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| Boilerpipe Text | 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 =
λ
v
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
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
= −5
x
1
+ 2
x
2
y
2
= −9
x
1
+ 6
x
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
= λ
v
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
+ 2
x
2
= λ
x
1
−9
x
1
+ 6
x
2
= λ
x
2
Rearranging gives:
−(5 −
λ
)
x
1
+ 2
x
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.`
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.`
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.`
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.`
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.`
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.`
`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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1. [Home](https://www.intmath.com/)
2. [Matrices and Determinants](https://www.intmath.com/matrices-determinants/matrix-determinant-intro.php)
3. 7\. Eigenvalues and Eigenvectors
On this page
- [Matrices and Determinants](https://www.intmath.com/matrices-determinants/matrix-determinant-intro.php)
- [1\. Determinants](https://www.intmath.com/matrices-determinants/1-determinants.php)
- [Systems of 3x3 Equations interactive applet](https://www.intmath.com/matrices-determinants/systems-equations-interactive.php)
- [2\. Large Determinants](https://www.intmath.com/matrices-determinants/2-large-determinants.php)
- [3\. Matrices](https://www.intmath.com/matrices-determinants/3-matrices.php)
- [4\. Multiplication of Matrices](https://www.intmath.com/matrices-determinants/4-multiplying-matrices.php)
- [4a. Matrix Multiplication examples](https://www.intmath.com/matrices-determinants/matrix-multiplication-examples.php)
- [4b. Add & multiply matrices applet](https://www.intmath.com/matrices-determinants/matrix-addition-multiplication-applet.php)
- [5\. Finding the Inverse of a Matrix](https://www.intmath.com/matrices-determinants/5-inverse-matrix.php)
- [5a. Simple Matrix Calculator](https://www.intmath.com/matrices-determinants/matrix-calculator.php)
- [5b. Inverse of a Matrix using Gauss-Jordan Elimination](https://www.intmath.com/matrices-determinants/inverse-matrix-gauss-jordan-elimination.php)
- [6\. Matrices and Linear Equations](https://www.intmath.com/matrices-determinants/6-matrices-linear-equations.php)
- [Matrices and Linear Transformations](https://www.intmath.com/matrices-determinants/matrices-linear-transformations.php)
- [Eigenvalues and eigenvectors - concept applet](https://www.intmath.com/matrices-determinants/eigenvalues-eigenvectors-concept-applet.php)
- 7\. Eigenvalues and Eigenvectors
- [8\. Applications of Eigenvalues and Eigenvectors](https://www.intmath.com/matrices-determinants/8-applications-eigenvalues-eigenvectors.php)
- [Eigenvalues and eigenvectors calculator](https://www.intmath.com/matrices-determinants/eigenvalues-eigenvectors-calculator.php)
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# 7\. Eigenvalues and Eigenvectors
### On this page
[Definitions](https://www.intmath.com/matrices-determinants/7-eigenvalues-eigenvectors.php#defs)
[Finding eigenvalues and eigenvectors summary)](https://www.intmath.com/matrices-determinants/7-eigenvalues-eigenvectors.php#summary)
[2x2 examples](https://www.intmath.com/matrices-determinants/7-eigenvalues-eigenvectors.php#ex2x2)
[How many eigenvalues and eigenvectors?](https://www.intmath.com/matrices-determinants/7-eigenvalues-eigenvectors.php#many)
[3x3 examples](https://www.intmath.com/matrices-determinants/7-eigenvalues-eigenvectors.php#ex3x3)
## Introduction
On the previous page, [Eigenvalues and eigenvectors - physical meaning and geometric interpretation applet](https://www.intmath.com/matrices-determinants/eigenvalues-eigenvectors-concept-applet.php) 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 =** λ**v**
>
> 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** − λ**I**\| = 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** − λ**I**
4. Choose a convenient value for *x*1, then find *x*2
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.
## Eigenvector Math Problem Solver
Loading...
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 = −5*x*1 + 2*x*2
>
> *y*2 = −9*x*1 + 6*x*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 + 2*x*2 = λ*x*1
>
> −9*x*1 + 6*x*2 = λ*x*2
Rearranging gives:
> −(5 − *λ*)*x*1 + 2*x*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](https://www.intmath.com/matrices-determinants/1-determinants.php), 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.\`
[1 2 3 -1 -2 -3 1 2 -1 -2 -3 x yOpen image in a new page](https://www.intmath.com/matrices-determinants/svg/svgphp-eigenvalues-eigenvectors-7-s0.svg)
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.\`
[5 10 15 20 5 10 15 20 25 30 35 40 x yOpen image in a new page](https://www.intmath.com/matrices-determinants/svg/svgphp-eigenvalues-eigenvectors-7-s1.svg)
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 \`\[(2,3), (2,1)\].\`
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.\`
[2 4 6 8 10 12 2 4 6 8 10 x yOpen image in a new page](https://www.intmath.com/matrices-determinants/svg/svgphp-eigenvalues-eigenvectors-7-s2.svg)
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.\`
[1 -1 1 -1 x yOpen image in a new page](https://www.intmath.com/matrices-determinants/svg/svgphp-eigenvalues-eigenvectors-7-s3.svg)
Graph indicating the transform y2 = Av2 = λ2x2
## 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.\`
[1 2 3 4 5 -1 1 2 3 4 5 6 -1 x yOpen image in a new page](https://www.intmath.com/matrices-determinants/svg/svgphp-eigenvalues-eigenvectors-7-s4.svg)
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.\`
[2 4 6 8 10 -2 -4 -6 x yOpen image in a new page](https://www.intmath.com/matrices-determinants/svg/svgphp-eigenvalues-eigenvectors-7-s5.svg)
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 \`\[(0,1,0),(1,-1,1),(0,1,0)\].\`
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=\[(1),(1),(1)\].\`
[Eigenvalues and eigenvectors - concept applet](https://www.intmath.com/matrices-determinants/eigenvalues-eigenvectors-concept-applet.php)
[8\. Applications of Eigenvalues and Eigenvectors](https://www.intmath.com/matrices-determinants/8-applications-eigenvalues-eigenvectors.php)
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| Readable Markdown | ## Introduction
On the previous page, [Eigenvalues and eigenvectors - physical meaning and geometric interpretation applet](https://www.intmath.com/matrices-determinants/eigenvalues-eigenvectors-concept-applet.php) 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 =** λ**v**
>
> 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** − λ**I**\| = 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** − λ**I**
4. Choose a convenient value for *x*1, then find *x*2
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.
Eigenvector Math Problem Solver
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Let's have a look at some examples.
## Example 1
We start with a system of two equations, as follows:
> *y*1 = −5*x*1 + 2*x*2
>
> *y*2 = −9*x*1 + 6*x*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 + 2*x*2 = λ*x*1
>
> −9*x*1 + 6*x*2 = λ*x*2
Rearranging gives:
> −(5 − *λ*)*x*1 + 2*x*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](https://www.intmath.com/matrices-determinants/1-determinants.php), 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.\`
#### 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.\`
## 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.\`
### 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.\`
## 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.\`
### 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.\`
## \`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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