Eigenvalues and Eigenvectors (Examples & Questions)
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.
If we apply the matrix A=(1320) to the entire ( ℝ 2 ) space (by multiplying every vector in the space by the matrix), you can see in the image below how it rotates, flips, and stretches the entire space.
However, eigenvectors preserve their direction (as shown in the image below), while all other vectors point in new directions.
Eigenvectors – Matrix Transformation
Finding Transformation Matrix
Let us do the transformation ourselves. We start with the unit vectors.
Multiplying A with the unit vector i = (1, 0)T:
Ai=(1320)(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 v1 = ( − 1, 1.5)T and v2 = (1, 1)T.
Av1=A(−11.5)=(2−3)
= − 2( − 1, 1.5)T = − 2v1
It is scaled by a factor of −2. Due to the negative sign of the scale factor, its direction gets reversed.
v1 is an eigenvector. The eigenvalue corresponding to this vector is −2.
Av2=A(11)
=(33)=3(11)=3v2
v2 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=(1320).
We use the following equation to calculate eigenvalues:
Determinant of (A − λI) = 0
Step 1. Form Matrix
A−λI=(1−λ320−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.
λ2 − 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 2. For eachλ, 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=(1320).
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)320−(−2))=(3322)
Step 3. Solve Equations
We now solve (A − Iλ)x = 0
(3322)(x1x2)=0
We get the equation:
3x1 + 2x2 = 0
x1=3−2x2
We can choose any value for x2 to calculate the eigenvector.
For x2 = 3 we get x1 = − 2
Eigenvector is ( − 2, 3)T
Any value of x2 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) =
(1−3320−3)=(−232−3)
We now solve (A − Iλ)x = 0
(−232−3)(x1x2)=0
We get the equation:
−2x1 + 2x2 = 0
x1 = x2
You may verify that the other equation is the same.
Again, we can choose any value for x2 to calculate an eigenvector.
For x2 = 1 we get x1 = 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.
_From Advanced Engineering Mathematics, by Erwin Kreyszig_
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 isA=(01−10)
Calculating Eigenvalues
A−λI=(−λ1−1−λ)
det(A − λI) = 0 gives the equation:
λ2 + 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)(x1x2)=0
We get the following result:
−ix1 − x2 = 0
−ix1 = x2
x1 = ix2
For x2 = 1 we get x1 = i
Eigenvector is (i, 1)T
Solving for eigenvectors of λ = − i:
(0+i1−10+i)(x1x2)=0
We get the following result:
ix1 − x2 = 0
x1=ix2
x1 = − ix2
For x2 = 1 we get x1 = − 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 A2 have the same eigenvectors. Eigenvalues of A2 = λ2.
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−1w = βw
AA−1w = βAw
Iw = w = βAw
β1w=Aw
Aw=β1w
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.
Can you find the roots of the equation det(A − λI) = 0? These are the eigenvalues.
Can you form linear equations using (A + 6I)x = 0? Can you solve these equations to get eigenvectors for -6?
Can you find eigenvectors using (A − 10I)x = 0?
Ans. −6, ( − 7, 9)T and 10, (1, 1)T
2 Question
The matrix M=(−1230) 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 M2. What is the relation between eigenvalues and eigenvectors of M and M2?
Answer
Can you find the matrix A = M2?
Can you find the roots of the equation det(A − λI) = 0? These are the eigenvalues.
Can you form linear equations using (A − 4I)x = 0? Can you solve these equations to get eigenvectors for 4?
Can you find eigenvectors using (A − 9I)x = 0?
Ans. 4, (1, 1)T and 9, ( − 3, 2)T
3 Question
The following matrix represents a rotation of a drone in three-dimensional space. Find its axis of rotation.
A=⎝⎛230−0.50100.5023⎠⎞
Answer
Will the points on the axis of a rotation change direction?
Can you find eigenvalues using the equation det(A − λI) = 0?
Which eigenvalues are relevant for calculating the axis of rotation?
Ans. (0, 1, 0)T. The drone is rotating about the y-axis.
4 Question
Given the matrices A=(0121) and A−1=(−0.50.510), 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?
Can you calculate the eigenvectors using the equation (A − λI)x = 0?
Can you find eigenvalues of A−1 and compare them with that of A?
Can you find eigenvectors of A−1 and compare with A?
Ans. Eigenvectors of A−1 = Eigenvectors of A. Eigenvalues of A−1 = Inverse of Eigenvalues of A.
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−λ971−λ)
Determinant(A − λI) = 0 gives the following equation:
(3 − λ)(1 − λ) − (7)(9) = 0
λ2 − (3 + 1)λ − 63 + (3)(1) = 0
λ2 − 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:
(9977)(x1x2)=0
∴ 9x1 + 7x2 = 0
x1 = ( − 7x2)/9
For x2 = 9 we get x1 = − 7
Eigenvector is ( − 7, 9)T
Let us calculate eigenvectors for λ = 10.
We form equations using (A − 10I)x = 0, which gives:
(−797−9)(x1x2)=0
We get the following equation:
−7x1 + 7x2 = 0
x1 = x2
For x2 = 1 we get x1 = 1
Eigenvector is (1, 1)T
2 Answer
Let us calculate the matrix A = M2.
M2=(−1230)(−1230)
=(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
λ2 − (7 + 6)λ − 6 + (7)(6) = 0
λ2 − 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)(x1x2)=0
We get the following equation:
3x1 − 3x2 = 0
x1 = x2
For x2 = 1 we get x1 = 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)(x1x2)=0
We get the following equation:
−2x1 − 3x2 = 0
−2x1 = 3x2
x1 = ( − 3x2)/2
For x2 = 2 we get x1 = − 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−λ0−0.501−λ00.5023−λ⎠⎞=0
Gives the equation:
−λ3 + (√3 + 1)λ2 − (√3 + 1)λ + 1 = 0
On factoring it we get:
−(λ − 1)(λ2 − √3λ + 1) = 0
This results in two equations:
λ − 1 = 0 Equation 1
λ2 − √3λ + 1 = 0 Equation 2
Solving the equations we get the following eigenvalues:
λ1 = 1
λ2 = (√3 − i)/2
λ3 = (√3 + i)/2
λ2, λ3 result in complex eigenvectors. These vectors are not in real 3-D space. So we find eigenvectors corresponding to λ1 = 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.
⎝⎛23−20−0.50000.5023−2⎠⎞⎝⎛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 λ1 = 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−λ121−λ)=0
Result is a quadratic equation:
( − λ)(1 − λ) − 2 = 0
−λ + λ2 − 2 = 0
λ2 − λ − 2 = 0
(λ − 2)(λ + 1) = 0
λ = 2 or λ = − 1
Eigenvalues: λ1 = 2, λ2 = − 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 λ1 = 2, we solve the equation (A − 2I)x = 0:
(−212−1)(x1x2)=0
We get the equation:
−2x1 = 2x2
x1 = x2
For x2 = 1 we get x1 = 1
Therefore eigenvector = (1, 1)T
Similarly we calculate the eigenvector for λ2 = − 1 as: ( − 2, 1)T
Let us find eigenvalues of A−1.
A−1−Iβ=(−0.5−β0.51−β)
det(A−1 − Iβ) = 0 gives:
( − β)( − 0.5 − β) − 0.5 = 0
0.5β + β2 − 0.5 = 0
2β2 + β − 1 = 0
(2β − 1)(β + 1) = 0
β = 0.5 or β = − 1
We can see that:
0.5 = β1 = 1/λ1 = 1/2
−1 = β2 = 1/λ2 = 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 β1(0.5) = (1, 1)T = Eigenvector for λ1(2)
Eigenvector for β2( − 1) = ( − 2, 1)T = Eigenvector for λ2( − 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.