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Suppose $a_{0},a_{1},a_{2}\in\mathbb{Q}$, such that the following determinant is zero, i.e.

$ \left |\begin{array}{cccc}\\ a_{0} &a_{1} & a_{2} \\ \\ a_{2} &a_{0}+a_{1} & a_{1}+a_{2} \\ \\ a_{1} &a_{2} & a_{0}+a_{1}\\ \end{array}\right| =0$

Problem. Show that $a_{0}=a_{1}=a_{2}=0.$

I think it's equivalent to show that the rank of the matrix is $0$, and it's easy to show the rank cannot be $1$. But I have no idea how to show that the case of rank 2 is impossible. So, is there any better idea?

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  • $\begingroup$ This is false. Try $a_1= a_0+a_2=0$. $\endgroup$
    – abx
    Feb 28, 2021 at 5:18
  • $\begingroup$ @abx Substituting $a_1=0$ and $a_2=-a_0$ gives determinant $-a_0^3$. So that isn't a counterexample. $\endgroup$
    – Brendan McKay
    Feb 28, 2021 at 5:22
  • $\begingroup$ @Brendan McKay: I find $a_0^3-a_0a_2^2$. $\endgroup$
    – abx
    Feb 28, 2021 at 5:29
  • $\begingroup$ @abx With two substitutions only one variable should be left. I'm using Maple. $\endgroup$
    – Brendan McKay
    Feb 28, 2021 at 5:32
  • $\begingroup$ There is no counterexample with any of the variables equal to 0. For example if $a_0=0$ and $a_1=ca_2$ then the determinant is $a_2^3(c^3-c^2+1)$. The cubic is irreducible so only $a_2=0$ makes this 0. There are 6 cases like this. $\endgroup$
    – Brendan McKay
    Feb 28, 2021 at 5:36

2 Answers 2

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If there is a rational nonzero solution, there is an integer nonzero solution by multiplying up. At least one of the integers can be assumed odd by dividing out a common power of two.

The determinant $a_0^3+2a_0^2a1+a_0a_1^2-3a_0a_1a_2-a_0a_2^2+a_1^3-a_1^2a_2+a_2^3$ is odd unless $a_0,a_1,a_2$ are all even.

This contradiction shows that $a_0=a_1=a_2=0$ is only rational solution.

Incidentally, the result does not hold modulo an arbitrary prime. For example $a_0=1, a_1=2, a_2=3$ works mod 5.

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    $\begingroup$ Alternatively, evaluate in Z/2. $\endgroup$
    – Jeff Strom
    Feb 28, 2021 at 6:21
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You can check that the determinant is the product of three terms $a_2+ma_1+a_0/m$ as $m$ runs over the three roots of the cubic $m^3-m-1$, of which one is real and the other two are complex conjugates. This does not immediately answer the question (which Brendan McKay has done anyway) but it may be useful context.

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