My book has an exercise:
"Suppose that $W$ is a subspace of a finite-dimensional vector space $V$.
a) Prove that there exists a subspace $W'$ and a function $T:V\longrightarrow V$ such that $T$ is a projection on $W$ along $W'$
b) Give an example of a subspace $W$ of a vector space $V$ such that there are two projections on W along two (distinct) subspaces."
And here is my answer:
a) I can choose $W'$ to satisfy $W' \oplus W = V$. Since then $W' \bigcap W = \{0\}$, $\forall v \in V, w\in W, w'\in W':v=w+w'$. Thus I can eliminate the $W'$-part by substraction such that $v - w' = w \in W$, and so obtain the $W$-part, i.e. project any $v$, that is, $V$, on $W$ along $W'$.
b) $V=\mathbb{R}, W=(a,0)$
1: $W' = (b,b) \longrightarrow (a,b) = (b,b) + (a-b,0) \longrightarrow T(a,b) = (a-b,0)$
2: $W' = (\frac{b}{2},b) \longrightarrow (a,b) = (\frac{b}{2},b) + (a-\frac{b}{2},0) \longrightarrow T(a,b) = (a-\frac{b}{2},0)$
Now my question is: is this legit? It seems so obvious. Or did I presume something that in turn again has to be proven (or some similar thing that still happens to me)?
Thank you
EDIT:
Apparently, this is not enough. According to my prof, for a) I will have to show that there exists $W'$ satisfying $W'\oplus W=V$. So the updated version looks like this:
a) Let $\beta = \{v_1, \dots, v_k\}$ be an ordered basis of $W$. Then since $W \subseteq V$ I can extend $\beta$ to a basis of $V$:
$\gamma = \{v_1,\dots ,v_k,v_{k+1},\dots ,v_n\}$
$W' = span(\{v_{k+1},\dots ,v_n\}) \subseteq V$
Since $v_1,\dots,v_n$ are linearly independent,
$W \cap W' = \{0\}$
$v \in V = \sum_{i=1}^n \alpha _i v_i = \sum_{i=1}^k \alpha _i v_i + \sum_{i=k+1}^n \alpha _i v_i = w + w' \Rightarrow V = W + W' $
Hence $W'\oplus W = V$.
Then, since $v = w + w' \forall v \in V, w\in W, w'\in W'$, and $v_1,\dots ,v_n$ are linearly independent, I can define $T$ like $T(v_i) = w_i, 1\le i\le k$ and $T(v_i)=0, k\lt i\le n$, which is a projection on $W$ along $W'$.
Since $T$ is linear, $T(v) = w$.
So to check: does this make sense to you now?