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I'm trying to work through this proof in Serre:

http://www.math.ualberta.ca/~schlitt/serretheorem.png

I'm confused how $W^{0}$ is different from $W$.

If I have a subspace $W$ of a vector space $V$, and a projection $p:V\to W$, I get a complement $W^{c}$ of $W$ such that $V = W^{c}\oplus W$. Intuitively I would just take $W^{c}$ to be the span of all basis vectors of $V$ not contained in $W$.

So in the proof we construct a new projection $p^{0}$, from $V$ onto $W$ given an arbitrary starting projection $p$, but I don't see why $W'$ ( the complement we start with ) would be any different than the one we end up with $W^{0}$. Can someone help me understand how complements are constructed? I think it's assumed as basic background understanding in the proof.

I think I may have clued into a partial answer, so the complement is just the kernel of the projection?

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    You can take "the span of all basis vectors of $V$ not contained in $W$" **if** you use a basis such that the span of the basis vectors contained in $W$ is $W$.2012-11-06

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If $W'$ is a complement to $W$, we can write each vector $x$ in $V$ in exactly one way as as sum $x=w+w'$, where $w\in W$ and $w'\in W'$. The map that sends $x$ to $w$ (for $x$ in $V$) is a projection onto $W$ and its kernel is $W'$. This shows that there is a bijection between complements to $W$ and projections onto $W$.

It is easy to construct complements. Choose a basis for $W$, then extend it to a basis for $V$. The extra vectors you've added in the extension process are a basis for a complement for $W$.

Changing projections is essentially changing complements.

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Perhaps a simple example would help. Take $V = {\mathbb R}^2$ and $W$ the span of $(1,0)^T$ (geometrically, the $x$ axis). A complement of $W$ could be the span of any vector $(a,b)^T$ with $b \ne 0$ (any line through the origin other than the $x$ axis). This would be the kernel of the projection $\pmatrix{1 & -a/b\cr 0 & 0\cr}$