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