# Dimension of a vector subspace of $F_2^n$

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?

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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

## 1 Answer

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 $|W| = 2^k$.

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