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So the linear map I'm talking about is defined as follows:

$l_1: P_1(mathbb{R}) rightarrow mathbb{R}^2, l_1(p):= begin{pmatrix}int_{0}^{1} p(x) dx\2p(1) end{pmatrix}$

My approach is this:

For injectivity:

Let $p,q in P_1(mathbb{R}),, p(x):= ax+b ; ,, q(x) = cx+d$
Now assume that $l_1(p) = l_1(q)$

Show that this possible if and only if $p = q Rightarrow a=c land b=d$

$l_1(p) = begin{pmatrix}frac{a}{2}+b\2a+2b end{pmatrix}$

$l_1(q) = begin{pmatrix}frac{c}{2}+d\2c+2d end{pmatrix}$

Because of my assumption from the beginning I can now write:

$begin{pmatrix}frac{a}{2}+b\2a+2b end{pmatrix} = begin{pmatrix}frac{c}{2}+d\2c+2d end{pmatrix} $

Two vectors are equal if their components in the same row are equal i.e.:

$frac{a}{2}+b = frac{c}{2}+d$ and $2a+2b = 2c+2d$

But now I'm struggling how to show that this means that a = c and b = d and therefore p = q or the opposite that $p neq q$

Note:
$P_1(mathbb{R})$ denotes the set of all polynomials with degree 1 which take reals as inputs.

You second equation implies $a+b = c+d$, which is equivalent to $frac{a}{2}+b+frac{a}{2} = frac{c}{2}+d+frac{c}{2}$. By the first equation, this gets you $a = c$ and hence also $b = d$.

By the way, since you have a linear map, it actually suffices to check that only $0$ is mapped to $0$, which would make the computation a little easier.

The spaces have the same dimension, so you just need to prove one of the properties and the other will follow.

For injectivity you need, given $p(x)=ax+b$ with $l_1(p)=0$,
$$
int_0^1(ax+b),dx=frac{a}{2}+b=0qquad 2p(1)=2a+2b=0
$$
This easily implies $a=b=0$. Thus the kernel of $l_1$ is ${0}$ and you're done.

It's not difficult to prove surjectivity as well: given $begin{pmatrix} s \ t end{pmatrix}$ you need to find $a$ and $b$ such that
$$
begin{pmatrix} a/2+b \ 2a+2b end{pmatrix}=begin{pmatrix} s \ t end{pmatrix}
$$
and this is a linear system.

More abstract method: find the matrix of $l_1$ with respect to the basis ${1,x}$ and the standard basis on $mathbb{R}^2$. Since
$$
l_1(1)=begin{pmatrix} 1 \ 2 end{pmatrix}
qquadtext{and}qquad
l_1(x)=begin{pmatrix} 1/2 \ 2 end{pmatrix}
$$
the matrix is
$$
begin{pmatrix}
1 & 1/2 \
2 & 2
end{pmatrix}
$$
which has rank $2$.

You have 2 equations (Note: I didn't check if your previous calculations were right)

$$
begin{cases}
frac{a}{2} + b = frac{c}{2} + d\
2a+2b = 2c + 2d
end{cases}$$

Now, multiply the first equation by $2$, and divide the second one by $2$

$$
begin{cases}
a + 2b = c + 2d\
a+b = c + d \
end{cases}$$

Substract second equation from the first one

$$
begin{cases}
b = d\
a+b = c + d \
end{cases}$$

Substract again:

$$
begin{cases}
b = d \
a = c
end{cases}$$

And you're done