Show determinant of matrix is non-zero

I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please. $$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3$$

Second question, what is the easiest way to argue that $\{1,\sqrt{2},(\sqrt{2})^2\}$ is linearly independent in $\mathbb{Q}$?

Motivation:
Prove that $\mathbb{Q}[\sqrt{2}] = \{a+b\sqrt{2}+c(\sqrt{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field.

Proof: Since $\mathbb{Q}[\sqrt{2}] \subset \mathbb{R}$, we prove $\mathbb{Q}[\sqrt{2}]$ is a subfield of $(\mathbb{R},+,\cdot)$

$\forall (a+b\sqrt{2}+c(\sqrt{2})^2)\in \mathbb{Q}[\sqrt{2}]\backslash\{0\}.$ We want to find $(a+b\sqrt{2}+c(\sqrt{2})^2)$ such that

$(a+b\sqrt{2}+c(\sqrt{2})^2)(d+e\sqrt{2}+f(\sqrt{2})^2) =$
$(ad+2ec+2bf)+(ae+bd+2cf)\sqrt{2}+(af+cd+be)(\sqrt{2})^2 = 1$

Since $\{1,\sqrt{2},(\sqrt{2})^2\}$ is linearly independent (?) over $\mathbb{Q}$, we show there is unique solution to:
$$\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix} \cdot \left[\begin{array}{l l} d\\e\\f \end{array}\right] = \left[\begin{array}{l l} 1\\0\\0 \end{array}\right]$$ Which is equivalent in showing the determinant is non-zero $$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3=(?)$$

By subfield test, 1)2)3)4) is enough to say that $(\mathbb{Q}[\sqrt{2}],+,\cdot)$ is a subfield of $(\mathbb{R},+,\cdot)$ therefore a field.

EDIT: If you have shorter way that prove the proposition without touching my 2 questions, that is even better.

• Your two questions are quite independent. Why not ask two questions? – Mariano Suárez-Álvarez Mar 11 '13 at 22:54
• @MarianoSuárez-Alvarez because they are inherently from the same problem where I need to show $\{a+b\sqrt{2}+c(\sqrt{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field. – mez Mar 11 '13 at 22:56
• If you just need to show it forms a field you could just verify its axioms. – user32847 Mar 11 '13 at 23:05
• If they are connected in some way, tell us how. Why make people guess? – Mariano Suárez-Álvarez Mar 11 '13 at 23:07
• @Alex Sure, but nevertheless these two things will still need to be shown along the way of verifying axioms, namely in the existence of multiplicative inverse. – mez Mar 11 '13 at 23:21

I'm not sure that's the best way to prove that $\mathbb{Q}[\sqrt{2}] = \{a+b\sqrt{2}+c(\sqrt{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ is a field.

I would argue instead that if $a+b\sqrt{2}+c(\sqrt{2})^2 \ne 0$, then the polynomial $f(x) = a+bx+c x^2$ is coprime to $x^{3} - 2$, so there are polynomials $u(x), v(x)$ such that $$1 = f(x) u(x) + (x^{3} - 2) v(x),$$ so that evaluating for $x = \sqrt{2}$, $$1 = f(\sqrt{2}) u(\sqrt{2}) = (a+b\sqrt{2}+c(\sqrt{2})^2) \cdot u(\sqrt{2}),$$ and $u(\sqrt{2}) \in \mathbb{Q}[\sqrt{2}]$ is the required inverse.

• That is short and nice. – mez Mar 12 '13 at 18:42
• What is the name of the theorem about $1 = px+qy$ in the context of UFD? – mez Mar 12 '13 at 21:50
• @mezhang, in the context of an Euclidean domain like $\Bbb{Q}[x]$, it's usually called Bezout's Lemma. For UFD it does not hold in general, think of $2$ and $x$ in $\Bbb{Z}[x]$, their $\gcd$ is $1$, but you cannot write it as a linear combination of them. – Andreas Caranti Mar 12 '13 at 21:55

Since $a,\ b$ and $c$ are rational, we may clear denominators in $$a^3 + 2b^3 -6abc +4c^3 = 0$$ The above equation is a homogenous equation of degree $3$ so we may cancel common factors. If there exists non-trivial solutions to the equation, we may therefore assume without loss of generality that $a,\ b$ and $c$ are integers with $\gcd(a,\ b,\ c)=1$.

Reducing modulo $2$, we find that $a\equiv 0\pmod 2$. Let $a=2\alpha$. Making the substitution and cancelling common factors, we arrive at $$4\alpha^3 +b^3 - 6\alpha bc + 2c^3 = 0$$ Reducing mod $2$ again, we get $b\equiv 0\pmod2$. So let $b=2\beta$ to obtain $$2\alpha^3 + 4\beta^3 - 6\alpha\beta c + c^3 = 0$$ Reducing modulo $2$ one last time gives $c\equiv 0\pmod 2$. This contradicts the fact that $\gcd(a,\ b,\ c)=1$. Therefore there are no non-trivial integer solutions to the above equation. It follows that the determinant is non-zero since $a,\ b$ and $c$ are not all zero.

To show the linear independence of $\left\{1,\ \sqrt{2},\ \left(\sqrt{2}\right)^2\right\}$ in $\mathbb{Q}$, suppose to the contrary that there exists some non-trivial rational linear combination such that $$r_0 + r_1\sqrt{2} + r_2\left(\sqrt{2}\right)^2 = 0$$ Then clearing denominators, there exists a non-trivial integral linear combination of the above set to $0$. Specifically, there exists an integral polynomial $p(x)$ of degree $2$ such that $\sqrt{2}$ is a root. But the minimal polynomial of $\sqrt{2}$ is $x^3 - 2$. This is a contradiction.

Edited in accordance with comment from Marc van Leeuwen:

Suppose $$a+b\root3\of2+c(\root3\of2)^2=0$$ with $a,b,c$ rational. Then $\root3\of2$ is a root of the polynomial $$f(x)=a+bx+cx^2$$ Now, $\root3\of2$ is also a root of $$g(x)=x^3-2$$ So $\root3\of2$ is a root of the gcd of $f$ and $g$. But $g$ is irreducible over the rationals, and $f$ has degree smaller than $g$ has, so $f$ is identically zero or the gcd is a nonzero constant. It isn't a nonzero constant, since it has to vanish at $\root3\of2$, so $f$ is the zero polynomial, so $$a=b=c=0$$ so $$\{{\,1,\root3\of2,(\root3\of2)^2\,\}}$$ is a linearly independent set over the rationals.

• Note however that since $f$ turned out to be $0$, one has $\gcd(f,g)=\gcd(0,x^3-2)=x^3-2$, it is not a constant after all! ($0$ is not a divisor of $x^3-2$ at all, so if you define $\gcd(0,x^3-2)$ at all, it has to be $x^3-2$.) The argument used that the degree of a $\gcd$ cannot exceed the degree of its operands is invalid if one of the operands in $0$. Of course the argument you give is basically valid, but you should not formulate it quite like you have done. – Marc van Leeuwen Mar 12 '13 at 14:33

Let $r=\root3\of2$. The other answerers have shown that $1,r,r^2$ are linearly independent over $\mathbb{Q}$. Eu Yu's answer has also nicely shown that the determinant in question is nonzero. I don't have a better answer than his. However, since this question is, after all, one about determinant, I can't resist the temptation to solve it in (guise of) a matrix theoretical way.

Let $\omega$ be a primitive cube root of unity. Then your matrix is $\mathbb{C}$-similar to $$A=\begin{pmatrix} a &r^2\omega ^2c &r\omega b\\ r\omega b &a &r^2\omega ^2c\\ r^2\omega ^2c &r\omega b &a \end{pmatrix}.$$ This is a circulant matrix. So, its eigenvalues are (see wikipedia): $$\begin{cases} \lambda_1 = a+r^2\omega^2c+r\omega b &= a+r^2\omega^2c+r\omega b,\\ \lambda_2 = a+r^2\omega^2c\,\omega +r\omega b\,\omega^2 &= a+r^2c+rb,\\ \lambda_3 = a+r^2\omega^2 c\,\omega^2 +r\omega b\,\omega &= a+r^2\omega c+r\omega^2 b. \end{cases}$$ Note that $\lambda_2\neq0$ because $1,r,r^2$ are linearly independent over $\mathbb{Q}$. It follows that if both $\lambda_1$ and $\lambda_3$ have nonzero imaginary parts, $\det A\neq0$. Yet, if one of them is real, by inspecting its imaginary part, we get $b=rc$. So $b=c=0$ and $\det A=a^3$ is still nonzero.

• This is very nice. Thank you. I didn't accept that answer yet thought it was complete in hope to see something like this. – mez Mar 12 '13 at 14:07

Just for the linear independence part over $\Bbb Q$. I suppose you know that $\alpha=\sqrt2$, which is by definition the (positive) real root of $X^3-2$, is irrational. But then $X^3-2$ has no rational roots, and (being of degree $3$) is irreducible over $\Bbb Q$, which means it is the minimal polynomial over $\Bbb Q$ of any of its (complex) roots, since such a minimal polynomial has to divide $X^3-2$. In particular the minimal polynomial of $\alpha$ has degree $3$, which implies that $1,\alpha,\alpha^2$ are $\Bbb Q$-linearly independent.