# 7. Eigenvalues and Eigenvectors

Definitions

How many eigenvalues and eigenvectors?

## Introduction

On the previous page, Eigenvalues and eigenvectors - physical meaning and geometric interpretation applet we saw the example of an elastic membrane being stretched, and how this was represented by a matrix multiplication, and in special cases equivalently by a scalar multiplication.

That example demonstrates a very important concept in engineering and science - eigenvalues and eigenvectors - which is used widely in many applications, including calculus, search engines, population studies, aeronautics and so on.

## Definition of eigenvalues and eigenvectors of a matrix

Let A be any square matrix. A non-zero vector v is an eigenvector of A if

Av = λv

for some number λ, called the corresponding eigenvalue.

NOTE: The German word "eigen" roughly translates as "own" or "belonging to". Eigenvalues and eigenvectors correspond to each other (are paired) for any particular matrix A.

The solved examples below give some insight into what these concepts mean. First, a summary of what we're going to do:

## How to find the eigenvalues and eigenvectors of a 2x2 matrix

1. Set up the characteristic equation, using |A − λI| = 0
2. Solve the characteristic equation, giving us the eigenvalues (2 eigenvalues for a 2x2 system)
3. Substitute the eigenvalues into the two equations given by A − λI
4. Choose a convenient value for x1, then find x2
5. The resulting values form the corresponding eigenvectors of A (2 eigenvectors for a 2x2 system)

There is no single eigenvector formula as such - it's more of a sset of steps that we need to go through to find the eigenvalues and eigenvectors.

Let's have a look at some examples.

## Example 1

y1 = −5x1 + 2x2

y2 = −9x1 + 6x2

We can write those equations in matrix form as:

[(y_1),(y_2)]=[(-5,2), (-9,6)][(x_1),(x_2)]

In general we can write the above matrices as:

y = Av

where

bb(y) = [(y_1),(y_2)],

bb(A) = [(-5,2), (-9,6)], and

bb(v) = [(x_1),(x_2)]

### Step 1. Set up the characteristic equation, using |A − λI| = 0

Our task is to find the eigenvalues λ, and eigenvectors v, such that:

y = λv

We are looking for scalar values λ (numbers, not matrices) that can replace the matrix A in the expression y = Av.

That is, we want to find λ such that :

−5x1 + 2x2 = λx1

−9x1 + 6x2 = λx2

Rearranging gives:

−(5 − λ)x1 + 2x2 = 0

−9x1 + (6 − λ)x2 = 0 (1)

This can be written using matrix notation with the identity matrix I as:

(bb(A) - lambdabb(I))bb(v) = 0, that is:

(bb(A) - lambda[(1,0),(0,1)])bb(v) = 0

(bb(A) - [(lambda,0),(0,lambda)])bb(v) = 0

Clearly, we have a trivial solution bb(v)=[(0),(0)], but in order to find any non-trivial solutions, we apply a result following from Cramer's Rule, that this equation will have a non-trivial (that is, non-zero) solution v if its coefficient determinant has value 0.

The resulting equation, using determinants, |bb(A) - lambdabb(I)| = 0 is called the characteristic equation.

### Step 2. Solve the characteristic equation, giving us the eigenvalues (2 eigenvalues for a 2x2 system)

In this example, the coefficient determinant from equations (1) is:

|bb(A) - lambdabb(I)| = | (-5-lambda, 2), (-9, 6-lambda) |

= (-5-lambda)(6-lambda) - (-9)(2)

= -30 - lambda + lambda^2 + 18

= lambda^2 - lambda - 12

= (lambda + 3)(lambda - 4)

Now this equals 0 when:

(lambda + 3)(lambda - 4) = 0

That is, when:

lambda = -3 or 4.

These two values are the eigenvalues for this particular matrix A.

### Step 3. Substitute the eigenvalues into the two equations given by A − λI

#### Case 1: lambda_1 = -3

When lambda = lambda_1 = -3, equations (1) become:

[-5-(-3)]x_1 + 2x_2 = 0

-9x_1 + [6-(-3)]x_2 = 0

That is:

-2x_1 + 2x_2 = 0

-9x_1 + 9x_2 = 0 (2)

Dividing the first line of Equations (2) by -2 and the second line by -9 (not really necessary, but helps us see what is happening) gives us the identical equations:

x_1 - x_2 = 0

x_1 - x_2 = 0

### Step 4. Choose a convenient value for x1, then find x2

There are infinite solutions of course, where x_1 = x_2. We choose a convenient value for x_1 of, say 1, giving x_2=1.

### Step 5. The resulting values form the corresponding eigenvectors of A (2 eigenvectors for a 2x2 system)

So the corresponding eigenvector is:

bb(v_1)=[(1),(1)]

NOTE: We could have easily chosen x_1=3, x_2=3, or for that matter, x_1=-100, x_2=-100. These values will still "work" in the matrix equation.

In general, we could have written our answer as "x_1=t, x_2=t, for any value t", however it's usually more meaningful to choose a convenient starting value (usually for x_1), and then derive the resulting remaining value(s).

### Is it correct?

We can check by substituting:

bb(Av)_1 = [(-5,2), (-9,6)][(1),(1)]

= [(-3),(-3)]

= -3[(1),(1)]

= lambda_1bb(v)_1

We have found an eigenvalue lambda_1=-3 and an eigenvector bb(v)_1=[(1),(1)] for the matrix bb(A) =[(-5,2), (-9,6)] such that bb(Av)_1 = lambda_1bb(v)_1.

Graphically, we can see that matrix bb(A) = [(-5,2), (-9,6)] acting on vector bb(v_1)=[(1),(1)] is equivalent to multiplying bb(v_1)=[(1),(1)] by the scalar lambda_1 = -3. The result is applying a scale of -3.

Graph indicating the transform y1 = Av1

#### Case 2: lambda_2 = 4

When lambda = lambda_2 = 4, equations (1) become:

(-5-(4))x_1 + 2x_2 = 0

-9x_1 + (6-(4))x_2 = 0

That is:

-9x_1 + 2x_2 = 0

-9x_1 + 2x_2 = 0

We choose a convenient value for x_1 of 2, giving x_2=9. So the corresponding eigenvector is:

bb(v)_2=[(2),(9)]

We could check this by multiplying and concluding [(-5,2), (-9,6)][(2),(9)] = 4[(2),(9)], that is bb(Av)_2 = lambda_2bb(v)_2.

We have found an eigenvalue lambda_2=4 and an eigenvector bb(v)_2=[(2),(9)] for the matrix bb(A) =[(-5,2), (-9,6)] such that bb(Av)_2 = lambda_2bb(v)_2.

Graphically, we can see that matrix bb(A) = [(-5,2), (-9,6)] acting on vector bb(v_2)=[(2),(9)] is equivalent to multiplying bb(v_2)=[(2),(9)] by the scalar lambda_2 = 4. The result is applying a scale of 4.

Graph indicating the transform y2 = Av2 = λ2x2

## How many eigenvalues and eigenvectors?

In the above example, we were dealing with a 2xx2 system, and we found 2 eigenvalues and 2 corresponding eigenvectors.

If we had a 3xx3 system, we would have found 3 eigenvalues and 3 corresponding eigenvectors.

In general, a nxxn system will produce n eigenvalues and n corresponding eigenvectors.

## Example 2

Find the eigenvalues and corresponding eigenvectors for the matrix [(2,3), (2,1)].

The matrix bb(A) = [(2,3), (2,1)] corresponds to the linear equations:

y_1 = 2x_1 + 3x_2

y_2 = 2x_1 + x_2

We want to find λ such that :

2x_1 + 3x_2 = lambda x_1

2x_1 + x_2 = lambda x_2

Rearranging gives:

(2-lambda)x_1 + 3x_2 = 0

2x_1 + (1-lambda)x_2 = 0 (3)

The characterstic equation |bb(A) - lambdabb(I)| = 0 for this example is given by:

|bb(A) - lambdabb(I)| = | (2-lambda, 3), (2, 1-lambda) |

= 2 - 3lambda + lambda^2 -6

= lambda^2 - 3lambda - 4

=0

This has value 0 when (lambda - 4)(lambda +1) = 0.

### Case 1: lambda = 4

With lambda_1 = 4, equations (3) become:

(2-4)x_1 + 3x_2 = 0

2x_1 + (1-4)x_2 = 0

That is:

-2x_1 + 3x_2 = 0

2x_1 -3x_2 = 0

We choose a convenient value for x_1 of 3, giving x_2=2. So the corresponding eigenvector is:

bb(v_1)=[(3),(2)]

Multiplying to check our answer, we would find:

[(2,3), (2,1)][(3),(2)] = 4[(3),(2)], that is bb(Av)_1 = lambda_1bb(v)_1.

Graphically, we can see that matrix bb(A) = [(2,3), (2,1)] acting on vector bb(v_1)=[(3),(2)] is equivalent to multiplying bb(v_1)=[(3),(2)] by the scalar lambda_1 = 4. The result is applying a scale of 4.

Graph indicating the transform y1 = Av1 = λ1x1

### Case 2: lambda = -1

With lambda_2 = -1, equations (3) become:

(2+1)x_1 + 3x_2 = 0

2x_1 + (1+1)x_2 = 0

That is:

3x_1 + 3x_2 = 0

2x_1 + 2x_2 = 0

We choose a convenient value x_1 = 1, giving x_2=-1. So the corresponding eigenvector is:

bb(v_2)=[(1),(-1)]

Multiplying to check our answer, we would find:

[(2,3), (2,1)][(1),(-1)] = -1[(1),(-1)], that is bb(Av)_2 = lambda_2bb(v)_2.

Graphically, we can see that matrix bb(A) = [(2,3), (2,1)] acting on vector bb(v_2)=[(1),(-1)] is equivalent to multiplying bb(v_2)=[(1),(-1)] by the scalar lambda_2 = -1. We are scaling vector bb(v_2) by -1.

Graph indicating the transform y2 = Av2 = λ2x2

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## Example 3

Find the eigenvalues and corresponding eigenvectors for the matrix [(3,2), (1,4)].

The matrix bb(A) = [(3,2), (1,4)] corresponds to the linear equations:

y_1 = 3x_1 + 2x_2

y_2 = x_1 + 4x_2

We want to find λ such that :

3x_1 + 2x_2 = lambda x_1

x_1 + 4x_2 = lambda x_2

Rearranging gives:

(3-lambda)x_1 + 2x_2 = 0

x_1 + (4-lambda)x_2 = 0 (4)

The characterstic equation |bb(A) - lambdabb(I)| = 0 for this example is given by:

|bb(A) - lambdabb(I)| = | (3-lambda, 2), (1, 4-lambda) |

= 12 - 7lambda + lambda^2 - 2

= lambda^2 - 7lambda +10

=0

This has value 0 when (lambda - 5)(lambda - 2) = 0.

### Case 1: lambda = 5

With lambda_1 = 5, equations (4) become:

(3-5)x_1 + 2x_2 = 0

x_1 + (4-5)x_2 = 0

That is:

-2x_1 + 2x_2 = 0

x_1 - x_2 = 0

We choose a convenient value x_1 = 1, giving x_2=1. So the corresponding eigenvector is:

bb(v_1)=[(1),(1)]

Multiplying to check our answer, we would find:

[(3,2), (1,4)][(1),(1)] = 5[(1),(1)], that is bb(Av)_1 = lambda_1bb(v)_1.

Graphically, we can see that matrix bb(A) = [(3,2), (1,4)] acting on vector bb(v_1)=[(1),(1)] is equivalent to multiplying bb(v_1)=[(1),(1)] by the scalar lambda_1 = 5. The result is applying a scale of 5.

Graph indicating the transform y1 = Av1 = λ1x1

### Case 2: lambda = 2

With lambda_2 = 2, equations (4) become:

(3-2)x_1 + 2x_2 = 0

x_1 + (4-2)x_2 = 0

That is:

x_1 + 2x_2 = 0

x_1 + 2x_2 = 0

We choose a convenient value x_1 = 2, giving x_2=-1. So the corresponding eigenvector is:

bb(v_2)=[(2),(-1)]

Multiplying to check our answer, we would find:

[(3,2), (1,4)][(2),(-1)] = 2[(2),(-1)], that is bb(Av)_2 = lambda_2bb(v)_2.

Graphically, we can see that matrix bb(A) = [(3,2), (1,4)] acting on vector bb(v_2)=[(2),(-1)] is equivalent to multiplying bb(v_2) by the scalar lambda_2 = 5. We are scaling vector bb(v_2) by 5.

Graph indicating the transform y2 = Av2 = λ2x2

## 3xx3 matrices and their eigenvalues and eigenvectors

The process for finding the eigenvalues and eigenvectors of a 3xx3 matrix is similar to that for the 2xx2 case.

## Example 4: 3xx3 case

Find the eigenvalues and eigenvectors for the matrix [(0,1,0),(1,-1,1),(0,1,0)].

We want to find lambda such that:

x_2 = lambda x_1

x_1 - x_2 + x_3 = lambda x_2

x_2 = lambda x_3

That is:

-lambda x_1 + x_2 = 0

x_1 - (x_2+lambda x_2) + x_3 = 0 (5)

x_2 - lambda x_3 = 0

The characterstic equation |bb(A) - lambdabb(I)| = 0 for this example is given by:

|bb(A) - lambdabb(I)| = | (0-lambda, 1,0), (1, -1-lambda, 1),(0,1,-lambda) |

= -lambda^3-lambda^2+2lambda

= -lambda(lambda^2 + lambda -2)

= -lambda(lambda + 2)(lambda -1)

=0

This occurs when lambda_1 = 0, lambda_2=-2, or lambda_3= 1.

### Case 1: lambda=0

Equations (5) become:

x_2 = 0

x_1 - x_2 + x_3 = 0

x_2 = 0

Clearly, x_2 = 0 and we'll choose x_1 = 1, giving x_3 = -1.

So for the eigenvalue lambda_1=0, the corresponding eigenvector is bb(v)_1=[(1),(0),(-1)].

### Case 2: lambda=-2

Equations (5) become:

2 x_1 + x_2 = 0

x_1 + x_2 + x_3 = 0

x_2 + 2 x_3 = 0

Choosing x_1 = 1 gives x_2 = -2 and then x_3 = 1.

So for the eigenvalue lambda_2=-2, the corresponding eigenvector is bb(v)_2=[(1),(-2),(1)].

### Case 3: lambda=1

Equations (5) become:

- x_1 + x_2 = 0

x_1 - 2x_2 + x_3 = 0

x_2 - x_3 = 0

Choosing x_1 = 1 gives x_2 = 1 and then x_3 = 1.

So for the eigenvalue lambda_3=1, the corresponding eigenvector is bb(v)_3=[(1),(1),(1)].

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