If some vectors in $\mathbb Q^n$ are linearly independent over $\mathbb Q$ , then are they also linearly independent over $\mathbb C$? Let $\vec v_1 , ..., \vec v_k $ be vectors in $\mathbb Q^n$ linearly independent over $\mathbb Q$ , then is it true that $\sum_{i=1}^ka_i\vec v_i=0, a_i\in \mathbb C, \forall 1\leq i\leq k \implies a_i=0 ,\forall 1\leq i\leq k $ ?
 A: Yes. We can prove that this is the case "by Gaussian elimination" as follows: 
Let $A$ be the matrix whose columns are $v_1,\dots,v_k$.  Since these columns are linearly independent, the problem
$$
A x = 0, \quad x \in \Bbb Q^k
$$
has the unique solution $x = \vec 0$. It follows that there is an invertible matrix $E$ (a sequence of rational elementary row-operations) such that $R = EA$ is in reduced row-echelon form, with a pivot in each column.
We then note that the solution to $Ax = 0$ with $x \in \Bbb C^n$ is the same as the solution to $Rx = 0$ for $x \in \Bbb C^n$.  Since $R$ has a pivot in each column, the problem $Rx = 0$ has the unique solution $x = 0$.
Thus, the only solution to 
$$
Ax = 0, \quad x \in \Bbb C^n
$$
is $x=0$, which is to say that the columns of $A$ are independent over $\Bbb C$.
A: Hint: I'll prove the case $k=2$, after which it's easy to generalize by induction on $k$ by associativity of sum.
Let $v_1, v_2\in \mathbb Q^n$ be $\mathbb Q$-linearly indepedent vectors.
Suppose $\lambda _1v_1+\lambda _2v_2=0_{n\times 1}$, for some complex scalars $\lambda _1, \lambda _2$.
Assume $\lambda _1\neq 0$. It follows that $-\dfrac{\lambda _2}{\lambda _1}v_2=v_1\in \mathbb Q$. Since all entries in $v_2$ are rational and at least one is not zero, it follows that $\dfrac {\lambda _2}{\lambda _1}$ is rational.
But this is saying that $\{v_1, v_2\}$ is $\mathbb Q$-dependent. Therefore the assumption that $\lambda _1\neq 0$ is false.
