# Prove subspace of finite dimensional vector space is finite dimensional [duplicate]

Prove that any subspace, W, of a finite-dimensional vector space V must also be finite dimensional.

-

## marked as duplicate by rschwieb, Davide Giraudo, N. F. Taussig, hardmath, John MaNov 24 '15 at 11:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Suppose for sake of contradiction that W isn't finite dimensional. That will probably contradict V being finite dimensional. – Mark S. Nov 18 '12 at 23:34
I'd rather not without first seeing what you've done. Mind showing us your work and where you got stuck? – Neal Nov 18 '12 at 23:34
@Neal Well my professor gave a hint saying we can use the Plus-minus theorem to prove this but i'm not sure how to go about proving this using plus-minus theorem. – kamron Nov 18 '12 at 23:54
You don't have to use the plus-minus theorem, but if you really want to, try to think of a set of vectors to apply the "plus" or the "minus" to yield a contradiction with W being finite dimensional. – Mark S. Nov 19 '12 at 0:10

## 2 Answers

Consider a basis $\beta$ of $W$, then $\beta$ is linearly independent in $V$ so you can extend $\beta$ to $\gamma$ a basis of $V$, so $\gamma$ must be finite, by hypothesis.

-

HINT: Let $n=\dim\ V$. What is the maximum size of any linearly independent subset of $V$?

-