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I have to find the kernel and range of each of the following linear operators on $P_{3}$

$a) L(p(x)) = xp'(x)$

$b) L(p(x)) = p(x) - p'(x)$

$c) L(p(x)) = p(0)x - p(1)$

I did already the $a)$ and $b)$ but in the $c)$ I have:

$L(p(x))=cx+(a+b+c)$ and I'm confuse in order to calculate it

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Welcome to MSE. I find some of your notation confusing. Is $p(x)=ax^2+bx+c$? Is it meant that $p(x)\in P_2\subset P_3$? –  Daryl Jun 30 '13 at 11:28
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Well you need $\forall x,cx + (a+b+c)=0$ so $c=0$ and $a+b+c=0$, that is $c=0$ and $b=-a$. –  xavierm02 Jun 30 '13 at 11:45
    
@xavierm02 Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Sep 15 '13 at 15:20

1 Answer 1

First of all, you probably have a mistake in what you did: if $L(p(x))=p(0)x-p(1)$ and $p(x)=ax^2+bx+c$, then $L(p(x))=cx-(a+b+c)$.

For the kernel you can proceed as xavierm02 did in the comments. A polynomial $p$ is in the kernel of $L$ if $L(p)=0$. Thus assuming that $p(x)=ax^2+bx+c$ you have that $c=0$ and $a+b+c=0$ by comparing coefficients. Now you can solve this system of linear equations to get a basis of the kernel.

For the range it is obvious that there is no polynomial with quadratic coefficient in the range. For the (at most) linear one's it should be pretty easy to find preimages either you do that generally, or you give a preimage of $x$ and $1$ and then argue with linearity, or you use the rank-nullity theorem and conclude it is two-dimensional so it has to be the whole space of (at most) linear polynomials.

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