# Maximal number of vectors in vector space

The vector space $V(n,d)$ has dimension $n$ and the vector coordinates are from set $\Sigma$.

$d:=|\Sigma|$

How many vectors can the vector space contain? Why?

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What do you mean by : how many vectors can the vector space contain. Are you asking about the no of elements in the basis! If not i think your question doesn't make any sense. –  anonymous Oct 10 '10 at 16:05
@Chandru1: You are right, I edited the question. –  tomp Oct 10 '10 at 16:08
Do you know what is Dimension of a vector space? –  anonymous Oct 10 '10 at 16:12
@Chandru1: Sorry, I was a bit confused. The dimension itself is obviously the number of vectors in the basis. I rollbacked to my original question. Can you please explain the reason why the question doesn't make sense? The vector space has dimension $n$ and only $d$ possible vector coordinates, it should be possible to count the maximal number of vectors in vector space like that. –  tomp Oct 10 '10 at 16:18
I don't understand. Is Sigma assumed to be a proper subset of the underlying field, or is it the underlying field itself? –  Qiaochu Yuan Oct 10 '10 at 16:52

If $\rm V$ is an $\rm n$-dimensional vector space over a field $\rm F$ then it has cardinality $\rm |V| = |F|^n$ if both $\rm |F|$ and $\rm n$ are finite; otherwise $\rm |V| = n\:|F| = max(n,|F|)\:$, as follows from basic properties of cardinal arithmetic. For example, this implies that the dimension of $\mathbb R$ over $\mathbb Q\:$ is $\rm |\mathbb R| > |\mathbb Q| = |\mathbb N|\:$.