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


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. – Tom Pažourek 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. – Tom Pažourek 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
up vote 2 down vote accepted

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 $\rm |\mathbb R| > |\mathbb Q| = |\mathbb N|\:$.

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V(n,d) contains dn vectors. As you know, each vector in V(n,d) can be described as an n-tuple (v1, ..., vn) where each coefficient vj is drawn from the coefficient-set Σ. It is straightforward to show that the number of such tuples is |Σ|n = dn.

In particular: if Σ is a finite field (such as the integers modulo 2), then the vector space has finitely many elements. Otherwise, it has infinitely many elements.

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