Suppose that $F$ is an $m$-dimensional vector space over a subfield $K$ of $F$. In Roman, Advanced Linear Algebra, there is a question: 
"Suppose that $F$ is an $m$-dimensional vector space over a subfield $K$ of $F$. If $V$ is an $n$-dimensional vector space over $F$, show that $V$ is also a vector space over $K$. What is the dimension of $V$ as a vector space over $K$?". Any help with this problem would be appreciated. I cannot really make sense of $F$ being an $m$-dimensional vector space and also a field. 
 A: I think the easiest way to understand the problem is to consider an example.  In particular, note that the complex numbers $F = \Bbb C$ contain the real numbers $K = \Bbb R$ as a subfield.  
$F$ is a vector space in the following sense: if we have any two numbers $z_1,z_2 \in F$ and any scalars $c_1,c_2 \in K$, then $c_1z_1 + c_2 z_2$ is another element of $K$.  Moreover, because the definition of scalar multiplication is a "natural" consequence of multiplication in $F$, we can see that $F$ will satisfy the vector-space axioms.  I suggest that you convince yourself that $\Bbb C$ is a vector space over $\Bbb R$ and then argue the more general statement along similar lines.  
So what is the dimension of $\Bbb C$ over $\Bbb R$? Because every complex number can be uniquely written as $a + bi$, the set $\{1,i\} \subset \Bbb C$ forms a basis of $\Bbb C$ as an $\Bbb R$-vector space.  So, $\Bbb C$ is a $2$-dimensional vector space over $\Bbb R$.  
Now, $V = \Bbb C^n$ is an $n$-dimensional vector space over $\Bbb C$.  What is its dimension over $\Bbb R$? In other words, how many real numbers are needed to uniquely specify an element of $\Bbb C^n$?  Use your conclusion here to get a more general answer.
