Does a Subspace with Finite Codimension Always Have a Complementary Subspace?

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Homework Statement


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 \left\{N+\alpha_i \right\} (1\leq i \leq n) be the basis of V/N, I want to show the set spanned by \alpha_i is the complementary subspace M.

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

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

Am I correct?

Thanks
 
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yifli said:
since \left\{N+\alpha_i \right\} are the basis, each v in V can be represented as \eta+\sum x_i \alpha_i, \eta \in N

there must be some element in N that can be represented as \sum x_i \alpha_i.

Why should such a representation exists? You only know that \{N+\alpha_i\} is a basis for V/N. This does not mean that \alpha_i is a basis for V! (which seems like you're using!)

For example: Take V=\mathbb{R}^2 and N=\mathbb{R}\times\{0\}. Then \{N+(0,1)\} is a basis for V/N. But (0,1) is not a basis for \mathbb{R}^2.
 
micromass said:
Why should such a representation exists? You only know that \{N+\alpha_i\} is a basis for V/N. This does not mean that \alpha_i 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 \alpha_i.


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

So suppose v \in M\cap N and v \neq 0, that's why I said v can be represented as \sum x_i \alpha_i
 
I see, I misunderstood your proof. It seems to be correct though!
 
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