Proving a linear map to be injective and/or surjective

Question

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.

Answer 1

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.

Answer 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

Answer 3

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

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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.

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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 $$ l_1(1)=\begin{pmatrix} 1 \\ 2 \end{pmatrix} \qquad\text{and}\qquad l_1(x)=\begin{pmatrix} 1/2 \\ 2 \end{pmatrix} $$ the matrix is $$ \begin{pmatrix} 1 & 1/2 \\ 2 & 2 \end{pmatrix} $$ which has rank $2$.