# How to prove that a transformation T is linear?

Suppose $\wp_n(\Re)$ is the space of polynomials with degree $n$. Let $T:\wp_n(\Re) \rightarrow \wp_n(\Re)$ being $T(p(x)) = xp'(x)$.

How can I prove that T is a linear transformation?

I know I should prove that:

• $T(u_1 + u_2) = T(u_1) + T(u_2)$, $\forall u_1, u_2 \in \wp_n(\Re)$
• $T(\lambda u) = \lambda T(u)$, $\forall \lambda \in \Re$ and $\forall u \in \wp_n(\Re)$

However, since $T(p(x))$ is a function with $p(x)$ as the argument, I'm not sure how to deal with de $x$ on $xp'(x)$.

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And how can I find a basis for the kernel and a basis for the image of this transformation? To find the kernel I need to find $T(f) = 0$, therefore $xf' = 0$. So ker T $= \{p(x) \in \wp_n(\Re) | p'(x) = 0\}$. How can I find a basis for this subspace? – João Daniel Apr 12 '12 at 19:49

The answer to your last question is: step by step and using the definitions.

Things work better when you discard $(x)$ and use just $p$. Suppose $f,g\in\wp_n$, then: $$T(f+g)=x(f+g)'=x(f'+g')=xf'+xg'=T(f)+T(g)$$

Similarly if $\alpha$ is a scalar then we have: $$T(\alpha f)=x(\alpha f)'=\alpha xf'=\alpha T(f)$$

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Ah, sure! I had in my mind that the $x$ was related to the argument of the function. Thanks! – João Daniel Apr 12 '12 at 19:29
@João: It is also related to the argument of the function. Since you want your result to be a polynomial of degree $n$ you need to correct the fact that $p'$ is of degree $n-1$. – Asaf Karagila Apr 12 '12 at 19:38

$T(p(x)+q(x))=x((p(x)+q(x))')=xp'(x)+xq'(x)=T(p(x))+T(q(x))$ (the second equality follows from the linearity of the derivative). The other is similar.

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You have $T(p(x)+q(x))=x(p'(x)+q'(x)) = xp'(x) + xq'(x)= T(p(x))+T(q(x))$. Does this help. Follow same procedure for checking the second condition as well

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