In-class Demonstration: Quadratic Forms; Orthogonal Diagonalization - Part 2

By Russell Blyth, modified by Mike May, S.J.

restart: with(LinearAlgebra): with(plots): with(plottools):

Example 3: The quadric surface 2xy + z = 0

C is the matrix of the quadratic form

C := Matrix(3,3,[[0,1,0],[1,0,0],[0,0,0]]);

Find the eigenvalues and eigenvectors of B

evC := [Eigenvectors(C)];

Normalize the eigenvectors:

u1 := Normalize(Column(evC[2],1),Euclidean); u2 := Normalize(Column(evC[2],2),Euclidean); u3 := Normalize(Column(evC[2],3),Euclidean);

Construct a matrix P which orthogonally diagonalizes A

PC := <u1|u2|u3>;

Compute the diagonal matrix (since A = PDNiMpJSJQRyUidEc=, we have D = NiMpJSJQRyUidEc=AP)

DC := Transpose(PC) . C . PC;

Make Maple compute the equation of the rotated conic:

LHSC := Transpose(<X,Y,Z>) . DC . <X,Y,Z> + Transpose(<0,0,1>) . PC . <X,Y,Z>;

The equation is LHSC = 0, which is the equation of a hyperboloid of one sheet.

Let's graph, first the quadric surface relative to the new axes:

implicitplot3d(x^2-z^2+y=0,x=-10..10,y=-10..10,z=-10..10,axes=normal,scaling=constrained,grid=[20,20,20]);

And then relative to the original axes: (the eigenspace basis vectors have been stretched to make them long enough to see clearly)

u15 := 5*u1; u25 := 5*u2; u35 := 5*u3;

u1g := line([0,0,0],convert(u15,list),color=red,thickness=4): u2g := line([0,0,0],convert(u25,list),color=red,thickness=4): u3g := line([0,0,0],convert(u35,list),color=red,thickness=4): conicB := implicitplot3d(2*x*y + z = 0, x=-10..10,y=-10..10,z=-10..10,axes=normal,grid=[20,20,20]): display([u1g,u2g,u3g,conicB],scaling=constrained);