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Suppose that $A$ is a ${m\times n}$ matrix with rational entries. Let $v$ be a $m\times 1$ vector in $\mathbb{Q}^m$. Then I want to conclude that if the system of equations $Ax=v$ admits real solution then it must also admit a rational solution.

My approach towards a solution is via row reduction. But I am not able to move forward. There is a simliar question on some other stackexhange page but the solution is not very explanatory.

The following seems to be an interesting aspect of the problem- It is clear that the solution set of the above system represents a lower dimensional affine space of $\mathbb{R}^n$, and to say that the system has a rational solution is to say that this affine space intersects $\mathbb{Q}^n$.

Parish
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  • Can you link to the other post? – dantopa Mar 24 '17 at 06:26
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    Row operations on a rational matrix give another rational matrix. – Ethan Alwaise Mar 24 '17 at 06:28
  • @dantopa the link is http://math.stackexchange.com/questions/1270946/prove-or-disprove-the-existence-of-solutions – Parish Mar 24 '17 at 06:47
  • I guess this question is asking if we can find a vector in the nullspace of the matrix that adds with the answer x in $\mathbb{R}$ to produce an answer in $\mathbb{Q}$. With a full rank matrix its not possible. – MNKY Mar 24 '17 at 10:09

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$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\R}{\mathbb{R}}$$\DeclareMathOperator{\rank}{rank}$If $A x = v$ has a solution $x \in \R^{n}$, then $\rank(A) = \rank(A \mid v)$, as $v$ is a linear combination (with coefficients in $\R$) of the columns of $A$.

Conversely, $\rank(A) = \rank(A \mid v)$ tells you that the vector subspace of $\Q^{n}$ generated by the columns of $A$ is the same as the vector subspace of $\Q^{n}$ generated by the columns of $A$ plus $v$. Therefore $v$ is a linear combination with coefficients in $\Q$ of the columns of $A$. In other words, $A x = v$ has a solution $x \in \Q^{n}$.

Andreas Caranti
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  • So, you are saying that as the $\mathbb{R}$ span of the columns of $A$ and $(A\mid v)$ are same, their $\mathbb{Q}$ span must also be same. Why is that? – Parish Mar 24 '17 at 09:29
  • Do you know that the rank of a matrix is the dimension of the vector subspace spanned by the columns? – Andreas Caranti Mar 24 '17 at 10:00
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    Thanks, I get it now. Please correct me if I am wrong. Say the rank of $A$ is $r$. then there are $r$ rational independent column vectors (over $\mathbb{R}$)which span the column space of $A$. But these $r$ column vectors will continue to remain independent over $\mathbb{Q}$, hence the $\mathbb{Q}$ span of the column space of $A$ is also of dimension $r$. Thus $v$ is in this column space. – Parish Mar 24 '17 at 10:06
  • @Parish, sounds good! – Andreas Caranti Mar 24 '17 at 10:13
  • So the only useful information conveyed by "Ax=v admits a real solution" is that Ax=v admits A solution. ie. the equation is consistent. – MNKY Mar 24 '17 at 10:34
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Cramer's Rule will give by your hypothesis real solutions, but the rule expresses the solutions as quotients of what here are rational determinants.

ancient mathematician
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If you can assume that $\mathbb{Q}$ is a vector space over $\mathbb{Q}$, then by definition of matrix multiplication:

$$ \sum_{j=1}^{n}a_{ij}x_j = v_i $$ for $i \in \{1,...,m\}$.

The existence of a solution means that there exists some $(x_1,...,x_m) \in \mathbb{R}^m$ such that these equations hold.

Since these equations are valid, and the $a_{ij}$'s and $v_i$'s are in $\mathbb{Q}$, it follows from the assumption that $\mathbb{Q}$ is a vector space over $\mathbb{Q}$ that the $x_j$'s are also in $\mathbb{Q}$ (or in the case with infinitely many solutions, there exists some choice of $(x_1,...,x_m)$ that is rational).

Ken Wei
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  • This does'nt make sense, since $\sqrt{2}- (\sqrt{2}-1)= 1$, i.e sum of irrational numbers may be rational. – Parish Mar 24 '17 at 08:38
  • @Widawensen: I am sorry, but can you be more elaborate. Having irrational solution does not lead in any natural way to a rational solution. – Parish Mar 24 '17 at 09:06
  • Your $a_{ij}$'s are rational by assumption from your problem, so using $\sqrt{2}$ is not a valid counterexample. Unless you are claiming to say that you reach a solution of $x=(\sqrt{2},\sqrt{2}-1)$ for when $\mathbf{a}_{ij} = (1,-1)$, in which case I claim that another rational solution exists using the assumed fact (e.g. in this case $x=(1,0)$). – Ken Wei Mar 25 '17 at 05:03