# How to show that an $n$-dimensional vector space over field $\mathbb{F}_p$ has $p^n$ elements.

Suppose $V$ is an $n$-dimensional vector space over the finite field $\mathbb{F}_p$ for some prime $p$. How do I show that $V$ has $p^n$ elements? I was thinking that considering $n$ basis elements one can show that each element of the basis spans its own subspace which doesn't have any other basis elements, but how do I show that there are exactly $p^n$ elements overall?

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You might consider how many $n$-tuples of elements of $\mathbb{F}_p$ there are. – Chris Leary Jan 23 '13 at 21:32
Count the number of linear combinations of the $n$ basis elements. And search this site! Your question has been answered here many times. At least in the context of showing that a finite field that is also an $n$-dimensional vector space over $\mathbb{F}_p$ has $p^n$ elements :-) – Jyrki Lahtonen Jan 23 '13 at 21:32
Yes, but how do I know that the elements of $V$ are $n$-tuples? – Jimmy R Jan 23 '13 at 21:32
Every vector space of dimension $n$ over $\mathbb{F}_p$ is isomorphic to $\mathbb{F}_p^n$. – andybenji Jan 23 '13 at 23:39

Let $(e_1,\dots,e_n)$ be a base of $V$ as a $\mathbb{F}_p$-vector space. By definition of a basis, the application $$\begin{array}{rcl}(\mathbb{F}_p)^n & \longrightarrow & V\\(x_1,\dots,x_n) & \longmapsto & x_1e_1 + \dots + x_n e_n\end{array}$$ yields an isomorphism $V \simeq (\mathbb{F}_p)^n$. In particular, $\text{card}V = \text{card}(\mathbb{F}_p)^n = p^n$.

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It is, as Julian says, a simple counting problem. Consider an "empty n-tuple" to be filled with elements from a field, F, with p elements. You have p choices for each of n spaces to fill. Therefore p^n elements.

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Since $V$ is $n$-dimensional over $\mathbb{F}_p$, any given element $v\in V$ can be written as $$v=\lambda _1 v_1 + \cdots +\lambda _nv_n$$ for some $\lambda _1,\dots ,\lambda _n \in \mathbb{F}_p$, where $\{v_1 ,\dots ,v_n\}$ is a basis of $V$ over $\mathbb{F}_p$. For each $v_i$ you have $p$ possibilities for $\lambda _i$, it follows that, since $\{v_1 ,\dots ,v_n\}$ is a basis, you have $p^n$ such $v$'s.

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Every $n$-dimensional vector space is isomorphic to $\mathbb{F}_p^n$ by choosing a basis. In particular it suffices to prove the statement for $\mathbb{F}_p^n$, where it follows by counting.

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