Let $L\colon P_3\to P_3$ be the linear operator defined as
$$L((p(x)) = p(0)x+p(1)\qquad \forall p \in P_3.$$
Looking for some help here. I'm close to the answer but I'm not quite getting the nuances of getting there.
Here's how I approached the problem:
$L(p(x)) = x(a + b(0) + c(0)) + (a + b(1) + c(1))= x(a) + (a+b+c)$
So, to find the Kernel set:
$a = 0$
$a + b + c = 0, or b + c = 0$
$a = 0$
$b = -c$
$c = \alpha$
So, $\alpha\,[0, -1, 1]$ is the Kernel, or $\operatorname{span}\{x^2, -x\}$. Answer says the span is $\operatorname{span}\{x^2- x\}$. How would I arrive here? Why are these answers different? Also, I'm not sure how the the $x$ multiplied by $p(0)$ comes into play. I assume it's important to the range. I guess the range is $P_2$ since there is only one vector in the kernel, but I'm not confident about these answers. Any help would be appreciated.
Best Answer
If $P(x)=a+bx+cx^2$, then $L\bigl(P(x)\bigr)=ax+a+b+c$. So$$P(x)\in\ker L\iff a=a+b+c=0\iff a=0\wedge c=-b.$$Therefore, $\ker L$ are the polynomials of the form $P(x)=bx-bx^2$. That is, $\ker L=\langle x-x^2\rangle$.
The range is $\bigl\langle L(1),L(x),L(x^2)\bigr\rangle=\langle x+1,1,1\rangle=\langle x+1,1\rangle=\langle x,1\rangle$.