# Linear Transformations - prove that $ST = 2TS$

There is linear transformation $S,T : \mathbb{R}_n[X] \to\mathbb{R}_n[X]$ . $T: T( p(x))=p(2x)$. $S:S(p(x)) = p'(x)$?

Where $\mathbb{R}_n[X]$ denotes polynomials with power up to $n$.

prove that $ST = 2TS$.

how can one prove this equality and does it exist ? tried using standard polynomial equation of $1 +x+x^2$ did not work. any help would be appreciated.

• It would be a good idea to have your question be self-contained in the body of your question and your title be descriptive. Placing half the question (only) in the title is poor practice. – Eric Towers Mar 27 '16 at 0:35
• Does your "$R$" mean "$\Bbb{R}$" ($\Bbb{R}$)? What is "$R[x]n$"? (Maybe you mean $(\Bbb{R}[x])^n$, the set of $n$-element vectors with elements from the polynomials on $x$ with real coefficients? If so, do you really mean to take component-wise vector derivatives?) – Eric Towers Mar 27 '16 at 0:40
• sorry if it wasnt clear, but by R[x]n i meant polynomials with power up to n. – Ancient Dragon Mar 27 '16 at 0:43

You have, for $p(x)=\sum_{k=0}^na_kx^k$, $$STp(x)=S\left(\sum_{k=0}^na_k(2x)^k\right)=S\left(\sum_{k=0}^n2^ka_kx^k\right) =\sum_{k=1}^nk\,2^ka_kx^{k-1}.$$ And $$TSp(x)=T\left(\sum_{k=1}ka_kx^{k-1}\right)=\sum_{k=1}^nka_k(2x)^{k-1}=\sum_{k=1}^nk\,2^{k-1}a_kx^{k-1}.$$ The two expressions differ by a factor of $2$, and $p$ was arbitrary. So $$ST=2TS.$$
• This is more targetted at OP: In what way is this $p(x)$ an element of "$R[x]n$"? – Eric Towers Mar 27 '16 at 0:42
This is a really cute problem! When you think about it some more it becomes clear that the seemingly curious fact $ST=2TS$ is more or less obvious, and it should be possible to provide an abstract nonsense proof. Here it is:
Write $p=\bigl(x\mapsto p(x)\bigr)$. Then $Tp=\bigl(x\mapsto p(2x)\bigr)$, and the chain rule then gives $$STp=\bigl(x\mapsto p(2x)\bigr)'=\bigl(x\mapsto 2p'(2x)\bigr)\ .\tag{1}$$ On the other hand, $Sp=\bigl(x\mapsto p'(x)\bigr)$, so that $$TSp=\bigl(x\mapsto Sp(2x)\bigr)=\bigl(x\mapsto p'(2x)\bigr)\ .\tag{2}$$ Since $(1)$ and $(2)$ hold for all differentiable $p$ ($p$ doesn't have to be a polynomial) it follows that $ST=2TS$.