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Let $A$ be an $m \times n$ matrix with integer entries. Show that, if the homogeneous system $Ax = 0$ has a nontrivial complex solution then it has a nontrivial integral solution.

I suppose an integral solution here means that all its entries are integers.

Here's what I did -

Let the non-trivial complex solution be $a + bi$, then it is easy to see that $a - bi$ is also a solution. Here $a$ and $b$ are vectors of size $n \times 1$. Any linear combination is also a solution, so $a$ and $bi$ are solutions! Consequently, if $a$ is a solution, so is $\lambda a$, i.e. any scalar multiple of $a$.

I'm stuck here - what do I do next? Clearly it's not always possible to find $\lambda \neq 0$ such that $\lambda a$ is integral, i.e. all its entries are integers. We don't really know much about $a$.

Any help is appreciated, thank you!

stoic-santiago
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  • In the argument you don't use that the matrix has integer entries. You have already shown complex solution $\Rightarrow$ real solution. Now suppose you have your matrix $A$ in the form $LDR$ where $L$ is lower triangular with $1$s on the main diagonal, $R$ is the same upper-triangular, and $D$ is a diagonal matrix. Then all the entries in $L,,D,,R$ are rationals so you have a rational solution, multiplying by common denominator makes it integer. – Alexey Burdin Sep 04 '20 at 12:06
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    Gaussian reduction shows that the solutions must be rational. Then multiply by the common denominator. – Chrystomath Sep 04 '20 at 12:50
  • essentially a duplicate of this https://math.stackexchange.com/questions/3777401/does-real-dimension-equal-rational-dimension/ – user8675309 Sep 04 '20 at 17:11

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