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I was given the following question:

Let $S = \{v_1, v_2, v_3\}$ be a set of vectors with $v_i \in \mathbb{Q}^4$. In particular, they are all vectors in the vector space $\mathbb{R}^4$ over $\mathbb{R}$. Is is true that $S$ is linearly independent as vectors in $\mathbb{Q}^4$ if and only if they are linearly independent as vectors in $\mathbb{R}^4$?

What varies in the linear combination of $S$ when considered as a set of vectors over one or another vector space is nothing but the coefficients of the linear combinations. The vectors themselves are stated to be the same.

It is trivially true that if $S$ is linearly independent when considered over $\mathbb{R}$ it will be linearly independent when considered over $\mathbb{Q}$. Indeed, if there are no non-null real coefficients that nullify the linear combination of $S$, there are no non-null rational coefficients that do, since every rational number is real. So the linear independence of $S$ over $\mathbb{R}$ implies its linear independence over $\mathbb{Q}$.

The question remains whether the linear independence of $S$ over $\mathbb{Q}$ implies its linear independence over $\mathbb{R}$. I have come to find no reason for assuming this is the case. However, I don't know how to prove it.

Assume there is no non-trivial multi-set $X = \{x_1, x_2, x_3 \} \in \mathbb{Q}$ such that

\begin{equation*} x_1v_1 + x_2v_2 +x_3v_3 = 0 \tag{$v_i \in \mathbb{Q}$} \end{equation*}

Assume as well there is some non-trivial multi-set $X' = \{x_1', x_2', x_3' \} \in (\mathbb{R} - \mathbb{Q})$ such that

\begin{equation*} x_1'v_1 + x_2'v_2 +x_3'v_3 = 0 \end{equation*}

This only means some $v_i$ is a linear combination of the others, when all of them are scaled by irrational scalars. No contradiction seems to be implied in suggesting this might be true. So it seems the linear independence of $S$ when considered over $\mathbb{Q}$ does not implie its linear independence when considered over $\mathbb{R}$. But then again, this reasoning seems too "soft".

The strongest proof would be to build a counter-example $S'$ with rational vectors that is linearly independent for scalars in $\mathbb{Q}$ but linearly dependent for scalars in $\mathbb{I}$. But I have failed to think of a good case where this is true. Any ideas on how one might settle this question?

lafinur
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    Hint: If you were given $v_1, v_2, v_3$ explicitly, how would you calculate whether the set is linearly independent? – Daniel Schepler Dec 20 '22 at 20:08
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    Hint: try finding a counterexample in $\Bbb{Q}$ instead of $\Bbb{Q}^4$. Then, embed $\Bbb{Q}$ in $\Bbb{Q}^4$ as $\Bbb{Q} \times {0}^3$. – Theo Bendit Dec 20 '22 at 20:08
  • It may (will) help to think of the following simplification: first convince yourself $\mathbb{R}$ is a $\mathbb{Q}$ vector space. Can you find two elements of $\mathbb{R}$ that are linearly independent over $\mathbb{Q}$? – Mr. Brown Dec 20 '22 at 20:16
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    @ZackFox Note that in the question, we're assuming all $v_i \in \mathbb{Q}^4$. So the corresponding question would be to try to find $x, y \in \mathbb{Q}$ which are linearly independent over $\mathbb{Q}$ but not over $\mathbb{R}$ -- but the first part is already impossible. – Daniel Schepler Dec 20 '22 at 20:23
  • Oh, sorry I read the question too fast! – Mr. Brown Dec 20 '22 at 21:28

1 Answers1

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Consider the equation $x_1v_1 + x_2v_2 +x_3v_3 = 0$ as an homogeneous linear system of equations in the variables $x_i$. All coefficients are rational.

If it has a solution in $\mathbb{Q}$ then it has automatically a solution in $\mathbb{R}$.

Therefore if the vectors $v_i$ are linearly dependent as vectors of $\mathbb{Q}^4$ they are linearly dependent as vectors of $\mathbb{R}^4$.

If it has a solution in $\mathbb{R}$ then it has one solution in $\mathbb{Q}$ too, since using Gauss Elimination method we arrive at an equivalent system of equations with rational coefficients and with possibly some free variables that can be chosen rational. This implies that the whole solution is rational (by Cramer's rule).

Therefore if the vectors $v_i$ are linearly dependent as vectors of $\mathbb{R}^4$ they are linearly dependent as vectors of $\mathbb{Q}^4$.

Petronella
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