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Thread: Vector Space - Invariant Subspaces

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    Vector Space - Invariant Subspaces

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

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

    Let $\displaystyle W \subseteq V$ be a subspace. Let $\displaystyle p\in W$ be a polynomial of maximum degree (no polynomials in $\displaystyle W$ have degree greater than degree of p). We know $\displaystyle p(x) = a_dx^d + \ldots + a_1x^1 + a_0$ for some integer $\displaystyle 0\le d \le n$ with $\displaystyle a_d \neq 0$. Then $\displaystyle \phi(p) = da_dx^{d-1} + \ldots + a_1\in W$. So, since $\displaystyle \phi(p) \in W$, it must be that $\displaystyle \phi(\phi(p)) \in W$. Keep doing this. Obviously, $\displaystyle \phi^{d+1}(p) = 0$. And, $\displaystyle \phi^d(p) \in \mathbb{R}$. Claim: $\displaystyle \left(\{p\} \cup \{\phi^k(p) \mid k \in \mathbb{N}\}\right) \setminus \{0\}$ is a basis for $\displaystyle W$ (where 0 is the polynomial $\displaystyle z(x) = 0$).
    Last edited by SlipEternal; Oct 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 $\displaystyle p(x)=a_nx^n + ... + a_1 + a_0$ has degree of at most n, let $\displaystyle Z=$power set of $\displaystyle \{p(x) \in V: \deg(p(x)) \le n\}$. 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 $\displaystyle n+1$ $\displaystyle \phi$-invariant subspaces of $\displaystyle V$.
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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 $\displaystyle n+1$ $\displaystyle \phi$-invariant subspaces of $\displaystyle V$.
    Sorry, that was a mistake. There will be a total of $\displaystyle n+2$ $\displaystyle \phi$-invariant subspaces of $\displaystyle V$. I forgot the trivial subspace.
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    Re: Vector Space - Invariant Subspaces

    Quote Originally Posted by SlipEternal View Post
    Let $\displaystyle W \subseteq V$ be a subspace. Let $\displaystyle p\in W$ be a polynomial of maximum degree (no polynomials in $\displaystyle W$ have degree greater than degree of p). We know $\displaystyle p(x) = a_dx^d + \ldots + a_1x^1 + a_0$ for some integer $\displaystyle 0\le d \le n$ with $\displaystyle a_d \neq 0$. Then $\displaystyle \phi(p) = da_dx^{d-1} + \ldots + a_1\in W$. So, since $\displaystyle \phi(p) \in W$, it must be that $\displaystyle \phi(\phi(p)) \in W$. Keep doing this. Obviously, $\displaystyle \phi^{d+1}(p) = 0$. And, $\displaystyle \phi^d(p) \in \mathbb{R}$. Claim: $\displaystyle \left(\{p\} \cup \{\phi^k(p) \mid k \in \mathbb{N}\}\right) \setminus \{0\}$ is a basis for $\displaystyle W$ (where 0 is the polynomial $\displaystyle z(x) = 0$).
    I follow the logic of defining the subspace and how $\displaystyle p, \phi(p), \phi(\phi(p)), $ etc. are in $\displaystyle W$, 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 $\displaystyle V$ is $\displaystyle \{1,x,x^2,\ldots, x^n\}$. Through linear combinations of those elements, you can get any polynomial of degree less than or equal to n. The dimension of $\displaystyle V$ is the number of elements in any of its bases (obviously $\displaystyle V$ is an $\displaystyle n+1$-dimensional vector space). There is only one $\displaystyle n+1$-dimensional subspace of $\displaystyle V$. How many $\displaystyle n$-dimensional subspaces are there? How many of them have polynomials with maximum degree $\displaystyle n-1$? If that claim is true, it tells you the dimension of $\displaystyle W$. It also gives you a very good idea of which subspaces of $\displaystyle V$ are invariant under $\displaystyle \phi$.
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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 $\displaystyle V$ is $\displaystyle \{1,x,x^2,\ldots, x^n\}$. Through linear combinations of those elements, you can get any polynomial of degree less than or equal to n. The dimension of $\displaystyle V$ is the number of elements in any of its bases (obviously $\displaystyle V$ is an $\displaystyle n+1$-dimensional vector space). There is only one $\displaystyle n+1$-dimensional subspace of $\displaystyle V$. How many $\displaystyle n$-dimensional subspaces are there? How many of them have polynomials with maximum degree $\displaystyle n-1$? If that claim is true, it tells you the dimension of $\displaystyle W$. It also gives you a very good idea of which subspaces of $\displaystyle V$ are invariant under $\displaystyle \phi$.
    I am not sure how to tell how many subspaces there are of $\displaystyle n$-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 $\displaystyle n$-dimensional subspaces of $\displaystyle V$, but there is only one $\displaystyle n$-dimensional subspace where the maximum degree of its elements is $\displaystyle n-1$.
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    Re: Vector Space - Invariant Subspaces

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

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

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

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