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I'm having my first linear algebra classes in college right now, and a few difficulties with the symbolism used. Missing some basics so to say.

So I have a few small questions I will just ask here:

  • what does the notation $C^\infty(a,b)$ as in $W = \{f \in C^\infty(a,b) \;/ \;\dfrac{d^2}{dx^2}f=0\}$ mean?

  • what does the notation $(u,v)\longmapsto u+v$ in the context of $V \times V \longrightarrow V$ mean?

  • what exactly is an 'additive unit'?

  • in the proof:

Prop.: If $W_1, W_2$ are two subspaces of V, then $W_1 \cap W_2$ is a subspace

(1) Since $0\in W_1$, $0\in W_2$, we have $0\in W_1 \cap W_2$.

(2) Want: If $u, v \in W_1 \cap W_2$ then $u+v \in W_1\cap W_2$.

If $u,v \in W_1 \cap W_2$ then $u,v \in W_i,\; i = 1,2$.

Since $W_i$ is a subspace we have $u,v \in W_i$.

Thus $u+v \in W_1 \cap W_2$.

(3) ...

How can he say that $W_i$ is a subspace, when that is what he is trying to prove? (or which part did i get wrong?) Also: how does the third part of the proof (if $u\in W,\alpha \in \mathbb{R}$ then $\alpha u \in W$) look, or why did my prof only write the first two?

Any help is very appreciated !

Thank you :)

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1 Answer 1

up vote 2 down vote accepted
  • $C^\infty(a,b)$ is usually the functions on the open intervall (a,b) which are infinitely many times differentiable.
  • $V\times V\to V$ is a map, let's call it $f$. For a map you need to know where an element is mapped to. The notation $(u,v)\mapsto u+v$ means that $f((u,v))=u+v$. Here $(u,v)$ is an element of $V\times V$ whereas $u+v$ is an element of $V$.
  • An additive unit (of $V$) is an element $o\in V$ such that $v+o=v$ for all $v\in V$.
  • For the proof: You have to distinguish between the assumptions and the claim. The assumptions are that $W_1$ and $W_2$ are already vector spaces. The claim is that the intersection $W_1\cap W_2$ is a vector space. So in the proof of (2) as you have that $W_1$ is already a subspace (by assumption) and $u,v\in W_1$ you also have $u+v\in W_1$. Similarly $u+v\in W_2$ since $W_2$ is a subspace. Hence $u+v\in W_1\cap W_2$.
  • (3) then follows similarly: As above $W_1$ and $W_2$ are already subspaces, hence $\alpha u\in W_1$ and $\alpha u\in W_2$. Hence $\alpha u\in W_1\cap W_2$.
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@compul: I don't know how much you are confident with the definition of a group, but note that to define a vector space you have to ask that your set $V$ be a commutative group w.r.t an operation of sum (to be introduced) between the elements of the set (the vectors). The map that takes $(u,v)$ to $u+v$ is the definition of that operation and you have to ask for associativity, commutativity, existence of the identity (the $o$, also indicated with $0$, s.t. $v+0 = v = 0+v$) and existence of an inverse. –  Andrea Orta Sep 14 '12 at 8:10
Hey Julian, thanks for your answer. I still have a question about the proof. I assume $W_1, W_2$ are subspaces, and claim $W_1 \cap W_2$ is one as well. Now i set u and v to be elements of $W_1 \cap W_2$. That means u and v are also elements of $W_1$, as well as $W_2$. Now that means $u+v\in W_1$ and $u+v\in W_2$. But how does that necessarily mean they are also in $W_1 \cap W_2$? –  foaly Sep 14 '12 at 8:15
If $u+v$ belongs to both the subspaces, then it must belong to their intersection (by definition of intersection of sets). –  Andrea Orta Sep 14 '12 at 8:18
oh .. well... ya i just took a short break and now it makes sense :D –  foaly Sep 14 '12 at 8:31
what exactly does 'the functions on the open interval' mean? –  foaly Sep 14 '12 at 10:39

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