Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

marked as duplicate by rschwieb, Davide Giraudo, N. F. Taussig, hardmath, Arctic Char Nov 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

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.

share|cite|improve this answer

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

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.