Calculation of Eigenvalues and Eigenvectors:

Motivating Example:

Let

.

Find the eigenvalues of A and their associated eigenvectors.

[solution:]

Let be the eigenvector associated with the eigenvalue . Then,

.

Thus,

is the nonzero (nontrivial) solution of the homogeneous linear system . is singular .

Therefore,

.

  1. As ,

.

  1. As ,

.

Note:

In the above example, the eigenvalues of A satisfy the following equation

.

After finding the eigenvalues, we can further solve the associated homogeneous system to find the eigenvectors.

Definition of the characteristic polynomial:

Let . The determinant

,

is called the characteristic polynomial of A.

,

is called the characteristic equation of A.

Theorem:

A is singular if and only if 0 is an eigenvalue of A.

[proof:]

.

A is singular has non-trivial solution There exists a nonzero vector x such that

.

x is the eigenvector of A associated with eigenvalue 0.

0 is an eigenvalue of A There exists a nonzero vector x such that

.

The homogeneous system has nontrivial (nonzero) solution.

A is singular.

Theorem:

The eigenvalues of Aare the real roots of the characteristic polynomial of A.

Let be an eigenvalue of A associated with eigenvector u. Also, let be the characteristic polynomial of A. Then,

The homogeneous system has nontrivial (nonzero) solution x is singular

.

is a real root of .

Let be a real root of is a singular matrix There exists a nonzero vector (nontrivial solution) v such that .

v is the eigenvector of A associated with the eigenvalue .

Procedure of finding the eigenvalues and eigenvectors of A:

  1. Solve for the real roots of the characteristic equation . These real roots are the eigenvalues of A.

2. Solve for the homogeneous system or , . The nontrivial (nonzero) solutions are the eigenvectors associated with the eigenvalues .

Example:

Find the eigenvalues and eigenvectors of the matrix

.

[solution:]

and 10.

  1. As ,

.

Thus,

,

are the eigenvectors associated with eigenvalue .

  1. As ,

.

Thus,

,

are the eigenvectors associated with eigenvalue .

Example:

.

Find the eigenvalues and the eigenvectors of A.

[solution:]

and 6.

  1. As ,

.

Thus,

,

are the eigenvectors associated with eigenvalue .

  1. As ,

.

Thus,

,

are the eigenvectors associated with eigenvalue .

Note:

In the above example, there are at most 2 linearly independent eigenvectors and for matrix A.

The following theorem and corollary about the independence of the eigenvectors:

Theorem:

Let be the eigenvectors of a matrix A associated with distinct eigenvalues , respectively, . Then, are linearly independent.

[proof:]

Assume are linearly dependent. Then, suppose the dimension of the vector space V generated by is

(i.e. the dimension of Vthe vector space generated by ). There exists jlinearly independent vectors of which also generate V. Without loss of generality, let be the j linearly independent vectors which generate V (i.e., is a basis of V). Thus,

,

are some real numbers. Then,

Also,

Thus,

.

Since are linearly independent,

.

Futhermore,

are distinct,

It is contradictory!!

Corollary:

If a matrix A has n distinct eigenvalues, then A has n linearly independent eigenvectors.

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