While I was studying some Coding Theory, I stumbled on a Linear Algebra over $\mathbb{F}_q$ issue.
In particular, this question generalises a question I asked in this post.
Let $q=p^n$, where $p$ is a prime number and $n\in \Bbb Z^+$ is a positive integer and we denote by $\mathbb{F}_q$ a finite field of order $q$.
Suppose that $V$ is a finite dimensional vector space over $\mathbb{F}_q$, $V_1,V_2\leq V$ are its subspaces and $\dim V_1=k<\infty$.
As one can see, the cardinality of $V_1$ is $|V_1|=q^k$. Now suppose moreover that $|V_2|=|V_1|=q^k$.
My question is if we can claim that in this case we also have $$\dim V_2=\dim V_1 = k.$$
If this is not correct, can I have a counterexample?
Best Answer
Yes, you can. If $v_1,\ldots,v_r$ is a basis, then every element can be expressed uniquely in the form $$\alpha_1v_1+\cdots + \alpha_rv_r$$ and since each $\alpha_i$ has $q$ choices, we see that the number of possible vectors is exactly $q^r$.
Thus, if $V$ is a vector space over a finite field with $q$ elements, then $\dim(V)=r$ if and only if $|V|=q^r$.
In particular, since two vector spaces over the same field are isomorphic if and only if they have the same dimension (in the sense of cardinality) it follows that for vector spaces $V$ and $W$ over a finite field, the following are equivalent: