# Invariant subspace

• Dec 11th 2012, 05:56 PM
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?
• Dec 11th 2012, 07:50 PM
Drexel28
Re: Invariant subspace
We have given you a lot of answers--let's see a little more work for this one.
• Dec 11th 2012, 08:52 PM
Re: Invariant subspace
Then how to calculate the invariant space?
• Dec 12th 2012, 09:21 AM
Deveno
Re: Invariant subspace
here's my idea: let U be a non-zero 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?
• Dec 12th 2012, 08:50 PM
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\}??\$
• Dec 13th 2012, 01:00 AM
Deveno
Re: Invariant subspace
suppose next that U is finite-dimensional, of dimension n. can you show that {1,x,...xn-1} 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 non-zero, 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 non-zero 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 non-zero. 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 P1.

now, use induction to show that if dim(U) = n, U = Pn-1.

now, all that is left is to show that U is not finite-dimensional, it must be all of P[x]. with a little cleverness you can use the same argument (what happens if xk is not in U, for some k in N?).