Problem:
The fact that a system of rational linear equations
$Ax=0$ has a nontrivial complex solution (not necessarily rational), does it imply that it also admit a nontrivial rational one?
One solution is as follows:
Write $Ax = 0$, where $x = (x_1,\cdots,x_n)^T$ is the complex solution. Let $V$ be the $\mathbb{Q}$-span of $\{x_1,\cdots,x_n\}$, and let $\{y_1,\cdots,y_m\}$ be a $\mathbb{Q}$-basis of $V$. Then we have $By =x$, where $B$ is a matrix of rational entries and $y = (y_1,\cdots,y_m)^T$. Then from $ABy = 0$ we deduce that $AB= 0$ since $y$ is a $\mathbb{Q}$-basis and $AB$ is a rational matrix. Thus any column of $B$ is a rational solution of the original equation.
Question:
Is it true for "inhomogeneous" linear systems? That is: if $A$ is a rational matrix and $b$ is a rational vector such that $Ax = b$ has a complex solution, does it imply that it also has a rational solution?
Best Answer
Let $\operatorname{rank}(A)=r$, that does not depend on the base field ($\mathbb{Q}$ or $\mathbb{C})$ and let (for instance) $c_1,\dots,c_r$ (the first $r$ columns of $A$) be a basis of $\operatorname{Im}(A)$. If $Ax=b$ admits a complex solution, then $b\in \operatorname{Im}(A)$ and $b$ is a linear combination of $c_1,\dots,c_r$. The vectors $c_1,\dots, c_r,b$ are in $\mathbb{Q}^n$ ; thus, by the Cramer formulas, $b=\sum_{i=1}^rs_ic_i$ where the $(s_i)_{1\leq i\leq r}$ are rational numbers. Finally a rational solution is $[s_1,\dots,s_r,0,\dots,0]^T$.