Given a conic equation $$H:= ax^2+by^2+cz^2+2fyz+2gzx+2hyx=0,$$

What is (are) the conditions on the coefficients that it represents a pair of planes?

I know that if this is true, then $H$ can be written as $$H=(px+qy+rz)(sx+ty+uz)$$ for some constants $p, q, r, s,t, u$. Expanding gives

$$ps x^2 + qt y^2 + ruz^2 + (pt+qs)xy+ (pu+ rs) xz+ (qu+rt) yz = H,$$

which gives us a system of $6$ equations \begin{align} ps &=a \\ qt &= b \\ ru &= c \\ pt+qs &= 2f \\ pu + rs &= 2g \\ qu+ rt &= 2h. \end{align}

The algebra gets a little bit too messy for me and I do not know how to move on. In the solution, it is given that $$f^2 \geq bc, \ \ g^2 \geq ac, \ \ h^2 \geq ab $$

and the determinant of some minor of \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} are zero.

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    $\begingroup$ Any thoughts of your own? $\endgroup$
    – amd
    Dec 20, 2017 at 0:01
  • $\begingroup$ Actually, I am ambiguous. In a text, it was given that some determinant is zero and $f^2\geq bc$, $g^2\geq ac$ and $h^2\geq ab$. But I think these conditions $f^2\geq bc$, $g^2\geq ac$ and $h^2\geq ab$. are obvious. $\endgroup$
    – user90533
    Dec 20, 2017 at 0:05
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    $\begingroup$ Here's a small hint: Why don't you start with the equation of a pair of planes and see what that will look like? How do you get a pair of planes? Start with an easy case, like the $xy$-plane and the $xz$-plane. $\endgroup$ Dec 20, 2017 at 0:06
  • $\begingroup$ I know the solution part, but are the conditions $f^2\geq bc$, $g^2\geq ac$ and $h^2\geq ab$ necessary? $\endgroup$
    – user90533
    Dec 20, 2017 at 0:08

2 Answers 2


We can confirm that

$$f^2 \geq bc, \ \ g^2 \geq ac, \ \ h^2 \geq ab$$

and the determinant of a certain matrix equal to $0$ are necessary conditions.

Here is how. The quadratic form can be written under a matrix form as follows :

$$\tag{1}\begin{bmatrix} x&y&z \end{bmatrix}\underbrace{\begin{bmatrix} a&h&g\\h&b&f\\g&f&c \end{bmatrix}}_Q\begin{bmatrix} x\\y\\z\\ \end{bmatrix}$$

If the quadratic form can be expressed as the product of two linear forms (when we will equal it to zero, we will get the equations of the two planes passing through the origin):


The equivalent matrix form of this expression is (many thanks to @Will Jagy) :

$$\begin{bmatrix} x&y&z \end{bmatrix}\begin{bmatrix} p\\q\\r\\ \end{bmatrix}\begin{bmatrix} s&t&u \end{bmatrix}\begin{bmatrix} x\\y\\z\\ \end{bmatrix}+\begin{bmatrix} x&y&z \end{bmatrix}\begin{bmatrix} s\\t\\u\\ \end{bmatrix}\begin{bmatrix} p&q&r \end{bmatrix}\begin{bmatrix} x\\y\\z\\ \end{bmatrix}=$$

$$\tag{2}\begin{bmatrix} x&y&z \end{bmatrix}\underbrace{\begin{bmatrix} 2ps&(pt+qs)&(pu+rs)\\(pt+qs)&2qt&(qu+rt)\\(pu+rs)&(qu+rt)&2ru\end{bmatrix}}_Q \begin{bmatrix} x\\y\\z\\ \end{bmatrix}$$

And yes, comparing (1) and (2), we check that necessarily, for example the "diagonal minor":

$$\begin{vmatrix} b&f\\f&c \end{vmatrix}=bc-f^2=4qurt-(qu+rt)^2=-(qu-rt)^2 \leq 0$$

the same for the other expressions...

Remark : As we can write :

$$Q=\begin{bmatrix} p&s\\q&t\\r&u \end{bmatrix} \begin{bmatrix} s&t&u\\p&q&r \end{bmatrix}, $$

the rank of $Q$ is at most 2.

Thus, $det(Q)=0$.

The rank of $Q$ falls to $1$ iff $\begin{bmatrix} p\\q\\r \end{bmatrix}$ and $ \begin{bmatrix} s\\t\\u \end{bmatrix} $ are proportional (the case of a double plane).

  • $\begingroup$ Jean, I got the symmetric matrix form of the product of two linears was $P S^T + S P^T,$ where $P$ and $S$ are yor column vectors of coefficients. In particular, symmetric matrix was rank two when $P,S$ not parallel. $\endgroup$
    – Will Jagy
    Dec 20, 2017 at 2:05
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    $\begingroup$ @Will Jagy I must leave now, but I will think a little more about your expression $PS^T+SP^T$... $\endgroup$
    – Jean Marie
    Dec 20, 2017 at 2:27
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    $\begingroup$ @Will Jagy One thing is sure, we haven't to consider the (degenerate) case of parallel planes because all planes here must contain the origin. $\endgroup$
    – Jean Marie
    Dec 20, 2017 at 2:37
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    $\begingroup$ Looks good to me. I upvoted a few minutes ago. It must be pretty late where you live. $\endgroup$
    – Will Jagy
    Dec 20, 2017 at 3:36
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    $\begingroup$ Very interesting. I scratch my head : this case of "two planes degeneracy" allows to interpret the matrix of our quadratic form as a certain Hessian with the form $PS^T+SP^T$, which, besides, has eigenvalues $(0, \lambda_1 \geq 0, \lambda_2 \leq 0)$ (I haven't a real proof, it is only based on extensive simulations). Does this constitutes a characterization of this case ? Hum, hum... $\endgroup$
    – Jean Marie
    Dec 20, 2017 at 23:09

This is a quadratic form, which is associated with the symmetric matrix $$Q_H=\begin{bmatrix} a&h&g\\h&b&f\\g&f&c \end{bmatrix}$$ and the quadric is degenerate if and only if the discriminant of $H$, i.e. $\det Q_H$ is $0$.

  • $\begingroup$ (ctd...) or example taking $a=h=1, b=2$ and all other coefficients set to $0$, you have $(x+y)^2+y^2$. Setting this expression to zero, you get a degenerate case which is a line (with equations $y=-x=0$), but not a pair of planes. $\endgroup$
    – Jean Marie
    Dec 20, 2017 at 1:52
  • $\begingroup$ I am not the downvoter ! I am fully aware that in this kind of situations, there are subtle differences and degeneracies of one kind that aren't degeneracies of another kind... $\endgroup$
    – Jean Marie
    Dec 20, 2017 at 2:02
  • $\begingroup$ I was completly wrong about rank=1 !!! (I have erased it because it would mislead a future reader, but I recognize my error). Thanks to a remark of @Will Jagy I have written a new version. I was right only on one thing: the remark about subtleties of this kind of situations... $\endgroup$
    – Jean Marie
    Dec 20, 2017 at 3:27
  • $\begingroup$ @Jean Marie: I focused only on the degeneracy criterion, not the type of degeneracy. $\endgroup$
    – Bernard
    Dec 20, 2017 at 9:40
  • $\begingroup$ That's right. And there are other degeneracy cases than those that have been considered. Regards. $\endgroup$
    – Jean Marie
    Dec 20, 2017 at 21:07

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