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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


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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1 Answer 1

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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

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