# Thread: Simple subspace proof using induction: STUCK

1. ## Simple subspace proof using induction: STUCK

Hi, I'm reviewing some linear algebra this summer as a refresher. I've come across the following proof problem, and have found myself stuck. Apparently, I'm supposed to use induction to solve this, but I'm not really sure exactly WHAT to use induction on. Anyway, here's the problem:

Prove that if W is a subspace of a vector space V and w(1), w(2),...,w(n) are in W, then a(1)w(1) + ... + a(n)w(n) is in W for any scalars a(1), ... ,a(n).

Any help on this? :-(

2. It will not be necessary to use induction in this problem. One simply notes that a subspace is a subset of a vector space that is also a vector space on its own. One of the properties of being a vector space is that it is closed under addition and scalar multiplication. That is all that is being used here. I suppose if one really wanted to be rigorous one could potentially use induction I suppose.

n=1
$a_1w_1 \in W$ by closure under scalar multiplication

Induction hypothesis for n=k
Assume $a_1w_1+a_2w_2+...+a_kw_k \in W$ for any $a_i \in \mathbb{F} w_i \in W$.

Show $a_1w_1+a_2w_2+...+a_kw_k +a_{k+1}w_{k+1} \in W$. By induction hypothesis, $a_1w_1+a_2w_2+...+a_kw_k \in W$ so say $a_1w_1+a_2w_2+...+a_kw_k =w'$.
Furthermore $a_{k+1}w_{k+1} = w''\in W$ by closure under scalar multiplication.
Then
$a_1w_1+a_2w_2+...+a_kw_k +a_{k+1}w_{k+1} = w' + w'' \in W$ by closure under addition.

I think it a little extreme, but you asked for the induction. Hope that helps.

3. The reason you should use induction is that the definition of subspace (closed under addition) only refers to the fact that au+ bv is in the space for u and v in the space, a, b, scalars.

4. Yeah, I guess it is just always understood that in vector spaces they are closed under linear combinations, guess I have never formally proved it because it is a pretty trivial exercise in induction. Like that's the whole point of having a basis for your vector space or free module.