Reduced equation of quadratic forms

  • #1
grimTesseract
4
0

Homework Statement

Given the following quadric surfaces:

1. Classify the quadric surface.
2. Find its reduced equation.
3. Find the equation of the axes on which it takes its reduced form.

Homework Equations



The quadric surfaces are:

(1) ##3x^2 + 3y^2 + 3z^2 - 2xz + 2\sqrt{2}(x+z)-2 = 0 ##

(2) ##2x^2 + 2y^2 -z^2 -2xy -2xz -4yz + 2x + 2y +10z -7 = 0 ##

(3) ##x^2 + y^2 -2 z^2 + 2xy -3 \sqrt{2}x +3 \sqrt{2}y = 0##

The Attempt at a Solution

I was able to classify (1) and (3) as an ellipsoid and a hyperbolic paraboloid, respectively. However, the characteristic polynomial of (2) has decimal roots and I wasn't able to continue past that... Wolfram Alpha tells me it should be a two-sheeted hyperboloid.

However, I do not know how to compute the 2nd or 3rd parts of my problem. I'm sure it will emerge from the work I've already done, so I include my methodology:

I found the eigenvalues of the matrix associated to the quadric form. In the case of (1) these were ##2, 3, 4## with the associated eigenvectors ##(1,0,1),(0,1,0),(-1,0,1)## respectively.

I normalized these, and my basis became ##(\frac{1}{\sqrt2}, 0, \frac{1}{\sqrt2}), (0, 1, 0), (\frac{-1}{\sqrt2}, 0, \frac{1}{\sqrt2})##. Not sure how to proceed from here. Knowing that the determinant of the original matrix was nonzero, I was able to deduce this was an ellipsoid from the signs of the eigenvalues and of the lower order terms.

In the case of (3), the eigenvalues are ##-2, 2, 0## and the normalized eigenvectors are ##(0, 0, 1), (\frac{1}{\sqrt2}, \frac{1}{\sqrt2}, 0), (\frac{-1}{\sqrt2}, \frac{1}{\sqrt2}, 0) ##. These are the columns of my invertible matrix ##Q## and I calculated ##Q^TB##, where ##B## is the matrix of the coefficients of my lower order terms. I was again able to deduce from a list of properties that this is a hyperbolic paraboloid.

And yet now I'm stuck. All help appreciated.
 
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  • #2
grimTesseract said:
(1) ##3x^2 + 3y^2 + 3z^2 - 2xz + 2\sqrt{2}(x+z)-2 = 0 ##
...

However, I do not know how to compute the 2nd or 3rd parts of my problem. I'm sure it will emerge from the work I've already done, so I include my methodology:

I found the eigenvalues of the matrix associated to the quadric form. In the case of (1) these were ##2, 3, 4## with the associated eigenvectors ##(1,0,1),(0,1,0),(-1,0,1)## respectively.

I normalized these, and my basis became ##(\frac{1}{\sqrt2}, 0, \frac{1}{\sqrt2}), (0, 1, 0), (\frac{-1}{\sqrt2}, 0, \frac{1}{\sqrt2})##. Not sure how to proceed from here. Knowing that the determinant of the original matrix was nonzero, I was able to deduce this was an ellipsoid from the signs of the eigenvalues and of the lower order terms.
.
This gives a matrix corresponding to a rotation about the ##\ y\ ## axis of ±45 ° .

## \left [ \begin{matrix} \displaystyle
\frac 1 {\sqrt{2}} & 0 & \displaystyle \frac 1{\sqrt{2}} \\
0 & 1 & 0 \\
\displaystyle \frac {-1} {\sqrt{2}} & 0 & \displaystyle \frac 1 {\sqrt{2}}
\end{matrix} \right ] ##

Use it to find x and z in terms of x' and z' . Plug them in. Do some algebra.
 

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