# 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.

## Eigenvector Math Problem Solver

This tool combines the power of mathematical computation engine that excels at solving mathematical formulas with the power of GPT large language models to parse and generate natural language. This creates math problem solver thats **more accurate than ChatGPT, more flexible than a calculator, and faster answers than a human tutor.**

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}

## 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)].`