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What counterexample can I use to prove that ($ \mathbb{R}_{[x]}$is any polynomial):

$L :\mathbb{R}_{[x]}\rightarrow\mathbb{R}_{[x]},(L(p))(x)=p(x)p'(x)$ is not linear transformation. I have already proven this using definition but it is hard to think about example. I would be grateful for any help.

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up vote 4 down vote accepted

Let $p$ be given by $p(x)=x$ and let $q=2p$. We note that $L(p)(x)=x$ but $L(q)(x)=2x\cdot 2=4x$ But if $L$ was linear then $L(q)(x)=L(2p)(x)=2L(p)(x)=2x\neq 4x$ and so we reach a contradiction. Hence $L$ is not linear.

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Thank you , I can not imagine simpler example. – Marcin Majewski Jan 12 '14 at 17:44

$L(x+1)=(x+1)\cdot1$, whereas $L(x)+L(1)= x\cdot1+1\cdot0$.

Edit: evidently $L(2x)\neq2L(x)$.

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It's good to know the map fails not one but both properties of linear maps. – Dan Rust Jan 12 '14 at 17:56

How about this? $$ L(x + x^2) = (x+x^2)(1 + 2x) = \dots $$ $$ L(x) + L(x^2) = \dots $$

As a side note, if you want to prove that $L$ is not linear, you just have to provide one example where one of the properties fail. You say that you have proved it "using the definition" (I am not sure what you mean here), but this is not necessary. By providing a counter example, you have shown that the axioms don't hold.

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I just use that it is linear transformation when $L(\alpha u + \beta v)(x)=\alpha L(u)(x) + \beta L(u)(x)$ (and do some transformation and show that it is not equal, never mind) I will try to provide only counterexample next time, it is much easier and faster. – Marcin Majewski Jan 12 '14 at 17:52

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