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Prove $Ax = \frac{1}{2}x$ only has the trivial solution where $A$ is a $n \times n$ matrix with integer entries and $x = (x_1, \ldots , x_n)$.

I am a bit rusty on my linear algebra and trying to review. I tried using the Invertible Matrix theorem. The problem was I couldn't seem to gain any traction with any of the equivalent statements.

Here is the link for anyone that needs a refresher: Invertible Matrix Theorem

Looking for hints rather than a specific solution. 
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    $\begingroup$ It means that $\frac12$ is an eigenvalue: a root of the characteristic equation. $\endgroup$ – Lord Shark the Unknown Aug 14 at 18:15
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    $\begingroup$ @LordSharktheUnknown: It means that $\frac12$ is not an eigenvalue. $\endgroup$ – Henning Makholm Aug 14 at 18:15
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Hint: The eigenvalues of $A$ are the roots of its characteristic polynomial, which is a monic polynomial with integer coefficients.

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  • $\begingroup$ I am probably missing something, but it seems to me that any matrix with no $1/2$ eigenvalue has only $x=0$ as the (trivial) solution. $\endgroup$ – jobe Aug 14 at 18:24
  • $\begingroup$ @jobe Of course, but how do you know that $\frac12$ is not an eigenvalue? $\endgroup$ – José Carlos Santos Aug 14 at 18:25
  • $\begingroup$ @JoséCarlosSantos By Gauss Lemma a monic polynomial with integer coefficients only has integer or irrational roots thus $\frac{1}{2}$ cannot be a root thus there is no non-zero solution to $Ax = \frac{1}{2}x$ $\endgroup$ – all.over Aug 14 at 18:27
  • $\begingroup$ That is correct! $\endgroup$ – José Carlos Santos Aug 14 at 18:28
  • $\begingroup$ I misunderstood the question... I thought that what was being asked is to prove that $Ax=\frac{1}{2}x$ has $x=0$ its only solution when $A$ has integer entries, what is obviously wrong. Now I see that $A$ having integer entries is assumed, not something to be proved. $\endgroup$ – jobe Aug 14 at 18:32
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Hint: Evaluate the characteristic polynomial of $A$ at $\lambda=\frac12$. Since all coefficients are integers and the leading term is $\lambda^n$, what can you say about the value?

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  • $\begingroup$ I take it this was headed in the same direction as the other answer using the Rational Root Theorem. We would need $2 \lvert \lambda^n = \frac{1}{2^n}$ and clearly this is not possible. $\endgroup$ – all.over Aug 14 at 18:35
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    $\begingroup$ @all.over: The Rational Root Theorem will do it, yes. I never recall how it goes, though, so what I really had in mind was that the first term is the single term that is not a multiple of $2^{-(n-1)}$. $\endgroup$ – Henning Makholm Aug 14 at 18:39
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Write it as $x=2Ax$. If this has a nonzero solution, it has a nonzero rational solution and indeed a nonzero integer solution. But if $x$ has integer entries, $x=2Ax$ has even entries. So then $x=2Ax$ has entries divisible by $4$, etc.....

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  • $\begingroup$ I'm sort of following this solution. Not quite sure what the divisibility by 4 does for it though. $\endgroup$ – all.over Aug 14 at 18:54
  • $\begingroup$ Well, then $x=2Ax$ will have entries divisible by $8$, etc. $\endgroup$ – Lord Shark the Unknown Aug 14 at 19:23
  • $\begingroup$ It seems there is something obvious I am missing here. $A$ has integer entries so $2A$ has even entries. If $x$ has integer entries then $2Ax$ has even integer entries (even if x=0) thus $x$ has even entries by $x = 2Ax = A2x$ so we can divide entries of $x$ by $4$ but then similar reasoning we can divide entries of $x$ by $8$, ..., $2^n$. Hence entries of $x$ are of the form $2^nk_i, 1\le i \le n, k_i \in \mathbb{Z}$ where $n$ grows indefinitely thus an absurdity that isn't possible unless $x = (0,0, \ldots, 0)$. $\endgroup$ – all.over Aug 14 at 20:08

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