# Let $a_0,a_1,a_2$ be rationals, $\det\begin{pmatrix}a_0&a_1&a_2\\a_2&a_0+a_1&a_1+a_2\\a_1&a_2&a_0+a_1\end{pmatrix}=0$, show $a_0=a_1=a_2=0$

Let $$a_0,a_1,a_2$$ be rationals, $$\det\begin{pmatrix}a_0&a_1&a_2\\a_2&a_0+a_1&a_1+a_2\\a_1&a_2&a_0+a_1\end{pmatrix}=0$$, show $$a_0=a_1=a_2=0$$.

This is a problem in PHD candidate to PKU (2019).

The matrix is not invariant under permutations of $$1,2,3$$. So I could not just let the last column being the linear combination of the first two. Soundly, I have no idea. Any hint is grateful.

• If any row has all its elements zero, the determinant is also zero. Aug 25 at 7:20
• @SubhamKarmakar Then what to do? We need to show $a_i=0$, $\forall\ i$.
– xldd
Aug 25 at 7:33
• @SubhamKarmakar Your comment is off target. There are certainly matrices with zero determinant which do not have any row with all elements zero. OP is only assuming that particular matrix of rational entries is zero, then trying to deduce from that the elements of its first (and therefore the others) row are all zero. Aug 25 at 7:41
• Could you stop making insignificant edits to the title of this question? Aug 25 at 10:20

HINT:

I suggest first to permute $$a_1$$ and $$a_2$$, getting the matrix $$A=\begin{pmatrix}a_0&a_2&a_1\\a_1&a_0+a_2&a_1+a_2\\a_2&a_1&a_0+a_2\end{pmatrix}$$

Now, consider the ring $$\mathbb{Q}[x]/(x^3-x-1)$$ Since the polynomial $$x^3 - x -1$$ is irreducible, we get in fact a field.

The multiplication by an element $$a_0 + a_1 x + a_2 x^2$$ has the above matrix in the basis $$1$$, $$x$$, $$x^2$$.

$$\bf{Added:}$$ Another solution in the style of linear algebra.

Consider the matrix $$X= \begin{pmatrix}0 & 0 & 1\\ 1 & 0& 1\\ 0 & 1 & 0\end{pmatrix}$$

( the companion matrix of the polynomial $$x^3 - x -1$$)

Then our matrix equals $$A = a_0 + a_1 X + a_2 X^2$$

Let's diagonalize $$X$$. Consider $$\alpha$$, $$\beta$$, $$\gamma$$ the roots of the polynomial $$x^3 - x -1$$. Then one can check that $$\begin{pmatrix}1 & \alpha & \alpha^2\\ 1 & \beta & \beta^2\\ 1 & \gamma & \gamma^2 \end{pmatrix} \cdot X = \operatorname{Diag}(\alpha, \beta, \gamma) \cdot \begin{pmatrix}1 & \alpha & \alpha^2\\ 1 & \beta & \beta^2\\ 1 & \gamma & \gamma^2 \end{pmatrix}$$

Therefore, $$X$$ is diagonalizable with eigenvalues $$\alpha$$, $$\beta$$, $$\gamma$$, and $$A$$ is diagonalizable with eigenvalues $$a_0 + a_1 \alpha + a_2 \alpha^2, \ldots$$. We conclude that $$\det A= (a_0 + a_1 \alpha + a_2 \alpha^2)(a_0 + a_1 \beta + a_2 \beta^2) ( a_0 + a_1 \gamma + a_2 \gamma^2)$$

Now we use that $$x^3 - x -1$$ is irreducible over $$\mathbb{Q}$$, so neither of $$\alpha$$, $$\beta$$, $$\gamma$$ is root of a rational polynomial of degree $$\le 2$$.

Let's note that the determinant equals the norm of the element $$a_0 + a_1 x + a_2 x^2$$ in the field $$K = \mathbb{Q}[x]/(x^3 - x -1)$$

$$N^K_{\mathbb{Q}}(a_0 + a_1 x + a_2 x^2) = \det A$$

$$\bf{Added:}$$ Another solution:

Note that we may assume $$a_0$$, $$a_1$$, $$a_2$$ integers (with $$\gcd = 1$$). We used the polynomial $$x^3 - x -1$$ which is irreducible over $$\mathbb{Q}$$. But the polynomial is also irreducible $$\mod 2$$. We could have worked over the basic field $$\mathbb{Z}/2$$ instead of $$\mathbb{Q}$$ and still get the same result. This suggests that we can show the result $$\mod 2$$. Therefore, we can show: if $$a_0$$, $$a_1$$, $$a_2$$ are integers, not all even then the above determinant is odd. Indeed the determinant equals $$\det A = a_0^3 + 2 a_0 ^3 a_3 - a_0 a_1^2 - 3 a_0 a_1 a_2 + a_0 a_2^2 + a_1^3 - a_1 a_2^2 + a_3^3$$

(see this WA calculation). Now, if the $$a_i$$ are in $$\mathbb{Z}/2$$ then the expression equals $$a_0 + a_1 + a_2 + a_0 a_1 + a_0 a_2 + a_1 a_2 + a_0 a_1 a_2 = (1+a_0)(1+ a_1)(1+a_2) + 1$$ and this equals $$1 \pmod 2$$ if at least one of the $$a_i$$ is $$1 \pmod 2$$. We are done.

• Is there any linear algebra proof?
– xldd
Aug 25 at 8:05
• @Eldar Sultanow: The expression factors into a product of $3$ linear forms, with coefficients involving the roots of $x^3 - x -1$. Now we can use the irreducibility of this polynomial. The trick is the factoring. Aug 25 at 8:54
• $X$ is the companion matrix of $x^3-x-1$ which is therefore its minimal polynomial. This is the reason for the ring $R=\{a_0+a_1X+a_2X^2:a_0,a_1,a_2\in\mathbb Q\}$ is isomorphic to $\mathbb Q[x]/(x^3-x-1)$. Aug 25 at 9:19
• @user26857: Yes, indeed!! added that. Aug 25 at 9:48
• @Eldar Sultanow: I mean that the determinant (in $x$, $y$, $z$) has no integral solutions non-zero solutions, and it is true in fact $\mod 2$. Aug 25 at 10:08