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

Let $W$ be a vector subspace of $F_2^n$. Then, $|W| = 2^k$ for $k \leq n$. Is it always true that $\text{dim}(W) = k$? If it is, where can I find a proof?

share|cite|improve this question
It may be easier to go the other way. If $W$ has a basis consisting $k$ vectors, how many elements does it have? Try it this way, if you are stuck! – Jyrki Lahtonen Aug 31 '11 at 21:09

Let $W$ be a vector subspace of $F_2^n$. Let $k$ = dim($W$). Let $w_1,\dots,w_k$ be a basis of $W$ over $F_2$. Every element $x$ of $W$ can be uniquely written as $x = a_1w_1 + \cdots + a_kw_k$, where $a_i \in F_2$. Hence as a vector space over $F_2$, $W$ is isomorphic to $F_2^k$. Hence $|W| = 2^k$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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