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Given the equation : $-3y^2 - 4xy + 2xz + 4yz - 2x - 2z + 1 = 0$. Check if the surface described by that equation has a center of symmetry and then by making the correct coordinate system change, find the type of the surface that this equation describes.

Attempt : For the symmetry part, we'll just plug in $x\to -x$, $y\to -y$, $z\to -z$ and we will conclude. My question is for the second part of the exercise. I've found some more like these, but ALL of them were described by a quadratic form, which I could write it's Matrix down and compute Eigenvalues-Eigenvectors and then Gram-Schmidt the eigenvectors to find an canonical form and see what surface it was. But in this particular example, I'd do the following :

$-3y^2 - 4xy + 2xz + 4yz = 2x + 2z - 1$

Now, we have that the equation $f(x,y,z) = -3y^2 - 4xy + 2xz + 4yz$ is a quadratic form that we can compute its Matrix - eigenvalues - eigenvectors and that we can find what surface it describes. Then, this particular surface will be "cut" by the plane $2x+2z - 1 =0$ and thus we can find the surface that the starting equation describes. Is this a correct approach ? If not, please give me some thorough help to understand how to work on these.

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Rebellos
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2 Answers2

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Your idea is good but note that the equation

$$ -3y^2 - 4xy + 2xz + 4yz = 2x + 2z - 1 $$

does not define the intersection of the surface described by the quadratic form with the plane. Such an intersection would have been described by two equations and not a single equation:

$$ -3y^2 - 4xy + 2xz + 4yz = a, \,\,\, 2x + 2z - 1 = b. $$

What works better is to try and write your equation as $(\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = c$. If $\mathbf{v} = \mathbf{0}$, you get the level set $q_A(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = c$ of the quadratic form associated to $A$ which you can analyze by your method. If $\mathbf{v} \neq 0$, the picture is the same with the "center" translated to $\mathbf{v}$ which is what I guess you call "the center of symmetry". Note that we have

$$ (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = \mathbf{x}^T A \mathbf{x} - \mathbf{v}^T A \mathbf{x} - \mathbf{x}^T A \mathbf{v} + \mathbf{v}^T \mathbf{A} \mathbf{v} = \mathbf{x}^T A \mathbf{x} - 2 \mathbf{v}^T A \mathbf{x} + \mathbf{v}^T \mathbf{A} \mathbf{v} $$

so the expression splits into a quadratic part in $\mathbf{x}$, a linear part and a constant part. In your case, quadratic part corresponds to:

$$ A = \begin{pmatrix} 0 & -2 & 1 \\ -2 & -3 & 2 \\ 1 & 2 & 0 \end{pmatrix}.$$

The linear part corresponds to

$$ -2 \mathbf{v}^T A \mathbf{x} = -2(v_1,v_2,v_3)A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = -2 (v_1, v_2, v_3) \begin{pmatrix} -2y + z \\ -2x - 3y + 2z \\ x + 2y \end{pmatrix} = \\ -2v_1(-2y + z) -2v_2(-2x - 3y + 2z) -2v_3(x + 2y) = \\ x(4v_2 - 2v_3) + y(4v_1 + 6v_2 - 4v_3) -z(2v_1 +4v_2) \stackrel{\mbox{?}}{=} -2x - 2z $$

and so

$$ 4v_2 - 2v_3 = -2, \\ 4v_1 + 6v_2 - 4v_3 = 0,\\ 2v_1 + 4v_2 = 2 $$

which implies that

$$ \mathbf{v} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \,\,\, \mathbf{v}^T A \mathbf{v} = 2. $$

Thus, we get

$$ (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = q_A(\mathbf{x} - \mathbf{v}) = -3y^2 -4xy + 2xz + 4yz -2x - 2z + 2. $$

To get your equation, we see that we need to choose $c = 1$ and then

$$ q_A(\mathbf{x} - \mathbf{v}) = -3y^2 -4xy + 2xz + 4yz -2x - 2z + 2 = 1 \iff \\ -3y^2 -4xy + 2xz + 4yz -2x - 2z + 1 = 0. $$

Since $A$ has eigenvalues $-5,1,1$, the surface is a one-sheeted circular hyperboloid (see here).

levap
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  • Well, hell, that's a hard exercise eventually. Thanks so much for the answer, going through it multiple times to understand it .. How do you separate the surfaces by their eigenvalues by the way ? – Rebellos Jun 22 '16 at 14:31
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    You're welcome - it is indeed relatively hard if you haven't seen the technique before. Regarding the surfaces, if $(\lambda_i)_{i=1}^3$ are the eigenvalues of $A$ then after a change of coordinates you can describe the level set $q_A(\mathbf{x}) = c$ as $\lambda_1 u_1^2 + \lambda_2 u_2^2 + \lambda_3 u_3^2 = c$. Depending on the signs and relative sizes of the $\lambda_i$'s, you can then determine which shape you get. If for example $\lambda_1 = \lambda_2 = \lambda_3$ and $c > 0$, you get a sphere. If $\lambda_1,\lambda_2,\lambda_3 > 0$ but they are not necessarily equal, you get an ellipsoid. – levap Jun 22 '16 at 14:42
  • You can find the description of all possible surfaces in https://en.wikipedia.org/wiki/Quadric. Note that the parabolic surfaces cannot be described by an equation of the form $q_A(\mathbf{x} - \mathbf{v}) = c$ we discussed above. – levap Jun 22 '16 at 14:43
  • BTW, according to the wiki article, what we get is called a "circular hyperboloid of one sheet". It might be a good exercise for you to find the axis of symmetry of this hyperboloid. – levap Jun 22 '16 at 14:50
  • Hey again, thanks so much for the answers. I am solving another similar exercise now to practice your method. – Rebellos Jun 22 '16 at 15:23
  • how do you get $(\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = q_A(\mathbf{x} - \mathbf{v}) = 1$ ? I mean the number $1$, how do you derive that ? – Rebellos Jun 22 '16 at 15:50
  • Let us continue this discussion in chat. – Rebellos Jun 22 '16 at 16:01
  • I can't talk right now but I added some details to my answer. – levap Jun 22 '16 at 18:58
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Some simplifications. The matrix of the quadratic form part is half the Hessian matrix of second partial derivatives, so indeed $$ A = \begin{pmatrix} 0 & -2 & 1 \\ -2 & -3 & 2 \\ 1 & 2 & 0 \end{pmatrix}.$$

As it happens, $\det A \neq 0.$ As a result, there really is a "center", call it column vector $C,$ given by $$ 2 A C = \left( \begin{array}{r} 2 \\ 0 \\ 2 \end{array} \right) $$ Now, as $A$ is invertible, we will succeed if solving by Gauss elimination, and this gives $$ C = \left( \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right) $$ This is the only critical point of the full function!!!!! We simply set the gradient vector to the zero vector, that is all.

We then make translated variables, $$ u = x-1, \; \; v = y, \; \; w = z-1 $$ to get $$ -3v^2 - 4 uv+2 u w + 4 v w - 1 = -3y^2 - 4 xy+2xz+4yz-2x-2z+1 $$

Next, it is not necessary to use eigenvalues here, see reference for linear algebra books that teach reverse Hermite method for symmetric matrices We can find a matrix $P$ with all rational entries, $\det P = 1,$ and $P^TAP=D$ diagonal with a fairly simple algorithm. In turn, the diagonal entries of $D$ are not the eigenvalues, but they must have the same $\pm$ signs as the eigenvalues; this is Sylvester's Law of Inertia. If desired, we could also mutilpy through by a common denominator to make everything integers, at the cost of changing the determinant of $P.$ Let me just typeset the one with all integers,

$$ Q = \left( \begin{array}{ccc} 28 & -20 & 7 \\ 28 & 8 & 14 \\ 0 & 0 & 28 \end{array} \right) $$ $$ Q^T A Q = \left( \begin{array}{ccc} -5488 & 0 & 0 \\ 0 & 448 & 0 \\ 0 & 0 & 980 \end{array} \right) $$

parisize = 4000000, primelimit = 500509
? a = [ 0, -2,1; -2,-3,2; 1,2,0]
%1 = 
[0 -2 1]

[-2 -3 2]

[1 2 0]

? a - mattranspose(a)
%2 = 
[0 0 0]

[0 0 0]

[0 0 0]

? pol = charpoly(a)
%3 = x^3 + 3*x^2 - 9*x + 5
? factor(pol)
%4 = 
[x - 1 2]

[x + 5 1]

? p1 = [ 1,0,0; 1,1,0; 0,0,1]
%5 = 
[1 0 0]

[1 1 0]

[0 0 1]

? a1 = mattranspose(p1) * a * p1
%6 = 
[-7 -5 3]

[-5 -3 2]

[3 2 0]

? p2 = [ 1, -5/7, 3/7; 0,1,0; 0,0,1]
%7 = 
[1 -5/7 3/7]

[0 1 0]

[0 0 1]

? a2 = mattranspose(p2) * a1 * p2
%8 = 
[-7 0 0]

[0 4/7 -1/7]

[0 -1/7 9/7]

? p3 = [ 1,0,0; 0,1,1/4; 0,0,1]
%9 = 
[1 0 0]

[0 1 1/4]

[0 0 1]

? a3 = mattranspose(p3) * a2 * p3
%10 = 
[-7 0 0]

[0 4/7 0]

[0 0 5/4]

? p = p1 * p2 * p3
%11 = 
[1 -5/7 1/4]

[1 2/7 1/2]

[0 0 1]

? a
%12 = 
[0 -2 1]

[-2 -3 2]

[1 2 0]

? d = mattranspose(p) * a * p
%13 = 
[-7 0 0]

[0 4/7 0]

[0 0 5/4]

? q = 28 * p
%14 = 
[28 -20 7]

[28 8 14]

[0 0 28]

? m = mattranspose(q) * a * q
%15 = 
[-5488 0 0]

[0 448 0]

[0 0 980]

?

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Will Jagy
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  • If you don't use eigenvalues, you won't be able to tell a sphere from an ellipsoid apart, etc. If you don't care about it, then you can get along with diagonalizing the quadratic form by similarity. If you do and you care about the axis, their orientation in space, etc, then you need to do orthogonal diagonalization... – levap Jun 22 '16 at 18:51
  • @levap yes. I have collected a number of posts where the students are asked to find $P^T AP= D$ when the eigenvalues of $A$ have no closed form, and the class is not in numerical methods either. Since the eigenvalues here are integers, your method seems best. However, compare this confused student: http://math.stackexchange.com/questions/395634/given-a-4-times-4-symmetric-matrix-is-there-an-efficient-way-to-find-its-eige It is fair to say that I have been trying to popularize the non-eigenvalue methods, not with much luck. – Will Jagy Jun 22 '16 at 18:58
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    I agree with you completely in the case one wants to understand the geometric shape up to affine equivalence but if one wants to understand the geometric shape up to euclidean equivalence, one can't avoid finding the eigenvalues (and eigenvectors) explicitly. If this turns out to be hard and there is no software available, then "diagonalization" using elementary row and column operators at least can give you some information about the shape. – levap Jun 22 '16 at 19:05