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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?

Antonio Alfieri
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mez
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    If we write the system in the form $AX=0$ with $X=(x_1,x_2,\ldots,x_n)^T$ is a column vector of the unknowns, then the system is equivalent to one, where we replace $A$ with its row reduced echelon form. Note that row reduction does not necessitate the use of non-rational numbers (only the four basic arithmetic operations are used). At that point the solutions of the system take the form where the values some of the unknowns can be chosen freely, and the rest are rational linear combinations of the freely chosen ones. – Jyrki Lahtonen Jul 29 '14 at 08:26
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    (cont'd) The latter group of variables consists of the variables corresponding to initial 1s on the row reduced echelon matrix. The same argument works over any field. If there is a solution to a linear homogeneous system over an extension field, then there is one over the field of definition, i.e. the smallest field containing the coefficients of the system. – Jyrki Lahtonen Jul 29 '14 at 08:27
  • @JyrkiLahtonen By your reasoning the same is true for inhomogeneous system of linear equations? – mez Jul 29 '14 at 12:51
  • Correct. Though in that case the field of definition must also contain the constants on the r.h.s., or, equivalently, the entries of the augmented matrix. – Jyrki Lahtonen Jul 29 '14 at 12:54
  • @JyrkiLahtonen I understood your first comment. But I don't see how does it exactly implies the existence of a rational solution if a complex one exists. Could you perhaps elaborate it in an answer? – mez Jul 29 '14 at 13:01
  • The argument goes as follows. Because there is a non-trivial solution over a bigger field it is possible to choose one of the variables freely. By selecting that variable to have value $1$, (and the other free variables, if any, to have value $0$) we get a solution with entries in the field of definition. I don't understand the part about matrix $B$, so I'm reluctant to post this as an answer. – Jyrki Lahtonen Jul 29 '14 at 14:48
  • @JyrkiLahtonen I don't think this argument works for inhomogeneous equations, because the solutions don't form a vector space any more. – mez Jul 29 '14 at 16:48
  • Related: http://math.stackexchange.com/questions/798071/a-solution-of-a-linear-system-in-some-extension-implies-a-solution-in-the-subfie – user26857 Aug 18 '14 at 06:53

1 Answers1

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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 $[s_1,\dots,s_r,0,\dots,0]^T$.

user26857
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  • @ user26857, from the matrix $[c_1,\cdots,c_r]$ we extract $r$ linearly independant rows. They are not necessarily the first $r$ ones. –  Aug 23 '14 at 22:48
  • @ user26857 , when I choose $r$ rows, the Cramer formula gives a vector $u=[s_1,\cdots,s_r]^T$ that satisfies the associated $r$ equations (a $r\times r$ system). Then the other equations are automatically satisfied and $u$ satisfies $[c_1,\cdots,c_r]^Tu=b$. Finally $A[s_1,\cdots,s_r,0,\cdots,0]^T=b$. Note also that $s_1,\cdots,s_r$ acts on the first $r$ COLUMNS of $A$. –  Aug 24 '14 at 09:41
  • Maybe it's worthy to mention that the problem (and this answer) work for any field extension instead of $\mathbb Q\subset\mathbb C$. – user26857 Aug 24 '14 at 13:20