Verification: Investigation of a linear map on surjectivity and injectivity

Question

The linear map I'm investigating is defined like this:

$l_2: P_3(\mathbb{R}) \rightarrow \mathbb{R}, l_2(p):= p'(1)$

And these are my calculations:

Injectivity:

$Let \,\, p,q \in P_3(\mathbb{R}), \,\, p(x) := 2x^3 \,\, and \,\, q(x):= x^3+x^2+x$

$\Rightarrow l_2(p) = l_2(q) = 6 \,\, but \,\, p \neq q \Rightarrow l_2 \,\, is \,\, not \,\, injective.$

Surjectivity:

$Let \,\, p \in P_3(\mathbb{R}), \,\, p(x):= ax^3+bx^2+cx+d \,\, and \,\, e \in \mathbb{R}$

Then

$l_2(p) = 3a+2b+c$

$3a+2b+c = e \,\, so \,\, for \,\, example \,\, a=\frac{e}{3}; b=\frac{e}{2} \,\, and \,\, c=e \,\, would\,\, be \,\,one\,\, possible\,\, solution\,\, s.t\,\, l_2(p) = -e.$

$\Rightarrow l_2 \,\, is \,\, surjective.$

Answer 1

For injectivity, what you have done is correct. However, your examples could have been much simpler. For example, just the constant polynomials $p_1 \equiv 0$ and $p_2 \equiv 1$ have derivatives identically zero, therefore $l_2(p_1) = l_2(p_2)$, although the two polynomials are different.

Similarly for surjectivity, you are okay, but simply note that if $p_c(x) := cx$, then $p_c'(1)=l_2(p_c)=c$.

Answer 2

Yes, this is correct. An easier way to see the lack of injectivity is that for any $f\in \mathcal{P}_3(\mathbb{R})$, $$f'(1)=\frac{d}{dx}(f(x)+c)\bigg|_{x=1}$$ for $c$ a constant. Surjectivity can be constructed using just the degree $1$ functions $g_a(x)=ax$. For instance, given any $a\in \mathbb{R}$, $g_a'(1)=a$.