0
$\begingroup$

Thm: If W is a subspace of a vector space V, and w1,w2,...,wn ∈ W, and a1,a2,...,an ∈ F (Field), then a1w1,a2w2,...,anwn ∈ W.

Comment: I believe this translates to the title "If W is a subspace of a vector space V, then span(w) is contained in W." If not, please correct me.

Proof: Since W is a subspace, and thus closed under scalar multiplication, it follows that a1,w1...,anwn ∈ W. Since W is also closed under addition, it follows that a1w1 + a2w2 + ... + anwn ∈ W.

Questions:

Am I allowed to claim that all of "a1,w1...,anwn ∈ W" in one sweep like this, or do I need to utilize a method of induction?

Are there any other problems with this proof?

If induction is required for this proof, can someone please show me how to go about it?

  • 0
    The proof, which is boringly trivial, is correct as it is, though an implicit induction is due there...but this is, imo, a minor point. In fact, the claim you want to prove follows almost immediately from the very definition of linear subspace. – 2017-01-14
  • 0
    Strictly speaking, any statement involving "$\ldots$" calls for a proof by induction - or the use of a result that itself as proved by induction. (Well, really strictly speaking, one should never use "$\ldots$" in math as it is always ambiguous). -- Alternatively, why don't you use the fact that if $W$ is a supspace of $V$ then $W$ is a vector space (with the restricted operations) and for vector spaces you ouhg tto know already that the span of a subset of it is contained in it? – 2017-01-14
  • 0
    @HagenvonEitzen I believe we used this proof to prove the result you suggested, though I could be mistaken. However, we definitely did not know that at the time this problem was assigned. If you have the time, would it be possible for you to show me how this would get setup for a proof by induction? – 2017-01-14

0 Answers 0