Let $F$ be a subfield of a field $K$ and let $n$ be a positive integer. Show that a nonempty linearly-independent subset $D$ of $F^n$ remains linearly independent when considered as a subset of $K^n$.
I'm not sure how to proceed, I tried to assume that $D$ is dependent in $F^n$ and then conclude that is also dependent in $K^n$.
It is Exercise 178 of Jonathan Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.
On
One way to expand on Mariano's hint:
Suppose $v_1,\dots,v_r$ are $n$-vectors with entries in a field $F$ and are linearly independent over $F$. Extend to a basis $v_1,\dots,v_r,\dots,v_n$ for $F^n$ over $F$. Then the $n\times n$ matrix whose entries are the components of the $v_i$ is nonsingular. This nonsingularity has nothing to do with the field in which the entries are seen to lie, as it's just the statement that the determinant is nonzero, and you don't have to know which field you are working in the compute the determinant. So the matrix is nonsingular when its entries are viewed as being in the bigger field $K$, so the vectors are linearly independent over $K$.
On
This result does not require determinants. Instead, we just need that $K$ is a vector space over $F$, so that it has some basis $B$ over $F$. Let $D \subseteq F^n$ be linearly independent over $F$, and suppose that $$ \sum_{\mathbf{x} \in D} \alpha_{\mathbf x} \mathbf x = 0$$ for $\alpha_{\mathbf x} \in K$. Then we may write uniquely $$\alpha_{\mathbf x} = \sum_{b \in B} \gamma^b_{\mathbf x} b$$ for $\gamma^b_{\mathbf x} \in F$. Combining these gives $$ \sum_{\mathbf x \in D} \sum_{b \in B} \gamma^b_{\mathbf x} b \mathbf x = \sum_{b \in B} \sum_{\mathbf x \in D} b \gamma^b_{\mathbf x} \mathbf x = 0.$$ Since $B$ is a basis for $K$ over $F$, taking these equations entry-wise in $K^n$ gives $$\sum_{\mathbf x \in D} \gamma_\mathbf{x}^b \mathbf{x} = 0$$ for all $b \in B$. Since $D$ is linearly independent, we conclude $\gamma^b_\mathbf x = 0$ for all $b \in B$ and $\mathbf x \in D$, which shows $\alpha_\mathbf x = 0$ for all $\mathbf x \in D$, as desired.
On
This is a comment. There is something interesting to consider here in the generalization of this question to the case of modules over commutative rings. That is, suppose $f : R \to S$ is a morphism of commutative rings, and suppose we are given a collection of linearly independent elements $v_1, \dots v_k \in R^n$. Do they remain linearly independent when considered as vectors in $S^n$ after applying $f$ componentwise?
The answer is no in general (e.g. consider $f : \mathbb{Z} \to \mathbb{F}_2$ the usual quotient map and vectors $(2, 2), (2, -2) \in \mathbb{Z}^2$). When is it yes? Here linear independence means that if $\sum r_i v_i = 0$ where $r_i \in \mathbb{R}$ then each $r_i = 0$; equivalently, the $v_i$ induce a map
$$g : R^k \ni (r_1, \dots r_k) \mapsto \sum r_i v_i \in R^n$$
and linear independence means that this map is injective / a monomorphism. We want to know when tensoring $(-) \otimes_R S$ preserves this property. A sufficient condition is that $S$ is flat over $R$: one way of characterizing flatness is precisely the condition that $(-) \otimes_R S$ always preserves monomorphisms. And $\mathbb{F}_2$ isn't flat as a $\mathbb{Z}$-module because it's torsion.
The general fact, which produces the desired result as a special case, is that every module over a field is flat; this means that if $R$ is a field then the result is always true even if we don't require that $S$ is a field. The most explicit proof of this proceeds via choosing a basis of $S$ exactly as done in Joshua Mundinger's answer.
But flatness also tells us what to look for in more general cases; for example, if $R = \mathbb{Z}$ then the condition we want is exactly that $S$ is torsion-free.
Hint: express the linear independence as the non-vanishing of some determinant.