1\tag{1}
\end{align*}
Now, (1) holds if and only if $a_0=b_0$ and $a_1=b_1$. Hence $T(f)=T(g)$ if and only if $f=g$ as required.
Alternatively, put
\begin{align*}
f_0(t) &= 1 & f_1(t) =t
\end{align*}
and note that $\{f_0,f_1\}$ is a basis for $P_1$ while 
\begin{align*}
e_1&=\begin{bmatrix}1\\0\end{bmatrix} &
e_2&=\begin{bmatrix}0\\1\end{bmatrix}
\end{align*}
is a basis for $\Bbb R^2$. The computations
\begin{array}{rcrcrcrcrc}
T(f_0) & = & \color{red}1\,e_1 &+& \color{red}1\,e_2 \\
T(f_1) &=& \color{green}0\,e_1&+&\color{green}1\,e_2
\end{array}
show that the matrix of $T$ relative to these bases is
$$
[T]=\begin{bmatrix}\color{red}1 & \color{green}0 \\ \color{red}1&\color{green}1\end{bmatrix}
$$
But $[T]$ is invertible so $T$ is one-to-one.
A: Yet another way is with the useful result that a map  $L$ between alebraic structures is injective iff the kernel of $L$ is trivial. This means that
$T(p)=0$ iff $p==0$, i.e., iff $p$ is the $0$ polynomial.
Assume then that $T(p)=(p(0),p(1))=(0,0)$. This means, for $p=ax+b$ : $$p(0)=a(0)+b=b=0 $$ ,together with, $$p(1)=a+b=a+0=0$$ It follows that $p==0$ necessarily, so the kernel of $T$ is trivial an the map is then injective..
