An isomorphism between $\mathbb R^2$ and the space of polynomials of degree at most $1$ 
Let $V$ be the vector space of real-valued polynomials $p(x)$ of degree at most $1$. Prove that   the map
$$T:V\to\mathbb R^2,\quad  p \mapsto [p(1), p(2)]$$
is an isomorphism.

Progress
For an isomorphism, the kernel must contain the zero vector. I tried saying $p(x) = a_0+a_1x$ and then finding $p(1)$ and $p(2)$ but it doesn't seem to be doing anything for me.
 A: To show isomorphism, you have to show the transformation is a bijection where addition and scalar multiplication are preserved. 
So consider $T(ax + b) = T(cx + d)$. Then $(a, b) = (c, d) \implies a = c$ and $b = d$. So we have that $T$ is one-to-one. To show onto, consider $(a, b) \in \mathbb{R}^{2}$. Such a polynomial in $P_{1}(\mathbb{R})$ is $ax + b$. And so $T$ is onto.
Now we show additivity is preserved: $T(p(x) + q(x)) = T(ax + b + cx + d) = T((a+c)x + b + d) = (a + c, b + d) = (a, b) + (c, d) = T(ax + b) + T(cx + d)$
To show scalar multiplication is preserved: $T(3(ax + b)) = T(3ax + 3b) = (3a, 3b) = 3(a, b) = 3T(ax + b)$.
And so $T$ is an isomorphism.
Note: You can show preservation of addition and multiplication in one step, and many people do so.
Also, an alternative to showing that the mapping is bijective is to show that $ker(T) = 0$, or that $T$ is of full rank. You can do this by constructing a transformation matrix from $T$ on the standard basis of $P_{1}(\mathbb{R})$. So $T(x) = (1, 0)$ and $T(1) = (0, 1)$. And so $det [T(x) T(1)] = 1$, and so $T$ is of full rank. Thus, $T$ is bijective. Then you just have to argue linearity.
A: Note you can write p as $p=ax+b$ then the image can be written as $v=(a+b, 2a+b)$, when $v=(0,0)$ we can directly get $a=b=0$ which means if and only if $p=0$, there can be $v=0$.
So the kernel is {0}.
