# How do complements relate to projections?

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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What do you mean by "basis vectors of $V$" if you haven't chosen a basis of $V$? You are correct that the complement is the kernel. An idempotent acting on a vector space is completely determined by its image and its kernel (exercise) and such idempotents are in natural bijection with direct sum decompositions given by their image and their kernel (another exercise). –  Qiaochu Yuan Nov 6 '12 at 20:22
Well I think my description of the complement in terms of basis vectors is independant of the choice, so I thought it would work. As for your answer, am I correct then in saying that $W^{0} = ker(p^{0})$ and $W' = ker(p)$ (in the notation of Serre's Proof). –  roo Nov 6 '12 at 20:28
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 $W$. –  Robert Israel Nov 6 '12 at 20:42

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 $W$.
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 $$\pmatrix{1 & -a/b\cr 0 & 0\cr}$$