# 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?

• 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, 2010 at 16:05
• @Chandru1: You are right, I edited the question. Oct 10, 2010 at 16:08
• Do you know what is Dimension of a vector space?
– anonymous
Oct 10, 2010 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. Oct 10, 2010 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? Oct 10, 2010 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 $\rm |\mathbb R| > |\mathbb Q| = |\mathbb N|\:$.