Exercise 5.5.9 (Quadratic form $dxy + eyz + fzx$)

Question

In diagonalizing a quadratic form by repeatedly completing the square, we encounter a problem if $a= b = c = 0$. Show that a quadratic form of the shape $dxy + eyz + fzx$ always takes the value $0$ nontrivially. Explain what happens if you put $x = u+v,\ y = u-v$. Similarly, show that any form of the shape $ax^2 + eyz$ takes the value $0$ nontrivially.

Richard Ganaye · Accepted Answer

\begin{proof} 
Put $f(x,y,z) = dxy + eyz +fzx$. Then $f(1,0,0) = 0$, so $f$ takes the value $0$ nontrivially.

Suppose that $f 
e 0$. Then $(d,e,f) 
e (0,0,0)$. Assume for instance that $d 
e 0$. The linear change of variables $x = u + v,\ y = u -v, z = w$ transforms the form $f$ in
\begin{align*}
g(u,v,w) &= f(u+v, u-v,w)\
&=d(u^2 - v^2) + e (u-v)w + fw(u+v)\
&= du^2 - dv^2 +(e + f)uw  + (-e+f) vw.
\end{align*}
Since $d 
e 0$, we can diagonalize $g$ by completing the square. Therefore we can diagonalize $f$.

Using formula (1) in Problem 3 (and Sagemath), this gives
\begin{align*}
g(u,v)  =
\frac{1}{4} \, d {\left(2 \, u + \frac{{\left(e + fight)} w}{d}ight)}^{2} - \frac{1}{4} \, d {\left(2 \, v + \frac{{\left(e - fight)} w}{d}ight)}^{2} - \frac{1}{4} \, {\left(\frac{{\left(e + fight)}^{2}}{d} - \frac{{\left(e - fight)}^{2}}{d}ight)} w^{2},
\end{align*}
thus (if $d 
e 0$)
\begin{align*}
f(x,y) &= 
\frac{1}{4} \, d {\left(x + y + \frac{{\left(e + fight)} z}{d}ight)}^{2} - \frac{1}{4} \, d {\left(x - y + \frac{{\left(e - fight)} z}{d}ight)}^{2} - \frac{1}{4} \, {\left(\frac{{\left(e + fight)}^{2}}{d} - \frac{{\left(e - fight)}^{2}}{d}ight)} z^{2}\
&=
\frac{1}{4} \, d {\left(x + y + \frac{{\left(e + fight)} z}{d}ight)}^{2} - \frac{1}{4} \, d {\left(x - y + \frac{{\left(e - fight)} z}{d}ight)}^{2} - \frac{ef}{d}z^2.
\end{align*}

Finally, if $f(x,y,z) = ax^2 + e yz$, then $f(0,0,1) = 0$, so $f$ takes the value $0$ nontrivially. 
\end{proof}

Exercise 5.5.9 (Quadratic form $dxy + eyz + fzx$)

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Exercise 5.5.9 (Quadratic form $dxy + eyz + fzx$)

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