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To which of the seventeen standard quadrics (https://mathworld.wolfram.com/QuadraticSurface.html) do these two equations reduce? \begin{equation} Q_1^2+3 Q_2 Q_1+\left(3 Q_2+Q_3\right){}^2 = 3 Q_2+2 Q_1 Q_3. \end{equation} \begin{equation} -9 Q_2-6 Q_3+3 \left(Q_1^2+\left(3 Q_2+4 Q_3-1\right) Q_1+9 Q_2^2+4 Q_3^2+6 Q_2 Q_3\right) = 0. \end{equation} Further, what are the associated transformations needed to accomplish the reductions?

This is a "distilled" form of a previous more expansive question https://mathoverflow.net/questions/359459/interpret-certain-expressions-in-terms-of-classical-quadratic-surfaces

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  • $\begingroup$ What are the $Q_i$s ? Ordinary real variables ? $\endgroup$
    – Jean Marie
    May 6, 2020 at 5:28
  • $\begingroup$ Yes, real--satisfying the constraint $Q_1\geq 0\land Q_2\geq 0\land Q_3\geq 0\land Q_1+3 Q_2+2 Q_3\leq 1$. $\endgroup$ May 6, 2020 at 12:42
  • $\begingroup$ These conditions mean that you restrict your attention to the interior of a tetraedron... $\endgroup$
    – Jean Marie
    May 6, 2020 at 15:01

2 Answers 2

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$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 3 }{ 2 } & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 \\ - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 9 } & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 6 & 0 & 0 & 0 \\ 0 & \frac{ 81 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 2 } & 2 & - \frac{ 1 }{ 2 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 9 } \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrrr} 6 & 9 & 12 & - 3 \\ 9 & 54 & 18 & - 9 \\ 12 & 18 & 24 & - 6 \\ - 3 & - 9 & - 6 & 0 \\ \end{array} \right) $$

This one has a nonzero constant involved, so no node.

$$ 3\left(x + \frac{3}{2}y +2z - \frac{1}{2}\right)^2 + \frac{81}{4} \left(y - \frac{1}{9}\right)^2 - 1 = 3x^2 + 27 y^2 + 12z^2 + 18yz +12zx+9xy-3x -9y -6z $$

This is an elliptic CYLINDER. Indeed, if we call the original polynomial $f(x,y,z),$ degree 2 but not homogeneous, we find $$ f( x_0 + 2t, y_0, z_0 - t) $$ is independent of $t$

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  • $\begingroup$ I just realized that you had an answer for each question ! $\endgroup$
    – Jean Marie
    May 6, 2020 at 14:51
  • $\begingroup$ @JeanMarie Hi! Yes. I wasn't going to bother with the second one until the OP gave some indication of what was really going on. Then I just did this one. I wrote a program to output in Latex code, including rational numbers, so I can just paste relevant parts of the algorithm directly on an MSE window. Oh, I bought the two volume English translation of Gantmacher, so I finally know what the students mean by Lagrange's method (repeated completing the square) for writing a homogeneous quadratic form as a sum of squares with coefficients. $\endgroup$
    – Will Jagy
    May 6, 2020 at 15:46
  • $\begingroup$ The second equation is that of an elliptical cylinder. $\endgroup$
    – Jean Marie
    May 6, 2020 at 17:04
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    $\begingroup$ Hi! Yes, one has to find different smart methods for not typing again and again latex boring parts... $\endgroup$
    – Jean Marie
    May 6, 2020 at 17:07
  • $\begingroup$ The final expression of the answer, $f(x_0+2 t),y_0,z-t)$ is intriguing, but "what does it tell us", and how/why did it come to your attention? $\endgroup$ May 6, 2020 at 21:53
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This is the simple way, uses rational numbers only. The alternative is to translate coordinates around the single critical point(if there is one and only one point where the gradient is the zero vector). Then, find the eigenvalues and construct an orthogonal matrix $P,$ leading to $P^T H P = D.$

your first one comes out, in letters xyz1, as $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 3 }{ 2 } & 1 & 0 & 0 \\ - 1 & \frac{ 2 }{ 3 } & 1 & 0 \\ 0 & - \frac{ 2 }{ 9 } & - \frac{ 1 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & \frac{ 27 }{ 2 } & 0 & 0 \\ 0 & 0 & - 6 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 2 } & - 1 & 0 \\ 0 & 1 & \frac{ 2 }{ 3 } & - \frac{ 2 }{ 9 } \\ 0 & 0 & 1 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 3 & - 2 & 0 \\ 3 & 18 & 6 & - 3 \\ - 2 & 6 & 2 & 0 \\ 0 & - 3 & 0 & 0 \\ \end{array} \right) $$

which means $$ \small \left(x + \frac{3}{2}y -z \right)^2 + \frac{27}{4} \left(y + \frac{2}{3}z - \frac{2}{9}\right)^2 - 3 \left(z - \frac{1}{3}\right)^2 = x^2 + 9 y^2 + z^2 + 6yz - 2zx+3xy -3y $$

This is an actual cone, cross section elliptic, vertex at $x = 1/3 \; , \; \; y=0 \; , \; \; z=1/3 \; . \;$

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    $\begingroup$ OK--so this result is for the first of the two equations in the question, From the classification table in mathworld.wolfram.com/QuadraticSurface.html this appears to be an "elliptic cone (real)". Right? What about the second equation in the question? According to the last three comments in mathematica.stackexchange.com/questions/220072/… this is simply a line. Can you state its equation? I will also try to obtain it. $\endgroup$ May 6, 2020 at 12:56
  • $\begingroup$ Yes, I now understand from Jean Marie's comment, that there is one answer for each of the quadrics in the question. $\endgroup$ May 6, 2020 at 18:06

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