a subspace has finite codimension n iff it has a complementary subspace of dim nuby yifli Tags: codimension, complementary, finite, subspace 

#1
Apr2811, 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 finitedimensional 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 finitedimensional. 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 



#2
Apr2811, 11:12 AM

Mentor
P: 16,565

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]. 



#3
Apr2811, 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] 


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