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Peter A. Brown 3-December, 2001 Math 308A

Ellipse and Linear Algebra

Abstract

Linear algebra can be used to represent conic sections, such as the ellipse. Before

looking at the ellipse directly symmetric matrices and the quadratic form must first be

considered. Then it can be shown, how to write the equation of an ellipse in terms of

matrices. For an ellipse that is not centered on the standard coordinate system an example

will show how to rotate the ellipse.

Symmetric Matrices

The symmetric matrix is a matrix in which the on ether side of the

diagonal, in corresponding positions are the same.

é1 5ù é 9 - 2ù éa bù ê ú ê ú ê ú ë5 8û ë- 2 7 û ëb cû Examples:

Quadratic Form

Now we have seen the symmetric matrices, we can move on to the quadratic

é ùé ù é ù T a b x1 ax1 + bx2 2 2 Q(x) = x Ax = [x1 x2 ]ê úê ú = [x1 x2 ]ê ú = ax1 + 2bx1x2 + cx2 ëb c ûëê x 2 ûú ëê bx1 + cx2 ûú

Page 1 of 5 3-December, 2001 Page 2 of 6 Peter A. Brown form. The quadratic form is the , Q(x) = xTAx, where x is a variable vector. 3-December, 2001 Page 3 of 6 Peter A. Brown

Ellipse

An ellipse has a the standard equation form:

x2 y2 + =1 a 2 c2

x2 y2 + = 1 10 2 4 2

Change Variable

Before we can rotate an ellipse we first need to see how to change the variable vector. By changing the variable ellipses in non standard form can be changed into 3-December, 2001 Page 4 of 7 Peter A. Brown

standard form. By using the function Q(x) (expanded) the x1x2 term, determines how much rotation an ellipse has relative to the standard coordinates, so by changing the variable in affect we are changing the coordinate system.

The Principal Axes Theorem:

Let Abe an n x n symmetric matrix. Then there is an orthogonal change

of variable, x=Py, that transforms the quadratic form xTAx into a

T quadratic from y Dy with no cross-product term (x1x2) (Lay, 453).

Example: Ellipse Rotation

Use the Principal Axes Theorem to write the ellipse in the quadratic form with no

x1x2 term. 3-December, 2001 Page 5 of 7 Peter A. Brown

2 2 5x1 - 4x1x2 + 5x2 = 1

Solution:

1. Write the matrix A for the equation: é 5 - 2ù A = ê ú 2 2 ë- 2 5 û 5x1 - 4x1x2 + 5x2 = 1

2. Find the eigenvalues for A. = 7,3

é- 1ù é1 ù ê ú = 7 ê ú = 3 ë 1 û ë1 û 3. Use the eigenvalues to find the eigenvectors. 3-December, 2001 Page 6 of 7 Peter A. Brown

4. Check to make sure the two eigenvectors are orthogonal, if the dot product equals

zero then the two vectors are orthogonal.

é1 ù [- 1 1]ê ú = 0 5. A = PDP-1: ë1 û é- 1 1ù é7 0ù é- 1/2 1/2ù P = ê ú D = ê ú P- 1 = ê ú ë 1 1û ë0 3û ë 1/2 1/2û

éx ù é y ù ê 1 ú ê 1 ú 6. x = Py, x = y = ëêx 2 ûú ëêy 2 úû

7. Compute

2 2 T T T T T 5x1 - 4x1x2 + 5x2 = x Ax = (Py) A(Py) = y P APy = y Dy

é ùé ù é ù 7 0 y1 7y1 2 2 = [y1 y2 ]ê úê ú = [y1 y2 ]ê ú = 7y1 + 3y2 ë0 3ûëê y ûú ëê3 y ûú 8. 2 2

9. Therefore

2 2 2 2 \ 5x1 - 4x1 x2 + 5x2 = 7y1 + 3y2 3-December, 2001 Page 7 of 7 Peter A. Brown

Bibliography

Banchoff, Thomas, and John Wermer. Linear Algebra Through Geometry, Springer-

Verlag, New York Inc., New York, 1993.

Lay, David C., Linear Algebra and Its Applications, 2nd ed. updated, Addison Wesley

Longman, Inc. New York, New York, 2000.