Are there nontrivial vector spaces with finitely many elements?

Solution 1:

Any vector space over an infinite field (like $\Bbb Q$ $\Bbb R$ or $\Bbb C$) will either be the space $\{0\}$, or will have infinitely many elements. In particular, for any non-zero vector $x$ and any field-elements $k_1,k_2$, the elements $k_1x, k_2 x$ are only equal when $k_1 = k_2$. So, there are infinitely many vectors of the form $k\,x$ as long as our vector space is non-zero over an infinite field.

There are, however, finite fields. For example, we can define multiplication and addition on $\Bbb F_2 = \{0,1\}$ modulo $2$. The vector space $\Bbb F_2^n$ is finite for any $n$ (and has, in particular, $2^n$ elements). For example, we have $$ \Bbb F_2^2 = \{(0,0),(0,1),(1,0),(1,1)\} $$

Solution 2:

There are tons of finite vector spaces. $\mathbb{R}^n$ is a finite vector space for any $n\in\mathbb{N}$.

Edit: that's not right, as has been pointed out; I feel like an idiot. Anyways, to expand upon an answer here with an example, consider the field $\mathbb{Z_3}$. I'll explain this under the assumption that you haven't seen much if any abstract algebra.

First I should tell you what a ring is. A ring is basically just a set in which we have a defined addition, subtraction, and multiplication (and a field is the same, but we also have division defined; therefore the real numbers, as you probably always see with vector spaces, are a field). So here I talk about this ring "modulo 3". This basically means we pretend that 0, 1, and 2 are the only 3 numbers that exist. Then let's say we want to add 2 and 2. Well that's 4, which is outside of our set. But, if you divide by 3 you then get a remainder of 1. So in this "ring", 4 is essentially the number 1. Same for 7, and 10, because if you divide those numbers by 3 you get a remainder 1.

So $\mathbb{Z}_3$ is what's called a ring of integers "modulo 3", and because 3 is prime, it is also a field (that's a theorem in ring theory, just trust it). So for example, in this ring we have $2+1=0$ because $2+1=3\equiv 0\pmod 3$, and $2^2=1$ because $2^2=4\equiv 1\pmod 3$. You can look up more on these rings if you like.

So we can take a ring of matrices over this field, and we have a vector space. So let's do it with $M_2(\mathbb{Z}_3)$. Then all of our entries have to be either a 0, 1, or 2. We may have matrices that look like this:

$$\pmatrix{0&1\\2&1}+\pmatrix{1&2\\0&2}=\pmatrix{1&0\\2&0}$$

because they follow the same addition we had above. Also we may have something like

$$2\pmatrix{0&1\\2&1}=\pmatrix{0&2\\1&2}$$

Now note that this field is finite. There are 3 options for each matrix index, and 4 indexes, so we have a vector space of size $3^4=81$.

I know you wanted a simple answer, and this seems long and drawn out, but if you take some time to try to understand what I've shown you, you should find it interesting as well as insightful into your own linear algebra studies.

Solution 3:

There are finite fields. The simplest one has two elements. Other examples include the ring of integers modulo a prime $p$.

Then take any matrix ring $M_n(F)$ over a finite field $F$.

On the other hand, there are no finite non-trivial vector spaces over an infinite field, such as $\mathbb R$, because if $v\in V$ and $v\ne0$, then the subspace defined by $v$ has the same cardinality as the field.