Linear transformation in $\mathcal{P}$ Let $T:\mathcal{P}\rightarrow \mathcal{P}$ be the function that maps each polynomial to its derivative
$$T(f)=f'$$
Let $S:\mathcal{P}\rightarrow \mathcal{P}$ be the function that multiplies its input polynomial by $x$.
$$S(f)=g \ \ \ \ \textrm{where} \ \ \ \ \ g=xf(x)$$
What I want to do is prove that they are both linear transformations, find $\textrm{ker}\ T$, $\textrm{ker}\ S$ and $\textrm{im}\ T$, $\textrm{im} \ S$. Determine whether they are injective and/or surjective and what $(T\circ S)-(S\circ T)$ does.
Proving that they are linear transformations by itself is easy by just utilizing the basic properties of differentiation, however finding the Kernel and Image aren't as easy for me. 
$T$ I am thinking that $$p_n(x)=\sum_{n=0}^k a_nx^n $$
Which gives $$\frac{d}{dx}p_n(x)=\sum_{n=0}^k na_nx^{n-1}$$
So $p_n(x)$ is spanned by $\{x^n,x^{n-1},x^{n-1},\cdots ,1\}$ where the image is spanned by $\{x^{n-1},x^{n-2},\cdots 1\}$
For $S$ I am completely lost. I know how to compute $(T\circ S)-(S\circ T)$  when I have a concrete matrix, but it becomes much harder to visualize how I do it in $\mathcal{P}$. Just for clarity, $\mathcal{P}$ is the vector space of all polynomials of an aribtrary degree. If someone could show me how it is done, and how to approach it I think I would learn a lot. 
While I am at it, how would I do the exact same things for $T$ if it was from $T:\mathcal{C}^\infty(\mathbb{R})\rightarrow \mathcal{C}^\infty(\mathbb{R})? $
 A: $T(S(f)) (x)-S(T(f))(x)=(xf(x))'-xf'(x)=[xf'(x)+f(x)]-xf'(x)=f(x)$. So $T \circ S-S \circ T$ is the identity function $f \to f$. 
$T$ is surjective but not injective. $S$ is injective but not surjective. 
A: The only functions from $\mathbb R$ into $\mathbb R$ whose derivative is the null function are the constant functions. Therefore, $\ker T$ is the set of all constant polynomials.
On the other hand, every polynomial function has a primitive which is also a polynomial function. Therefore, $T$ is surjective.
And, clearly, $S$ is injective. In other words, $\ker S=\{0\}$.
Finally,\begin{align}T\bigl(S\bigl(P(x)\bigr)\bigr)-S\bigl(T\bigl(P(x)\bigr)\bigr)&=T\bigl(xP(x)\bigr)-S\bigl(P'(x)\bigr)\\&=xP'(x)+P(x)-xP'(x)\\&=P(x).\end{align}
A: $$Ker(S)=\{f \in \mathscr{P}: xf(x)=0\}$$ So if $f\in Ker(S)$ then $xf(x)=0$ the zero polynomial hence $f$ must have been the zero polynomial to begin with so $Ker(S)=0$ and S is injective.
S cannot be surjective since the constant polynomials do not lie in its range.  
A: $\textrm{ker}\ T$ is the set of Polynomials whose derivative is $0$, i.e. constant polynomials.
$\textrm{im}\ T$ is the set of Polynomials which are the derivatives of some other polynomials, which is $\mathcal{P}$ because each polynomial is the image of its primitive.
$\textrm{ker}\ S$ is $0$, because $xf(x) = 0 \implies \sum_{n=0}^k a_n x^{n+1} = 0 \implies a_n = 0$ for all n.
$\textrm{im}\ S$ is simply $x.\mathcal{P}$.
$(ToS - SoT)(P) = (XP)' - XP' = P \implies ToS - SoT = Id$
