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I am having a little trouble with this problem -

Let $A$ be a $m\times n$ matrix and $v$ be a $n\times 1$ matrix, both of which only has rational entries. It is known that the equation $Ax=v$ has a solution in $\mathbb R^n$. Does this imply that the given equation will have a solution with rational entries?

I think that the result is true, but I am unable to prove it.Thanks for any help.

user26857
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Ester
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2 Answers2

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If you set the system up as an augmented matrix $A\mid v$, and carry out row reductions, the row reduced form will not be inconsistent, because of the assumption that there is a solution in $\mathbb{R}^n$. Then if the row reduced augmented matrix is used in the usual way by solving for each basic variable in terms of the augmented part of the row reduced matrix and rational combinations of non-basic variables ("parameters"), the previously assumed real solution will occur as one possible choice of the parameters. However all coefficients involved in the expressions of the solved for basic variables will be rational, so that if you assign rational values to any such parameters, the result will be a solution in $\mathbb{Q}^n.$

user26857
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coffeemath
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I would answer in the following steps:

  1. From the fact that you know a solution exists, you have $v$ in the range of $A$, and $x$ can be assumed in the orthogonal complement of its null space.

  2. Let $B$ the matrix that describes the mapping induced by $A$ from the orthogonal complement of its null space to its range. $B$ is a quadratic, non-singular matrix.

  3. Cramer's rule can be applied to $B$, and it only involves addition, multiplication and division.

Thus, your solution will be rational.

Guido Kanschat
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