Vector Space - Invariant Subspaces

Let V be a vector space (over ) of all polynomials with real coefficients whose degrees do not exceed ( is a nonnegative integer). Let be a linear transformation that sends every polynomial to its derivative. Find all vector subspaces of that are invariant with respect to . (A vector subspace of a vector space is said to be invariant with respect to a mapping if .)

I am not even sure where to start....

Re: Vector Space - Invariant Subspaces

Let be a subspace. Let be a polynomial of maximum degree (no polynomials in have degree greater than degree of p). We know for some integer with . Then . So, since , it must be that . Keep doing this. Obviously, . And, . Claim: is a basis for (where 0 is the polynomial ).

Re: Vector Space - Invariant Subspaces

Since [tex]p(x)=a_nx^n + ... + a_1 + a_0[\tex] has degree of at most n, let [tex]Z=[\tex]power set of [tex]{p(x) in V: degree (p(x)) less than or equal to n}[\tex]. Then Z has to be the set of all phi-invariant subspaces of V.

Re: Vector Space - Invariant Subspaces

Quote:

Originally Posted by

**abscissa** Since

has degree of at most n, let

power set of

. Then Z has to be the set of all phi-invariant subspaces of V.

Most of the sets of Z will not be vector spaces. In fact, almost none of them will be. There will be a total of -invariant subspaces of .

Re: Vector Space - Invariant Subspaces

Quote:

Originally Posted by

**SlipEternal** Most of the sets of Z will not be vector spaces. In fact, almost none of them will be. There will be a total of

-invariant subspaces of

.

Sorry, that was a mistake. There will be a total of -invariant subspaces of . I forgot the trivial subspace.

Re: Vector Space - Invariant Subspaces

Quote:

Originally Posted by

**SlipEternal** Let

be a subspace. Let

be a polynomial of maximum degree (no polynomials in

have degree greater than degree of p). We know

for some integer

with

. Then

. So, since

, it must be that

. Keep doing this. Obviously,

. And,

. Claim:

is a basis for

(where 0 is the polynomial

).

I follow the logic of defining the subspace and how etc. are in , but I do not understand what the claim means or its significance?

Re: Vector Space - Invariant Subspaces

I will give you a bigger hint. The standard basis for is . Through linear combinations of those elements, you can get any polynomial of degree less than or equal to n. The dimension of is the number of elements in any of its bases (obviously is an -dimensional vector space). There is only one -dimensional subspace of . How many -dimensional subspaces are there? How many of them have polynomials with maximum degree ? If that claim is true, it tells you the dimension of . It also gives you a very good idea of which subspaces of are invariant under .

Re: Vector Space - Invariant Subspaces

Quote:

Originally Posted by

**SlipEternal** I will give you a bigger hint. The standard basis for

is

. Through linear combinations of those elements, you can get any polynomial of degree less than or equal to n. The dimension of

is the number of elements in any of its bases (obviously

is an

-dimensional vector space). There is only one

-dimensional subspace of

. How many

-dimensional subspaces are there? How many of them have polynomials with maximum degree

? If that claim is true, it tells you the dimension of

. It also gives you a very good idea of which subspaces of

are invariant under

.

I am not sure how to tell how many subspaces there are of -dimension. Also how do I tell the maximum degree of its elements?

Re: Vector Space - Invariant Subspaces

The answer is, there are potentially infinite -dimensional subspaces of , but there is only one -dimensional subspace where the maximum degree of its elements is .

Re: Vector Space - Invariant Subspaces

Re: Vector Space - Invariant Subspaces

No, I am saying that is the only -dimensional subspace of that is invariant. Next, if you look for dimensional subspaces of that are invariant, there will only be 1. Etc.

Re: Vector Space - Invariant Subspaces

Re: Vector Space - Invariant Subspaces

Pretty close, but itself is a -invariant subspace of itself, as is the trivial subspace (the space consisting only of the zero polynomial).

Re: Vector Space - Invariant Subspaces

Quote:

Originally Posted by

**SlipEternal** Pretty close, but

itself is a

-invariant subspace of itself, as is the trivial subspace (the space consisting only of the zero polynomial).

Right I knew that would be a -invariant subspace of itself, which is why I said n -invariant subspaces. But I did forget the trivial subspace so I guess there are total.

Re: Vector Space - Invariant Subspaces

Here are the bases for the -invariant subspaces: . So, there are total. (The empty set is the basis for the trivial subspace.)