MATH 323 Lecture 25

From Notes
Jump to navigation Jump to search

« previous | Thursday, November 29, 2012 | next »

Eigenvalues and Eigenvectors


Therefore, the eigenvalues are ,

To find eigenvectors, find basis for :

Therefore, is an eigenvector belonging to , and and are linearly independent eigenvectors belonging to .

Complex Eigenvalues

For an arbitrary matrix, the characteristic polynomial will be .

Polynomials of degree always have exactly complex roots:

is the constant term of .

The coefficient of the remaining is given by . This is called the trace of .


Therefore, is an eigenvector belonging to . If we multiply this eigenvector by , we get , which is a much nicer form.

Theorem 6.1.1

Let and be matrices.

If is similar to , then and have the same characteristic polynomials and the same eigenvalues (not eigenvectors).


Recall that

Review: Complex Numbers

Every complex number has a conjugate .

A polynomial that has a root will also have the conjugate as a root

corollary: if is an eigenvalue of a real matrix , then is also an eigenvalue.
if is an eigenvector of complex eigenvalue , then is an eigenvector of the conjugate eigenvalue .

multiplying a complex by its conjugate will always yield a real number

Conjugate of a Product of complex numbers is just the product of conjugates

Solving Systems of Linear Differential Equations

A linear differential equation is of the form

Where are functions in .

Which can be written in the form .

Suppose is of the form .

Then and .

If is an eigenvector of , then

The set of solutions to is an -dimensional subspace, and if solutions are linearly independent solutions, then any linear combination of 's are also solutions.


For initial value problem , and .