# Prove $Ax = \frac{1}{2}x$ only has the trivial solution where $A$ has all integer entries.

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.

• It means that $\frac12$ is an eigenvalue: a root of the characteristic equation. – Lord Shark the Unknown Aug 14 at 18:15
• @LordSharktheUnknown: It means that $\frac12$ is not an eigenvalue. – Henning Makholm Aug 14 at 18:15

Hint: The eigenvalues of $$A$$ are the roots of its characteristic polynomial, which is a monic polynomial with integer coefficients.

• 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. – jobe Aug 14 at 18:24
• @jobe Of course, but how do you know that $\frac12$ is not an eigenvalue? – José Carlos Santos Aug 14 at 18:25
• @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 $Ax = \frac{1}{2}x$ – all.over Aug 14 at 18:27
• That is correct! – José Carlos Santos Aug 14 at 18:28
• 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. – jobe Aug 14 at 18:32

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?

• 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. – all.over Aug 14 at 18:35
• @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 $2^{-(n-1)}$. – Henning Makholm Aug 14 at 18:39

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.....

• I'm sort of following this solution. Not quite sure what the divisibility by 4 does for it though. – all.over Aug 14 at 18:54
• Well, then $x=2Ax$ will have entries divisible by $8$, etc. – Lord Shark the Unknown Aug 14 at 19:23
• 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 $x = (0,0, \ldots, 0)$. – all.over Aug 14 at 20:08