Let $S$ be a linear operator on $W=P_3(\mathbb R)$ defined by $S(p(x)) = p(x) - \frac {dp(x)} {dx}$ Let S be a linear operator on W = $P_3(\Bbb R)$ defined by S(p(x)) = $p(x) - $$\frac {dp(x)} {dx}$.
(a) Find nullity (S) and rank (S)
(b) Is S an isophormism? If so, write down a formula for $S^{-1}$.
Please help me correct my working: 
(a) W = $P_3(\Bbb R)$ = a$x^3$+b$x^2$+c$x$+d
$p(x) - $$\frac {dp(x)} {dx}$ = a$x^3$+b$x^2$+c$x$+d-3a$x^2$-2b$x$-c = a$x^3$+ (b-3a)$x^2$+ (c-2b)x + (d-c).
Then equating S(p(x))=0, a=0, b=0, c=0, d=0. 
ker(S)=0. 
Therefore, nullity(S)=0, rank(S)=4 since there are 4 column in the matrix.
(b) For any h + e$x$ + f$x^2$ + g$x^3$, S(p(x) = h + e$x$ + f$x^2$ + g$x^3$. 
Through gaussian elimination, a = g, b = f+3g, c = e+2f+6g, d = h+e+2f+6g. S is surjective and from (a), S is injective. 
Thus, S is isomorphism.
But i got stuck here, how to find $S^{-1}$?

(a) Looks fine, but instead of saying "since there are 4 column in the matrix" (what matrix?), I would say "since $\dim P_3(\mathbb{R})=4$".
(b) You have already found $S^{-1}$, haven't you? According to your calculation, $S^{-1}(h+ex+fx^2+gx^3)=gx^3 + (f+3g)x^2 + (e+2f+6g)x + (h+e+2f+6g)$.
BTW, you can show that $S$ is an isomorphism without finding $S^{-1}$: the result in (a) that the nullity of $S$ is zero already implies that $S$ is injective. As $S$ is linear and $P_3(\mathbb{R})$ is finite dimensional, it follows that $S$ is an isomorphism. But surely the Gaussian elimination part is still useful because you have to find $S^{-1}$.
