Eigenvalues and eigenvectors help solve varied problems. What are they, how do we find and use them?

**The vectors that are scaled by matrix multiplication are called eigenvectors of a matrix. The scale factor is called the eigenvalue.**

We can multiply a square matrix with a vector to get a new vector. We say the matrix has transformed the old vector into a new vector. Usually, the new vector has a different direction, except for a few special ones. These are the eigenvectors. They get scaled by a factor called the eigenvalue. If the scale factor is negative, it reverses the direction of the eigenvector.

## Finding Transformation Matrix

The matrix $A=(13 20 )$ rotates, flips and stretches the entire space ( ℝ ^{2} ). Only eigenvectors preserve their direction. All other vectors point to a new direction.

Let us do the transformation ourselves. We start with the unit vectors.

Multiplying A with the unit vector i = (1, 0)^{T}:

$Ai=(13 20 )(10 )=(13 )$.

You can verify the transformation of the unit vector j as:

(0, 1)^{T} → $(20 )$

Observe that **transformed unit vectors and columns of the matrix are identical**.

### Eigenvector Transformation

Let us transform vectors v_{1} = ( − 1, 1.5)^{T} and v_{2} = (1, 1)^{T}.

$Av_{1}=A(−11.5 )=(2−3 )$

= − 2( − 1, 1.5)^{T} = − 2v_{1}

It is scaled by a factor of −2. Due to the negative sign of the scale factor, its direction gets reversed.

v_{1} is an eigenvector. The eigenvalue corresponding to this vector is −2.

$Av_{2}=A(11 )$

$=(33 )=3(11 )=3v_{2}$

v_{2} is also an eigenvector. It is scaled by a factor of 3. Its eigenvalue is 3.

## Finding Eigenvalue

The eigenvalue is the amount by which a square matrix scales its eigenvector.

If x is an eigenvector of a matrix A, and λ its eigenvalue, we can write:

Ax = λx where A is an n × n matrix.

We want to solve this equation for λ and x( ≠ 0). Rewriting the equation:

Ax − λx = 0

(A − λI)x = 0

This multiplication will result in n linear equations. Since there is no constant term, this is a homogeneous system of equations (you will see this in the examples below). It has a solution if and only if:

**Determinant of (A − λI) = 0**

### Steps for calculating Eigenvalues

- Step 1. Form the matrix B = (A − λI)
- Step 2. Create an equation using: Determinant of B = 0. It is a polynomial equation in λ, p(λ) = 0
- Step 3. Solve p(λ) = 0. These are the eigenvalues of the matrix A

On calculating the determinant you get a polynomial p(λ) of degree n. So at most n possible roots exist or a maximum of n different eigenvalues exist for an n × n matrix.

Let us see an example.

#### Example

Find eigenvalues of the 2 x 2 matrix $A=(13 20 )$.

We use the following equation to calculate eigenvalues:

Determinant of (A − λI) = 0

**Step 1. Form Matrix**

$A−λI=(1−λ3 20−λ )$

**Step 2. Create Equation**

det(A − λI) = 0 gives the equation:

(1 − λ)( − λ) − 6 = 0

The 2 × 2 matrix, resulted in a second-degree polynomial in our equation.

λ^{2} − λ − 6 = 0

**Step 3. Solve**

(λ − 3)(λ + 2) = 0

λ = 3 or λ = − 2

Eigenvalues are 3 and -2

## Finding Eigenvector

**Eigenvector or characteristic vector of a matrix is a unique vector that does not change its direction (except reversal) under matrix multiplication.**

- Matrix multiplication transforms any vector.
- w = Av is the transformed vector v. Generally v and w point in different directions.
- v and w are in the same direction if v is an eigenvector.

### Steps for Calculating Eigenvectors

We can easily solve the original equation Ax = λx for eigenvectors using the eigenvalue.

- Step 1. Find eigenvalues λ of A
- Step 2. For
**each**λ, form homogeneous system of linear equations (A − Iλ)x = 0. - Step 3. Solve the above equations to get eigenvectors for λ

#### Example

Find eigenvectors of $A=(13 20 )$.

**Step 1. Find Eigenvalues**

We have already calculated the eigenvalues as -2 and 3. First we find the eigenvector for λ = − 2.

**Step 2. Form Equations for λ = − 2**

(A − Iλ) =

$(1−(−2)3 20−(−2) )=(33 22 )$

**Step 3. Solve Equations**

We now solve (A − Iλ)x = 0

$(33 22 )(x_{1}x_{2} )=0$

We get the equation:

3x_{1} + 2x_{2} = 0

$x_{1}=3−2 x_{2}$

We can choose any value for x_{2} to calculate the eigenvector.

For x_{2} = 3 we get x_{1} = − 2

Eigenvector is ( − 2, 3)^{T}

Any value of x_{2} results in an eigenvector. They are all scaled by a factor. **Any scalar multiple of an eigenvector is also an eigenvector.**

Eigenvector for λ = 3:

Again we calculate (A − Iλ) =

$(1−33 20−3 )=(−23 2−3 )$

We now solve (A − Iλ)x = 0

$(−23 2−3 )(x_{1}x_{2} )=0$

We get the equation:

−2x_{1} + 2x_{2} = 0

x_{1} = x_{2}

You may verify that the other equation is the same.

Again, we can choose any value for x_{2} to calculate an eigenvector.

For x_{2} = 1 we get x_{1} = 1

Eigenvector is (1, 1)^{T}

### Example

Find eigenvalues and eigenvectors corresponding to counterclockwise rotation through the angle π/2 about the origin in ℝ ^{2}.

#### Finding Rotation Matrix

Where do the unit vectors get mapped? Counterclockwise rotation will transform them as:

- $i=(10 )→(01 )$. This is the first column of our matrix.
- $j=(01 )→(−10 )$. This is the second column of our matrix.

Our rotation matrix is$A=(01 −10 )$

#### Calculating Eigenvalues

$A−λI=(−λ1 −1−λ )$

det(A − λI) = 0 gives the equation:

λ^{2} + 1 = 0

$λ=±−1 $

λ = i or λ = − i

The matrix was real, but eigenvalues are complex!

#### Calculating Eigenvectors

Solving for eigenvectors of λ = i:

$(0−i1 −10−i )(x_{1}x_{2} )=0$

We get the following result:

−ix_{1} − x_{2} = 0

−ix_{1} = x_{2}

x_{1} = ix_{2}

For x_{2} = 1 we get x_{1} = i

Eigenvector is (i, 1)^{T}

Solving for eigenvectors of λ = − i:

$(0+i1 −10+i )(x_{1}x_{2} )=0$

We get the following result:

ix_{1} − x_{2} = 0

$x_{1}=ix_{2} $

x_{1} = − ix_{2}

For x_{2} = 1 we get x_{1} = − i

Eigenvector is ( − i, 1)^{T}

There are no real eigenvectors. This transformation rotates all the points in ℝ ^{2} (except origin). No vector preserves its direction. The absence of a real eigenvector is expected.

## Properties

Eigenvectors (and eigenvalues) have interesting properties.

Ax = λx, where λ is an eigenvalue and x is an eigenvector of the matrix A.

- Any scalar multiple of an eigenvector is also an eigenvector.
- Real matrices can have complex eigenvalues and eigenvectors.
- Complex eigenvectors (and eigenvalues) appear in conjugate pairs.
- Matrix A and A + I have the same eigenvectors. Eigenvalues of (A + I) = λ + 1.
- Matrix A and A
^{−1}have the same eigenvectors. Eigenvalues of A^{−1}= $λ1 $. - Matrix A and A
^{2}have the same eigenvectors. Eigenvalues of A^{2}= λ^{2}. - Eigenvalue of the identity matrix is 1

### Example

What is the relationship between the eigenvalues of a matrix and its inverse? What about eigenvectors?

Suppose eigenvalue of the matrix A^{−1} = β. Let the corresponding eigenvector be w.

We can write:

A^{−1}w = βw

AA^{−1}w = βAw

Iw = w = βAw

$β1 w=Aw$

$Aw=β1 w$

So $β1 $ is the eigenvalue of matrix A and w is its eigenvector.

A matrix and its inverse have the same eigenvectors. Their eigenvalues are the multiplicative inverse of each other.

## Questions

Apart from the questions below, you can attempt Problem Set 6.1 on this page at the MIT site.

### 1 Question

Find the eigenvalues and the eigenvectors of:

$A=(39 71 )$

#### Answer

Can you find the roots of the equation det(A − λI) = 0? These are the eigenvalues.

### 2 Question

The matrix $M=(−12 30 )$ has eigenvalues -3 and 2.

For -3 the eigenvector is ( − 3, 2)^{T} and for 2 it is (1, 1)^{T}.

Find the eigenvalues and eigenvectors of M^{2}. What is the relation between eigenvalues and eigenvectors of M and M^{2}?

#### Answer

Can you find the matrix A = M^{2}?

### 3 Question

The following matrix represents a rotation of a drone in three-dimensional space. Find its axis of rotation.

$A=⎝⎛ 23 0−0.5 010 0.5023 ⎠⎞ $

#### Answer

Will the points on the axis of a rotation change direction?

### 4 Question

Given the matrices $A=(01 21 )$ and $A_{−1}=(−0.50.5 10 )$, find their eigenvalues and the eigenvectors. How are the trace and the eigenvalues of a matrix related?

What is the relationship between the eigenvalues of A and A^{−1}? What about the eigenvectors?

#### Answer

Can you find eigenvalues using the equation det(A − λI) = 0?

### 5 Question

A matrix represents a pure rotation in three-dimensional space. Can you guess one of its eigenvalues?

#### Answer

What can we conclude about eigenvectors, given that the axis of a rotation is invariant?

### Answers

#### 1 Answer

To calculate eigenvalues we use the polynomial equation Determinant(A − λI) = 0.

So let us form the matrix A − λI.

$A−λI=(3−λ9 71−λ )$

Determinant(A − λI) = 0 gives the following equation:

(3 − λ)(1 − λ) − (7)(9) = 0

λ^{2} − (3 + 1)λ − 63 + (3)(1) = 0

λ^{2} − 4λ − 60 = 0

Factoring it we get:

(λ + 6)(λ − 10) = 0

λ = − 6 or λ = 10

Eigenvalues are -6 and 10

To calculate eigenvectors for λ = − 6, we form equations using (A − λI)x = 0.

(A − λI)x = (A + 6I)x = 0

This give the following equation:

$(99 77 )(x_{1}x_{2} )=0$

∴ 9x_{1} + 7x_{2} = 0

x_{1} = ( − 7x_{2})/9

For x_{2} = 9 we get x_{1} = − 7

Eigenvector is ( − 7, 9)^{T}

Let us calculate eigenvectors for λ = 10.

We form equations using (A − 10I)x = 0, which gives:

$(−79 7−9 )(x_{1}x_{2} )=0$

We get the following equation:

−7x_{1} + 7x_{2} = 0

x_{1} = x_{2}

For x_{2} = 1 we get x_{1} = 1

Eigenvector is (1, 1)^{T}

#### 2 Answer

Let us calculate the matrix A = M^{2}.

$M_{2}=(−12 30 )(−12 30 )$

$=(7−2 −36 )$

= A

To calculate eigenvalues we use the polynomial equation Determinant(A − λI) = 0.

Matrix (A − λI) = .

$(7−λ−2 −36−λ )$

Determinant(A − λI) = 0 gives the following equation:

(7 − λ)(6 − λ) − ( − 3)( − 2) = 0

λ^{2} − (7 + 6)λ − 6 + (7)(6) = 0

λ^{2} − 13λ + 36 = 0

Factoring it we get:

(λ − 4)(λ − 9) = 0

λ = 4 or λ = 9

Eigenvalues are 4 and 9

These are squares of the eigenvalues of M.

To calculate eigenvectors for λ = 4, we form equations using (A − λI)x = 0.

(A − λI)x = (A − 4I)x = 0

$(3−2 −32 )(x_{1}x_{2} )=0$

We get the following equation:

3x_{1} − 3x_{2} = 0

x_{1} = x_{2}

For x_{2} = 1 we get x_{1} = 1

Eigenvector is (1, 1)^{T}

This is same as the eigenvector of M for λ = 2

Let us calculate eigenvectors for λ = 9 using the equation (A − λI)x = (A − 9I)x = 0:

$(−2−2 −3−3 )(x_{1}x_{2} )=0$

We get the following equation:

−2x_{1} − 3x_{2} = 0

−2x_{1} = 3x_{2}

x_{1} = ( − 3x_{2})/2

For x_{2} = 2 we get x_{1} = − 3

Eigenvector is ( − 3, 2)^{T}

This is same as the eigenvector of M for λ = − 3

#### 3 Answer

The points on the rotation axis maintain direction. For any transformation matrix, we know that its eigenvectors preserve orientation. So the axis must align with an eigenvector.

Let’s first calculate the eigenvalues of the 3 x 3 matrix.

det(A − λI) = 0

det$⎝⎛ 23 −λ0−0.5 01−λ0 0.5023 −λ ⎠⎞ =0$

Gives the equation:

−λ^{3} + (√3 + 1)λ^{2} − (√3 + 1)λ + 1 = 0

On factoring it we get:

−(λ − 1)(λ^{2} − √3λ + 1) = 0

This results in two equations:

λ − 1 = 0 Equation 1

λ^{2} − √3λ + 1 = 0 Equation 2

Solving the equations we get the following eigenvalues:

λ_{1} = 1

λ_{2} = (√3 − i)/2

λ_{3} = (√3 + i)/2

λ_{2}, λ_{3} result in complex eigenvectors. These vectors are not in real 3-D space. So we find eigenvectors corresponding to λ_{1} = 1. A pure rotation matrix always has 1 as an eigenvalue, and its vector aligns with the axis.

To find the eigenvector we solve (A − I)v = 0.

$⎝⎛ 23 −2 0−0.5 000 0.5023 −2 ⎠⎞ ⎝⎛ xyz ⎠⎞ =0$

Multiplying the vector with the first row of A we get:

x(√3 − 2)/2 + 0.5z = 0

z = − (√3 − 2)x Equation 1

On multiplying the vector with the third row of the A we get:

−0.5x + z(√3 − 2)/2 = 0

z = x/(√3 − 2) Equation 2

Only values of x and z to satisfy both these equations are x = 0, z = 0.

y can have any value, so let it be 1.

Eigenvector for λ_{1} = 1 is:

(0, 1, 0)^{T}

The drone is rotating around the y-axis.

#### 4 Answer

We calculate the eigenvalues using the det(A − λI) = 0:

$(0−λ1 21−λ )=0$

Result is a quadratic equation:

( − λ)(1 − λ) − 2 = 0

−λ + λ^{2} − 2 = 0

λ^{2} − λ − 2 = 0

(λ − 2)(λ + 1) = 0

λ = 2 or λ = − 1

Eigenvalues: λ_{1} = 2, λ_{2} = − 1

Trace of matrix A = Sum of its diagonal elements

= 0 + 1 = 1

Sum of eigenvalues of A = 2 + ( − 1) = 1 = Trace of A

To find eigenvectors of λ_{1} = 2, we solve the equation (A − 2I)x = 0:

$(−21 2−1 )(x_{1}x_{2} )=0$

We get the equation:

−2x_{1} = 2x_{2}

x_{1} = x_{2}

For x_{2} = 1 we get x_{1} = 1

Therefore eigenvector = (1, 1)^{T}

Similarly we calculate the eigenvector for λ_{2} = − 1 as: ( − 2, 1)^{T}

Let us find eigenvalues of A^{−1}.

$A_{−1}−Iβ=(−0.5−β0.5 1−β )$

det(A^{−1} − Iβ) = 0 gives:

( − β)( − 0.5 − β) − 0.5 = 0

0.5β + β^{2} − 0.5 = 0

2β^{2} + β − 1 = 0

(2β − 1)(β + 1) = 0

β = 0.5 or β = − 1

We can see that:

0.5 = β_{1} = 1/λ_{1} = 1/2

−1 = β_{2} = 1/λ_{2} = 1/( − 1)

Eigenvalues of A^{−1} = Inverse of Eigenvalues of A.

Trace of matrix A^{−1} = − 0.5 + 0 = − 0.5

Sum of eigenvalues of A^{−1} = 0.5 + ( − 1) = − 0.5

= Trace of A^{−1}

If we compute eigenvectors of A^{−1} the results are:

Eigenvector for β_{1}(0.5) = (1, 1)^{T} = Eigenvector for λ_{1}(2)

Eigenvector for β_{2}( − 1) = ( − 2, 1)^{T} = Eigenvector for λ_{2}( − 1)

A and A^{−1} have the same eigenvectors.

#### 5 Answer

For any transformation matrix, its eigenvectors preserve direction. For a rotation matrix, one of the eigenvectors must align with the axis (as the axis is invariant). In pure rotation, the points on the axis remain unchanged. So a rotation matrix maps points on the axis to themselves. They are neither stretched nor reversed in direction. Therefore the corresponding eigenvalue is 1.