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

[yifli]

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

 

[micromass]

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 $$\mathbb{R}^2$$.

 

[yifli]

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 $$\sum x_i \alpha_i$$

 

[micromass]

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