a subspace has finite codimension n iff it has a complementary subspace of dim nu


by yifli
Tags: codimension, complementary, finite, subspace
yifli
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#1
Apr28-11, 10:35 AM
P: 70
1. The problem statement, all variables and given/known data
A subspace N of a vector space V has finite codimension n if the quotient space V/N is finite-dimensional with dimension n. Show that a subspace N has finite codimension n iff N has a complementary subspace M of dimension n. Do not assume V to be finite-dimensional.


2. The attempt at a solution
Let [tex]\left\{N+\alpha_i \right\}[/tex] ([tex]1\leq i \leq n[/tex]) be the basis of V/N, I want to show the set spanned by [tex]\alpha_i[/tex] is the complementary subspace M.

First I show V=N+M:
since [tex]\left\{N+\alpha_i \right\}[/tex] are the basis, each v in V can be represented as [tex]\eta+\sum x_i \alpha_i, \eta \in N[/tex]

Next I prove [tex]N\bigcap M[/tex] = {0}:
if this is not the case, there must be some element in N that can be represented as [tex]\sum x_i \alpha_i[/tex]. Since N is a subspace, this means [tex]\alpha_i[/tex] must be in N. Therefore, [tex]\left\{N+\alpha_i \right\}[/tex] cannot be a basis for V/N

Am I correct?

Thanks
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micromass
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#2
Apr28-11, 11:12 AM
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Quote Quote by yifli View Post
since [tex]\left\{N+\alpha_i \right\}[/tex] are the basis, each v in V can be represented as [tex]\eta+\sum x_i \alpha_i, \eta \in N[/tex]
there must be some element in N that can be represented as [tex]\sum x_i \alpha_i[/tex].
Why should such a representation exists? You only know that [tex]\{N+\alpha_i\}[/tex] is a basis for V/N. This does not mean that [tex]\alpha_i[/tex] is a basis for V! (which seems like you're using!)

For example: Take [tex]V=\mathbb{R}^2[/tex] and [tex]N=\mathbb{R}\times\{0\}[/tex]. Then [tex]\{N+(0,1)\}[/tex] is a basis for V/N. But (0,1) is not a basis for [tex]\mathbb{R}^2[/tex].
yifli
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#3
Apr28-11, 12:36 PM
P: 70
[QUOTE=micromass;3272539]Why should such a representation exists? You only know that [tex]\{N+\alpha_i\}[/tex] is a basis for V/N. This does not mean that [tex]\alpha_i[/tex] is a basis for V! (which seems like you're using!)

I try to prove the complementary subspace M in question is the space spanned by [tex]\alpha_i[/tex].


In orde to do this, I need to show [tex]M\cap N[/tex]={0}.

So suppose [tex]v \in M\cap N[/tex] and [tex]v \neq 0[/tex], that's why I said v can be represented as [tex]\sum x_i \alpha_i[/tex]

micromass
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#4
Apr28-11, 12:41 PM
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a subspace has finite codimension n iff it has a complementary subspace of dim nu


I see, I misunderstood your proof. It seems to be correct though!!


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