# Vector Space - Invariant Subspaces

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• Oct 20th 2013, 11:18 AM
vidomagru
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....
• Oct 20th 2013, 02:28 PM
SlipEternal
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 $z(x) = 0$).
• Oct 20th 2013, 06:15 PM
abscissa
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.
• Oct 20th 2013, 06:29 PM
SlipEternal
Re: Vector Space - Invariant Subspaces
Quote:

Originally Posted by abscissa
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 $V$.
• Oct 20th 2013, 07:43 PM
SlipEternal
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 $n+1$ $\phi$-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.
• Oct 21st 2013, 05:44 PM
vidomagru
Re: Vector Space - Invariant Subspaces
Quote:

Originally Posted by SlipEternal
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?
• Oct 21st 2013, 08:36 PM
SlipEternal
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 $\phi$.
• Oct 23rd 2013, 05:19 PM
vidomagru
Re: Vector Space - Invariant Subspaces
Quote:

Originally Posted by SlipEternal
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?
• Oct 23rd 2013, 05:28 PM
SlipEternal
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 $n-1$.
• Oct 23rd 2013, 05:48 PM
vidomagru
Re: Vector Space - Invariant Subspaces
Quote:

Originally Posted by SlipEternal
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?
• Oct 23rd 2013, 06:53 PM
SlipEternal
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.
• Oct 23rd 2013, 07:29 PM
vidomagru
Re: Vector Space - Invariant Subspaces
Quote:

Originally Posted by SlipEternal
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?
• Oct 23rd 2013, 07:32 PM
SlipEternal
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).
• Oct 23rd 2013, 08:02 PM
vidomagru
Re: Vector Space - Invariant Subspaces
Quote:

Originally Posted by SlipEternal
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.
• Oct 23rd 2013, 08:07 PM
SlipEternal
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.)
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