Vector space from field

Show that if $\mathbb{F}_q$ is a field with $q$ elements, then $\mathbb{F}_q^n$ is a vector space of dimension $n$ over $\mathbb{F}_q$.

Why $\mathbb{F}_q^n$ is a vector space over $\mathbb{F}_q$ Take the set of all $n$-tuples of elements of $\mathbb{F}_q$:

$$ \mathbb{F}_q^{\,n}=\{(a_1,\dots,a_n)\mid a_i\in\mathbb{F}_q\}. $$

Define the usual component-wise operations

• Addition: $(a_1,\dots,a_n)+(b_1,\dots,b_n)=(a_1+b_1,\dots,a_n+b_n)$.
• Scalar multiplication: for $\lambda\in\mathbb{F}_q$, $\lambda\,(a_1,\dots,a_n)=(\lambda a_1,\dots,\lambda a_n)$.

Because $\mathbb{F}_q$ is already a field, its addition “$+$” and multiplication “$\cdot$” satisfy all field axioms. Applying them component-wise:

• Closure, associativity, commutativity, additive identity $0=(0,\dots,0)$, and additive inverses $-(a_1,\dots,a_n)=(-a_1,\dots,-a_n)$ follow immediately from the corresponding properties in $\mathbb{F}_q$.
• For scalar multiplication, distributivity, associativity, and the identity element $1\in\mathbb{F}_q$ also transfer component-wise.

Hence the standard 8 vector-space axioms hold, so $\mathbb{F}_q^{\,n}$ is indeed a vector space over $\mathbb{F}_q$.

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A basis and the dimension

Consider the $n$ canonical unit vectors

$$ e_1=(1,0,\dots,0),\;e_2=(0,1,0,\dots,0),\;\dots,\;e_n=(0,\dots,0,1) \in\mathbb{F}_q^{\,n}. $$

• Spanning: Any vector $v=(a_1,\dots,a_n)$ can be written

  $$ v=a_1e_1+\dots+a_ne_n, $$

  so $\{e_1,\dots,e_n\}$ spans $\mathbb{F}_q^{\,n}$.

• Linear independence: If $\lambda_1e_1+\dots+\lambda_ne_n=0$ then each coordinate is $\lambda_i=0$ in $\mathbb{F}_q$, so the only linear relation is the trivial one.

Therefore $\{e_1,\dots,e_n\}$ is a basis. A basis has exactly $n$ vectors, so the dimension of $\mathbb{F}_q^{\,n}$ over $\mathbb{F}_q$ is $n$.

$$ \boxed{\dim_{\mathbb{F}_q}\bigl(\mathbb{F}_q^{\,n}\bigr)=n.} $$