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(All real numbers)
Show that
$$ax^2+2hxy+by^2+2gx+2fy+c=0$$ can be factored as $$(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=0$$ iff $$\begin{vmatrix} a&h&g\\ h&b&f\\g&f&c \end{vmatrix} = 0 $$


I've seen people using this as a fact without any proof. For example, in this video (sorry, not english) he states it as a fact w/o any explanation, which is really frustrating. I'm wondering if there is a way to make sense of this using linear algebra or calculus. Highly appreciate any help. Thanks!

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    $\begingroup$ FWIW, the left-hand side is $$ax^2+2hxy+by^2+2gx+2gy+c=\left(\matrix{x & y & 1}\right)\left(\begin{matrix} a&h&g\\ h&b&f\\g&f&c \end{matrix}\right)\left(\matrix{x\\y\\1}\right)$$ $\endgroup$ – Arnaud Mortier Aug 17 at 14:24
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    $\begingroup$ Yes, and due to the $1$ it looks related to the projective plane. $\endgroup$ – Arnaud Mortier Aug 17 at 14:27
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    $\begingroup$ In the OP, it should be $2fy$ instead of $2gy$. $\endgroup$ – Arnaud Mortier Aug 17 at 14:36
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    $\begingroup$ I think it's saying that a degenerate conic is the union of two (not necessarily distinct) lines. $\endgroup$ – saulspatz Aug 17 at 14:44
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    $\begingroup$ The video is missing some conditions. See here. $\endgroup$ – A.Γ. Aug 17 at 19:31
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This is wrong.

To see this, observe that the solution space to $$(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=0$$ is

  • either the union of two straight lines (if $a_1, b_1$ are not both zero and $a_2, b_2$ are not both zero),
  • or one straight line (happens for instance if $a_1, b_1$ are not both zero but $a_2=b_2=0$ and $c_2\neq 0$),
  • or the whole plane (for instance if $a_2=b_2=c_2=0$)
  • or empty (if $a_1=a_2=b_1=b_2=0$ and $c_1\neq 0$, $c_2\neq 0$).

Now consider $x^2+y^2=0$, whose solution set is a single point, i.e. not falling into one of the above cases. It corresponds to $a=b=1$, all other parameters are $0$, and the determinant is zero since $g=f=c=0$.


Note that as was mentioned by @saulspatz in the comments, this is related to degenerate conic sections. Degenerate conic sections are defined as solution sets to quadratic equations like this one, such that the determinant of the matrix you mention is zero. A single point is the only case of a degenerate conic section that cannot be factored into two (possibly degenerate as well) linear forms. In other words, this counterexample is the only counterexample.


Finally, note that the above is due to the fact that you set $\Bbb R$ as your base field. If you were to work over the complex numbers, then $x^2+y^2$ would factor into two linear forms, as $(x+iy)(x-iy)$.

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