
Invariant subspace
Let $\displaystyle P[x]$ be the polynomials on the field $\displaystyle P$, and let $\displaystyle D[f(x)]=f'(x)$ be the differentiating operator. Find all the invariant subspace of $\displaystyle D$.
I think the answer is $\displaystyle P_n(x)$, the polynomial with degree less than or equal to $\displaystyle n$. But I could not prove. Would you help me?

Re: Invariant subspace
We have given you a lot of answerslet's see a little more work for this one.

Re: Invariant subspace
Then how to calculate the invariant space?

Re: Invariant subspace
here's my idea: let U be a nonzero invariant subspace. since P[x] has a countable basis, so does U. pick an element of a basis for U of minimal degree. under what condition will that basis element be in D(U)? can you generalize?

Re: Invariant subspace
I do know that the polynomial with least degree is constant. However, I could not see anymore from this.
Is it all the invariant subspaces are of the form $\displaystyle span\{1,x,x^2,\cdots,x^n\}??$

Re: Invariant subspace
suppose next that U is finitedimensional, of dimension n. can you show that {1,x,...x^{n1}} must all be in U?
we've already (at least you claim you have) shown that if U has dimension 1, then it must be span{c} for c nonzero, that is: span{1}, so let's suppose dim(U) = 2.
then U = span({p(x),q(x)} for 2 LI polynomials p and q. thus D(U) = {ap'(x)+bq'(x): a,b in P}. without loss of generality assume deg(p) ≤ deg(q).
clearly p(x) = c, for some nonzero c (or else deg(p') = deg(p)  1, and then D(U) is not contained in U). so deg(q) > 0 (since p and q are assumed LI).
now suppose q'(x) is in U (which it must be, for U to be invariant). so q'(x) = ap(x) + bq(x) = ac + bq(x).
thus q'(x)  ac = bq(x).
comparing leading terms, we see that b = 0, so q'(x) = ac, a constant. hence q(x) is linear, say rx + s, and r is nonzero. since 1 = (1/c)c + 0(rx + s), 1 is in U.
since x = (s/(rc))(c) + (1/r)(rx + s), we see that {1,x} is contained in U, and since dim(U) = 2, U must be P_{1}.
now, use induction to show that if dim(U) = n, U = P_{n1}.
now, all that is left is to show that U is not finitedimensional, it must be all of P[x]. with a little cleverness you can use the same argument (what happens if x^{k} is not in U, for some k in N?).