Let $v_1,v_2$ be vectors in $\mathbb{Q}^3$. Prove that they are linearly independent on $\mathbb{Q}$ if and only if they are on $\mathbb{R}$. Let $v_1,v_2$ be vectors in $\Bbb Q^3$ (and each one of them is a vector in the $\Bbb R$-vectorial space $\Bbb R^3$ $\Bbb R$). Is it true that these vectors are linearly independent as vectors in $\Bbb Q^3$ if and only if they are linearly independent in the $\Bbb R$-vectorial space $\Bbb{R}^3$?
I tried to do it by the definition of linearly Independent (if $a_1v_1+a_2v_2 =0$ then $a_1=a_2=0$, no matter if I'm in $\Bbb Q^3$ or in $\Bbb R^3$), but I feel like the problem maybe has more ideas that I should use.
 A: You can do it like this:
Take your vectors $\{v_1,v_2\}\subset\mathbb{Q}^3$ and consider the $3\times2$ matrix $A=(v_1|v_2)$ with columns $v_1$ and $v_2.$ Now this matrix has linearly independent columns if and only if there is a $2\times 2$ submatrix whose determinant does not vanish. But this condition is the same whether we consider $A$ as a matrix over $\mathbb{Q},\mathbb{R}$ or even $\mathbb{C}.$
A: First, the non-independence over $\mathbb{R}$ implies the same over $\mathbb{Q}$.
Now, if $v_1=av_2$ with $a\in \mathbb{R}\setminus \mathbb{Q}$, then component a component we have that
a real non-zero number times a rational number is rational, i.e., $av_{2j}=v_{1j}$. But, it creates a contradiction because since some component is non-zero, $a=\frac{v_{1j}}{v{2j}}\in \mathbb{Q}$.
A: Recall the following procedure for determining whether a set of $m$ vectors in $F^n$ are linearly independent: put the vectors as the columns of an $m \times n$ matrix, and then perform Gaussian elimination on that matrix to get it into row echelon form.  If you get $m$ pivots at the end, then the original vectors are linearly independent; while if you get strictly less than $m$ pivots, then the original vectors are linearly dependent.
Now in your situation, suppose you perform that procedure with scalars in $\mathbb{Q}$.  Then the Gaussian elimination you performed is also a valid Gaussian elimination with scalars in $\mathbb{R}$.  Therefore, whatever answer you get for linear independence over $\mathbb{Q}$, you will get the same answer for linear independence over $\mathbb{R}$.
A: $$p\Rightarrow q \Leftrightarrow \not \not q\Rightarrow \not p:$$
$$v_1, v_2\in \mathbb Q^3 \text{ linearly dependent on }\mathbb R^3 \Leftrightarrow \exists r\ne 0 \in \mathbb R: v_1=r\cdot v_2\Leftrightarrow \left(\underbrace{v_{1i}}_{\in \mathbb Q}=r\underbrace{v_{2i}}_{\in \mathbb Q}\right )_{1\le i\le 3}\Rightarrow r\in \mathbb Q\Rightarrow v_1, v_2 \text { linearly dependent on }\mathbb Q^3$$
$$q\Rightarrow p:$$
$$v_1, v_2\in \mathbb Q^3 \text{ linearly independent on } \mathbb R^3 \Leftrightarrow (a_1v_1+a_2v_2=0\Rightarrow a_1=a_2=0)\land 0 \in \mathbb Q\Leftrightarrow v_1,v_2 \text{ linearly independent on } \mathbb Q^3$$
$$(p\Rightarrow q \land q\Rightarrow p) \Leftrightarrow (p\Leftrightarrow q) (\therefore)$$
