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Eigenvectors and Eigenvalues

Module by: Michael Haag, Justin Romberg. E-mail the authors

Summary: This module defines eigenvalues and eigenvectors and explains a method of finding them given a matrix. These ideas are presented, along with many examples, in hopes of leading up to an understanding of the Fourier Series.

In this section, our linear systems will be n×n matrices of complex numbers. For a little background into some of the concepts that this module is based on, refer to the basics of linear algebra.

Eigenvectors and Eigenvalues

Let A A be an n×n matrix, where A A is a linear operator on vectors in Cn n .

Ax=b A x b
(1)
where x x and b b are n×1 vectors (Figure 1).

Figure 1: Illustration of linear system and vectors.
(a)
Figure 1(a) (eigv_f1.png)
(b)
Figure 1(b) (eigv_f2.png)
Definition 1: eigenvector
An eigenvector of A A is a vector vCn v n such that
Av=λv A v λ v
(2)
where λ λ is called the corresponding eigenvalue. A A only changes the length of v v, not its direction.

Graphical Model

Through Figure 2 and Figure 3, let us look at the difference between Equation 1 and Equation 2.

Figure 2: Represents Equation 1, Ax=b A x b .
Figure 2 (eigv_f3.png)

If v v is an eigenvector of A A, then only its length changes. See Figure 3 and notice how our vector's length is simply scaled by our variable, λλ, called the eigenvalue:

Figure 3: Represents Equation 2, Av=λv A v λ v .
Figure 3 (eigv_f4.png)

Note:

When dealing with a matrix A A, eigenvectors are the simplest possible vectors to operate on.

Examples

Exercise 1

From inspection and understanding of eigenvectors, find the two eigenvectors, v 1 v 1 and v 2 v 2 , of A=( 30 0-1 ) A 3 0 0 -1 Also, what are the corresponding eigenvalues, λ 1 λ 1 and λ 2 λ 2 ? Do not worry if you are having problems seeing these values from the information given so far, we will look at more rigorous ways to find these values soon.

Solution

The eigenvectors you found should be: v 1 =10 v 1 1 0 v 2 =01 v 2 0 1 And the corresponding eigenvalues are λ 1 =3 λ 1 3 λ 2 =-1 λ 2 -1

Exercise 2

Show that these two vectors, v 1 =11 v 1 1 1 v 2 =1-1 v 2 1 -1 are eigenvectors of A A, where A=( 3-1 -13 ) A 3 -1 -1 3 . Also, find the corresponding eigenvalues.

Solution

In order to prove that these two vectors are eigenvectors, we will show that these statements meet the requirements stated in the definition. A v 1 =( 3-1 -13 )11=22 A v 1 3 -1 -1 3 1 1 2 2 A v 2 =( 3-1 -13 )1-1=4-4 A v 2 3 -1 -1 3 1 -1 4 -4 These results show us that A A only scales the two vectors (i.e. changes their length) and thus it proves that Equation 2 holds true for the following two eigenvalues that you were asked to find: λ 1 =2 λ 1 2 λ 2 =4 λ 2 4 If you need more convincing, then one could also easily graph the vectors and their corresponding product with AA to see that the results are merely scaled versions of our original vectors, v 1 v 1 and v 2 v 2 .

Figure 4: video from Khan Academy - Introduction to Eigenvectors and Eigenvalues - 7:43 min.
Khan Lecture on Eigenvectors

Calculating Eigenvalues and Eigenvectors

In the above examples, we relied on your understanding of the definition and on some basic observations to find and prove the values of the eigenvectors and eigenvalues. However, as you can probably tell, finding these values will not always be that easy. Below, we walk through a rigorous and mathematical approach at calculating the eigenvalues and eigenvectors of a matrix.

Finding Eigenvalues

Find λC λ such that v0 v 0 , where 0 0 is the "zero vector." We will start with Equation 2, and then work our way down until we find a way to explicitly calculate λ λ. Av=λv A v λ v Avλv=0 A v λ v 0 (AλI)v=0 A λ I v 0 In the previous step, we used the fact that λv=λIv λ v λ I v where II is the identity matrix. I=( 100 010 00 01 ) I 1 0 0 0 1 0 0 0 0 1 So, AλI A λ I is just a new matrix.

Example 1

Given the following matrix, AA, then we can find our new matrix, AλI A λ I . A=( a1,1a1,2 a2,1a2,2 ) A a 1 1 a 1 2 a 2 1 a 2 2 AλI=( a1,1λa1,2 a2,1a2,2λ ) A λ I a 1 1 λ a 1 2 a 2 1 a 2 2 λ

If (AλI)v=0 A λ I v 0 for some v0 v 0 , then AλI A λ I is not invertible. This means: det(AλI)=0 A λ I 0 This determinant (shown directly above) turns out to be a polynomial expression (of order nn). Look at the examples below to see what this means.

Example 2

Starting with matrix AA (shown below), we will find the polynomial expression, where our eigenvalues will be the dependent variable. A=( 3-1 -13 ) A 3 -1 -1 3 AλI=( 3λ-1 -13λ ) A λ I 3 λ -1 -1 3 λ det(AλI)=3λ2-12=λ26λ+8 A λ I 3 λ 2 -1 2 λ 2 6 λ 8 λ=24 λ 2 4

Example 3

Starting with matrix AA (shown below), we will find the polynomial expression, where our eigenvalues will be the dependent variable. A=( a1,1a1,2 a2,1a2,2 ) A a 1 1 a 1 2 a 2 1 a 2 2 AλI=( a1,1λa1,2 a2,1a2,2λ ) A λ I a 1 1 λ a 1 2 a 2 1 a 2 2 λ det(AλI)=λ2(a1,1+a2,2)λa2,1a1,2+a1,1a2,2 A λ I λ 2 a 1 1 a 2 2 λ a 2 1 a 1 2 a 1 1 a 2 2

If you have not already noticed it, calculating the eigenvalues is equivalent to calculating the roots of det(AλI)= c n λn+ c n 1 λn1++ c 1 λ+ c 0 =0 A λ I c n λ n c n 1 λ n 1 c 1 λ c 0 0

Conclusion:

Therefore, by simply using calculus to solve for the roots of our polynomial we can easily find the eigenvalues of our matrix.

Finding Eigenvectors

Given an eigenvalue, λ i λ i , the associated eigenvectors are given by Av= λ i v A v λ i v A v 1 v n = λ 1 v 1 λ n v n A v 1 v n λ 1 v 1 λ n v n set of nn equations with nn unknowns. Simply solve the nn equations to find the eigenvectors.

Figure 5: video from Khan Academy - Example Deriving Eignevectors and Eigenvalues - 5:39 min.
Khan Lecture on Deriving Eigenvectors and Eigenvalues

Main Point

Say the eigenvectors of AA, v 1 v 2 v n v 1 v 2 v n , span Cn n , meaning v 1 v 2 v n v 1 v 2 v n are linearly independent and we can write any xCn x n as

x= α 1 v 1 + α 2 v 2 ++ α n v n x α 1 v 1 α 2 v 2 α n v n
(3)
where α 1 α 2 α n C α 1 α 2 α n . All that we are doing is rewriting xx in terms of eigenvectors of AA. Then, Ax=A( α 1 v 1 + α 2 v 2 ++ α n v n ) A x A α 1 v 1 α 2 v 2 α n v n Ax= α 1 A v 1 + α 2 A v 2 ++ α n A v n A x α 1 A v 1 α 2 A v 2 α n A v n Ax= α 1 λ 1 v 1 + α 2 λ 2 v 2 ++ α n λ n v n =b A x α 1 λ 1 v 1 α 2 λ 2 v 2 α n λ n v n b Therefore we can write, x=i α i v i x i α i v i and this leads us to the following depicted system:

Figure 6: Depiction of system where we break our vector, xx, into a sum of its eigenvectors.
Figure 6 (eigv_sys.png)

where in Figure 6 we have, b=i α i λ i v i b i α i λ i v i

Main Point:

By breaking up a vector, xx, into a combination of eigenvectors, the calculation of Ax A x is broken into "easy to swallow" pieces.

Practice Problem

Exercise 3

For the following matrix, AA and vector, xx, solve for their product. Try solving it using two different methods: directly and using eigenvectors. A=( 3-1 -13 ) A 3 -1 -1 3 x=53 x 5 3

Solution

Direct Method (use basic matrix multiplication) Ax=( 3-1 -13 )53=124 A x 3 -1 -1 3 5 3 12 4 Eigenvectors (use the eigenvectors and eigenvalues we found earlier for this same matrix) v 1 =11 v 1 1 1 v 2 =1-1 v 2 1 -1 λ 1 =2 λ 1 2 λ 2 =4 λ 2 4 As shown in Equation 3, we want to represent xx as a sum of its scaled eigenvectors. For this case, we have: x=4 v 1 + v 2 x 4 v 1 v 2 x=53=411+1-1 x 5 3 4 1 1 1 -1 Ax=A(4 v 1 + v 2 )= λ i (4 v 1 + v 2 ) A x A 4 v 1 v 2 λ i 4 v 1 v 2 Therefore, we have Ax=4×211+41-1=124 A x 4 2 1 1 4 1 -1 12 4 Notice that this method using eigenvectors required no matrix multiplication. This may have seemed more complicated here, but just imagine AA being really big, or even just a few dimensions larger!

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