# What is the cardinal of a field F_5 vector space of dimension 3?

What is the cardinal of a field F_5 vector space of dimension 3? The mark scheme says since F_5 = { 0,1,2,3,4 } there are 5 possibilities. so it is 5^3. So the card(v) = 125. But in the lecture notes it says the cardinal = dimension for a finite space. Is this a finite space? Why is the cardinal not equal to the dimension in this case? Thanks!

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My guess is that the lecture notes are meant to say: "the cardinality of a basis = the dimension of a (finitely generated vector) space." Anyway, you calculated the cardinality of the space itself (=the number of distinct vectors) correctly. – Jyrki Lahtonen Jul 22 '14 at 14:14

Let $V = (F_5)^3$ be a vector space of dimension $3$ over $F_5$, the field with $5$ elements (which is $\mathbb{Z}/5\mathbb{Z} = \{0,1,2,3,4\}$ working with mod 5 arithemetic).

Now the dimension of $V$ is $3$ (by assumption).

Then cardinality of $V$ is the number of elements of $V$, i.e. the number of different vectors $v \in V$.

Now what does it take to get a vector $v \in V$? Well, pick a basis for $V$, e.g. $e_1 = (1,0,0), e_2 = (0,1,0), e_3 = (0,0,1)$ now to get all vectors in $V$ means to write linear combinations $\lambda_1 e_1 + \lambda_2 e_2 + \lambda_3 e_3$ with $\lambda_1,\lambda_2,\lambda_3 \in F_5$. Each $\lambda_i$ may be one of $5$ different choices $\{0,1,2,3,4\}$. So there are $5$ choices for $\lambda_1$, $5$ choices for $\lambda_2$ and $5$ choices for $\lambda_3$. Moreover, each choice gives a unique vector (since $e_1,e_2,e_3$ make up a basis). Thus there are $5*5*5$ different vectors. I.e. the cardinality of $V$ is $5^3$.

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thankyou! great explanation :D – user135688 Jul 22 '14 at 14:46

The cardinality of a finitely generated vector space is never equal to the dimension.

The case of the trivial vector space is readily done: $\{0\}$ has one element and dimension $0$.

This is also clear if the base field is infinite, because in this case every non trivial vector space is infinite, so its cardinality is larger than the (finite) dimension.

If the base field is finite, say of cardinality $q$, then an $n$-dimensional vector space has cardinality $q^n$ and $n=q^n$ is impossible, because $q\ge2$ and $n<2^n$ for all $n$.

What is true is that the dimension equals the cardinality of a basis, which is just the definition of dimension.

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thankyou! now i understand what my lecture notes should write! – user135688 Jul 22 '14 at 14:44

Your notes are either wrong or badly imprecise. The cardinality of a vector space is the cardinality of its underlying set. It is not equal to the dimension of the vector space.

For example, $\mathbb{R}$ is a one-dimensional vector space over itself, but it has cardinality of the continuum.

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i noticed my notes weren't precise, probably why i didnt know what a cardinal was :p – user135688 Jul 22 '14 at 14:44