# Is this proof legit?

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."

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?

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Your new proof looks OK, but in the last part you shouldn't say "Since $T$ is linear, $T(v)=w$". What you mean is: "$\gamma$ is a basis of $V$, and therefore I can define a linear map by specifying values on the basis vectors and extending linearly". Your example also looks fine. The main idea there is that the choice of $W'$ is not unique (unless $W=V$), so you will end up with different projections.

Also, be careful with how you use $\forall$ (in fact, you shouldn't have to use it in this proof at all). You claim $\forall v \in V,w\in W,w' \in W'$ we have $v=w+w'$, but that is obviously not true if I take $v=w=0$ and $w'$ to be some nonzero element.

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Thank you, and that's exactly what i mean! Still fighting with how to express things. With the $\forall$ i actually meant that for all $v$ there exist $w$ and $w'$ such that $v = w + w'$. Gonna be more careful with my notations i guess :P. One question: Why do I need to show that there exists such a $W'$? If $W \subseteq V$, then isn't it obvious that there is W either $= \{0\}$ or that what is missing between $W$ and $V$? I would have never guessed that i actually have to show that... – foaly Oct 19 '12 at 12:35
@foaly: It's certainly a basic fact, but I wouldn't say it's obvious. If you learn about modules later on, you will find that a lot of these properties that seem "obvious" do not hold! – wj32 Oct 19 '12 at 20:54