Finding the dimension of a subspace

  • #1
fishturtle1
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Homework Statement:
Prove that if ##y## is a linear functional on an ##n##-dimensional vector space ##V##, then the set of all those vectors ##x## such that ##[x, y] = 0## is a subspace of ##V##; what is the dimension of this subspace?
Relevant Equations:
A function ##y : V \rightarrow \mathbb{R}## is a linear functional on ##V## if for all ##u, v \in V## and ##\alpha, \beta \in F##, we have ##[\alpha u + \beta v , y] = \alpha [u, y] + \beta [v, y]##.

We have the notation ##[v, y]## to mean ##y(v)##.
I am stuck on finding the dimension of the subspace. Here's what I have so far.

Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v, y] = \alpha \cdot 0 + \beta \cdot 0 = 0##. So, ##\alpha u + \beta v \in W##. This shows ##W## is a subspace of ##V##.

Next, we will find the dimension of ##W##. Let ##x_1, x_2, \dots, x_n## be a basis of ##V##. We can find a maximal subset of this basis ##z_1, \dots, z_k## such that ##[z_i, y] = 0## for all ##i##. We claim this is a basis for ##W##. Since ##u \in V##, there exists scalars ##\alpha_1, \dots, \alpha_n## such that ##u = \alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n##. Then,
$$[u, y] = [\alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n, y] = \alpha_{k+1} [x_{k+1}, y] + \dots + \alpha_n [x_n, y] = 0$$.
We want to somehow conclude ##\alpha_{k+1} = \dots = \alpha_n = 0##, but I'm not sure how to proceed. Can I have a hint, please?
 

Answers and Replies

  • #2
fresh_42
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I don't see what you are trying to do with the dimension proof. Do you know the rank formula for linear functions ##\varphi : V\longrightarrow W##?
 
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  • #3
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Homework Statement:: Prove that if ##y## is a linear functional on an ##n##-dimensional vector space ##V##, then the set of all those vectors ##x## such that ##[x, y] = 0## is a subspace of ##V##; what is the dimension of this subspace?
Relevant Equations:: A function ##y : V \rightarrow \mathbb{R}## is a linear functional on ##V## if for all ##u, v \in V## and ##\alpha, \beta \in F##, we have ##[\alpha u + \beta v , y] = \alpha [u, y] + \beta [v, y]##.

We have the notation ##[v, y]## to mean ##y(v)##.
Why not use a more standard notation?
I.e., ##f: V \rightarrow \mathbb{R}##
and ##f(v)## instead of [v, y]
fishturtle1 said:
I am stuck on finding the dimension of the subspace. Here's what I have so far.
I don't think the work below is going to get you anywhere. You're not going to be able to find the dimension of the nullspace directly, because you don't know how many vectors are in a basis for the nullspace. A better option is to use the rank-nullity theorem - see Rank–nullity theorem - Wikipedia .
fishturtle1 said:
Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v, y] = \alpha \cdot 0 + \beta \cdot 0 = 0##. So, ##\alpha u + \beta v \in W##. This shows ##W## is a subspace of ##V##.

Next, we will find the dimension of ##W##. Let ##x_1, x_2, \dots, x_n## be a basis of ##V##. We can find a maximal subset of this basis ##z_1, \dots, z_k## such that ##[z_i, y] = 0## for all ##i##. We claim this is a basis for ##W##. Since ##u \in V##, there exists scalars ##\alpha_1, \dots, \alpha_n## such that ##u = \alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n##. Then,
$$[u, y] = [\alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n, y] = \alpha_{k+1} [x_{k+1}, y] + \dots + \alpha_n [x_n, y] = 0$$.
We want to somehow conclude ##\alpha_{k+1} = \dots = \alpha_n = 0##, but I'm not sure how to proceed. Can I have a hint, please?
 
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  • #4
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Thank you for your replies. Yes, I think we can use Rank-Nullity theorem, I will try it.

And yeah, I'm happy to use a more standard notation if that's better; I was trying to mimic the textbook.
 
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Thank you for your replies. Yes, I think we can use Rank-Nullity theorem, I will try it.

And yeah, I'm happy to use a more standard notation if that's better; I was trying to mimic the textbook.
You could also use ##\langle y, .\rangle\, : \,v\longmapsto \langle y,v\rangle## in case it is an inner product space.
 
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  • #6
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You could also use ##\langle y, .\rangle\, : \,v\longmapsto \langle y,v\rangle## in case it is an inner product space.
I don't think that's applicable here, about an inner product space. My objection to the notation [x, y] was that it hinted at being an inner product, but wasn't.
 
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  • #7
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We know ##y : V \rightarrow \mathbb{R}## is a linear transformation. By Rank-Nullity theorem we have ##\dim V = rank(y) + null(y)##. We note that ##null(y) = \dim W##. If ##rank(y) = 0##, then ##\dim V = \dim W## and so ##\dim W = n##.

If ##rank(y) \neq 0##, then ##rank(y) = 1##. And so ##\dim W = n - 1##.
 
  • #8
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I don't think that's applicable here, about an inner product space. My objection to the notation [x, y] was that it hinted at being an inner product, but wasn't.
In case we have finite dimensional vector space, there is a one-to-one correspondence between ##V## and ##V^*##, and if it is a real vector space, then any ##v\in V^*## can be written as some ##\langle y_v,-\rangle##, i.e. the bijection can be realized by the inner product. This should even hold for other fields of characteristic zero.

However, you were right that the brackets were problematic. They are usually used for commutators, in group as well as in algebra theory.
 

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