# 7. 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

Abe any square matrix. A non-zero vectorvis aneigenvectorofAifAv =λv

for some number λ, called the correspondingeigenvalue.

**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

- Set up the
**characteristic equation**, using |**A**− λ**I**| = 0 **Solve**the characteristic equation, giving us the**eigenvalues**(2 eigenvalues for a 2x2 system)**Substitute**the eigenvalues into the two equations given by**A**− λ**I**- Choose a convenient value for
*x*_{1}, then find*x*_{2} - 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

We start with a system of two equations, as follows:

y_{1}= −5x_{1}+ 2x_{2}

y_{2}= −9x_{1}+ 6x_{2}

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 :

−5

x_{1}+ 2x_{2}= λx_{1}−9

x_{1}+ 6x_{2}= λx_{2}

Rearranging gives:

−(5 −

λ)x_{1}+ 2x_{2}= 0−9

x_{1}+ (6 −λ)x_{2}= 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 *x*_{1}, then find *x*_{2}

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 y_{1} = Av_{1}

#### 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 y_{2} = Av_{2} = λ_{2}x_{2}

## 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)].`

Answer

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 y_{1} = Av_{1} = λ_{1}x_{1}

### 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 y_{2} = Av_{2} = λ_{2}x_{2}

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## Example 3

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

Answer

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 y_{1} = Av_{1} = λ_{1}x_{1}

### 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 y_{2} = Av_{2} = λ_{2}x_{2}

## `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)].`

Answer

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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