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Math Help - Vector Space - Invariant Subspaces

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    Vector Space - Invariant Subspaces

    Let V be a vector space (over \mathbb{R}) of all polynomials with real coefficients whose degrees do not exceed n ( n is a nonnegative integer). Let \phi : V \rightarrow V be a linear transformation that sends every polynomial to its derivative. Find all vector subspaces of V that are invariant with respect to \phi. (A vector subspace W of a vector space V is said to be invariant with respect to a mapping \phi : V \rightarrow V if \phi(W) \subseteq W.)

    I am not even sure where to start....
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    Re: Vector Space - Invariant Subspaces

    Let W \subseteq V be a subspace. Let p\in W be a polynomial of maximum degree (no polynomials in W have degree greater than degree of p). We know p(x) = a_dx^d + \ldots + a_1x^1 + a_0 for some integer 0\le d \le n with a_d \neq 0. Then \phi(p) = da_dx^{d-1} + \ldots + a_1\in W. So, since \phi(p) \in W, it must be that \phi(\phi(p)) \in W. Keep doing this. Obviously, \phi^{d+1}(p) = 0. And, \phi^d(p) \in \mathbb{R}. Claim: \left(\{p\} \cup \{\phi^k(p) \mid k \in \mathbb{N}\}\right) \setminus \{0\} is a basis for W (where 0 is the polynomial z(x) = 0).
    Last edited by SlipEternal; October 20th 2013 at 01:30 PM.
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    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.
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by abscissa View Post
    Since p(x)=a_nx^n + ... + a_1 + a_0 has degree of at most n, let Z=power set of \{p(x) \in V: \deg(p(x)) \le n\}. 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 n+1 \phi-invariant subspaces of V.
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by SlipEternal View Post
    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 n+1 \phi-invariant subspaces of V.
    Sorry, that was a mistake. There will be a total of n+2 \phi-invariant subspaces of V. I forgot the trivial subspace.
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by SlipEternal View Post
    Let W \subseteq V be a subspace. Let p\in W be a polynomial of maximum degree (no polynomials in W have degree greater than degree of p). We know p(x) = a_dx^d + \ldots + a_1x^1 + a_0 for some integer 0\le d \le n with a_d \neq 0. Then \phi(p) = da_dx^{d-1} + \ldots + a_1\in W. So, since \phi(p) \in W, it must be that \phi(\phi(p)) \in W. Keep doing this. Obviously, \phi^{d+1}(p) = 0. And, \phi^d(p) \in \mathbb{R}. Claim: \left(\{p\} \cup \{\phi^k(p) \mid k \in \mathbb{N}\}\right) \setminus \{0\} is a basis for W (where 0 is the polynomial z(x) = 0).
    I follow the logic of defining the subspace and how p, \phi(p), \phi(\phi(p)), etc. are in W, but I do not understand what the claim means or its significance?
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    Re: Vector Space - Invariant Subspaces

    I will give you a bigger hint. The standard basis for V is \{1,x,x^2,\ldots, x^n\}. Through linear combinations of those elements, you can get any polynomial of degree less than or equal to n. The dimension of V is the number of elements in any of its bases (obviously V is an n+1-dimensional vector space). There is only one n+1-dimensional subspace of V. How many n-dimensional subspaces are there? How many of them have polynomials with maximum degree n-1? If that claim is true, it tells you the dimension of W. It also gives you a very good idea of which subspaces of V are invariant under \phi.
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by SlipEternal View Post
    I will give you a bigger hint. The standard basis for V is \{1,x,x^2,\ldots, x^n\}. Through linear combinations of those elements, you can get any polynomial of degree less than or equal to n. The dimension of V is the number of elements in any of its bases (obviously V is an n+1-dimensional vector space). There is only one n+1-dimensional subspace of V. How many n-dimensional subspaces are there? How many of them have polynomials with maximum degree n-1? If that claim is true, it tells you the dimension of W. It also gives you a very good idea of which subspaces of V are invariant under \phi.
    I am not sure how to tell how many subspaces there are of n-dimension. Also how do I tell the maximum degree of its elements?
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    Re: Vector Space - Invariant Subspaces

    The answer is, there are potentially infinite n-dimensional subspaces of V, but there is only one n-dimensional subspace where the maximum degree of its elements is n-1.
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by SlipEternal View Post
    The answer is, there are potentially infinite n-dimensional subspaces of V, but there is only one n-dimensional subspace where the maximum degree of its elements is n-1.
    So what you are saying is that the only invariant subspace of V is W with the basis { 1,x,x^2,...,x^{n-1}} because it is the only subspace of V which contains x^{n-1} which is required for any polynomial in V that has ax^n because \phi(ax^n) = x^{n-1}. Is that what you are saying?
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    Re: Vector Space - Invariant Subspaces

    No, I am saying that is the only n-dimensional subspace of V that is \phi invariant. Next, if you look for n-1 dimensional subspaces of V that are \phi invariant, there will only be 1. Etc.
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by SlipEternal View Post
    No, I am saying that is the only n-dimensional subspace of V that is \phi invariant. Next, if you look for n-1 dimensional subspaces of V that are \phi invariant, there will only be 1. Etc.
    Oh ok. So of n-dimensional subspaces of V there is only one subspace which is \phi invariant because it requires every element from 0 to x^{n-1}. Now if you look at n-1 dimensional subspaces there is just one for the same reason from 0 to x^{n-2}. And so it follows that this is case for all n until n=0 which gives n total subspaces of V that are \phi invariant.
    Is that the right direction?
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    Re: Vector Space - Invariant Subspaces

    Pretty close, but V itself is a \phi-invariant subspace of itself, as is the trivial subspace (the space consisting only of the zero polynomial).
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by SlipEternal View Post
    Pretty close, but V itself is a \phi-invariant subspace of itself, as is the trivial subspace (the space consisting only of the zero polynomial).
    Right I knew that V would be a \phi-invariant subspace of itself, which is why I said n \phi-invariant subspaces. But I did forget the trivial subspace so I guess there are n+1 total.
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    Re: Vector Space - Invariant Subspaces

    Here are the bases for the \phi-invariant subspaces: \emptyset, \{x^0\}, \{x^0,x^1\}, \ldots, \{x^0,x^1,\ldots, x^{n-1}, x^n\}. So, there are n+2 total. (The empty set is the basis for the trivial subspace.)
    Thanks from vidomagru
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