Last updated on July 16th, 2025
In linear algebra, eigenvectors help us understand how matrices transform space. In this article, we will learn about eigenvectors and eigenvalues, how to find them, their properties, and examples.
The eigenvectors are the non-vectors that do not change direction when a linear transformation is applied to them. In linear algebra, an eigenvector helps in complex transformations of a matrix. For a square matrix A and a vector v, the eigenvector is represented as:
Av = λv, where λ is the eigenvalue.
In linear algebra, the eigenvector and eigenvalues are fundamental concepts. Eigenvectors are the non-zero vectors that satisfy the equation Av = λv. Where A is the square matrix, v is the vector, and λ is the eigenvalue. The eigenvalue is the scalar λ associated with the eigenvector.
Geometrically, eigenvectors are used to find the directions that remain unchanged after transformation. For example, in a rotation matrix, the eigenvector lies along the axis of rotation. Algebraically, an eigenvector satisfies the equation (A - λI)v = 0, where I is the identity matrix.
Eigenvectors are non-zero vectors when transformed into matrix, it does not change direction. It only changes the length. In this section, we will discuss the difference between the eigenvector and other vectors.
Eigenvector |
Other Vector |
Eigenvectors do not change directions under a transformation |
The other vectors may change their directions when the same transformation is applied. |
It is scaled by a specific factor called the eigenvalue |
It can be transformed, that is, it can be stretched, rotated, or skewed |
Eigenvalues satisfy the equation Av = λv |
They do not follow a specific relationship with the matrix |
In matrix transformation, it represents special and invariant directions |
Without any scaling behavior, it represents the general direction |
Eigenvectors and eigenvalues are used to simplify the matrix operations and to understand the linear transformation. In this section, let’s learn how to find eigenvectors and eigenvalues.
Step 1: Finding the characteristic equation
To find the eigenvalue λ of a square matrix A, we use the formula det(A - λI) = 0.
Step 2: Substitute eigenvalues
After finding the value of λ, solve (A - λI)v = 0 and find the value of v.
Step 3: Normalize the eigenvector
If unit vectors are required, we divide each eigenvector by its magnitude to normalize.
For example, let’s find the eigenvalues and eigenvectors of the matrix A = 1221
Finding the value of λ,
det(A - λI) = 12 - λ 2 - λ1
= (2 - λ)2 - 1
= λ2 - 2λ + 4 - 1
= λ2 - 2λ + 3
Solving λ2 - 2λ + 3 = 0
Using the quadratic equation to find the value of λ
λ = -b ± b2 - 4ac2a
Substituting the value:
λ = -(-4) ± (-4)2 - 4 × 1 × 32 × 1
λ = 4 ± 16 - 122
λ = 4 ± 22
λ = 4 + 22 λ = 4 - 22
λ = 3, λ = 1
Finding the eigenvector using the equation (A - λI)v
For λ = 1:
(A - 1I)v = 1221 - 01 10 = 1111
1111 yx = 0
x + y = 0
y = -x
So, any vector of the form is:
v = -x x is an eigenvector
If x = 1 v becomes
v1 = -1 1
For λ = 3:
(A - 3I)v = 1221 - 03 30 = 1-1-1 1
1-1-1 1 yx = 0
-x + y = 0
y = x
So, v2 = xx
If x = 1, v2 become
v2 = 11
So, the eigenvector corresponding to the eigenvalues λ = 1 and λ = 3 are:
λ = 1 → -1 1
λ = 2 → 11
The eigenvectors and eigenvalues follow many properties that help to understand the matrix behavior. In this section, we will learn some properties of eigenvectors and eigenvalues.
Based on the type of matrices, the eigenvalues and eigenvectors follow a specific pattern. Let’s understand some types of matrices and their eigenvectors and eigenvalues.
In real life, eigenvectors are used in fields like principal component analysis, physics, engineering, and computer graphics. In this section, we will discuss some real-world applications of eigenvectors.
When finding the eigenvectors of a matrix, students make errors, and they mostly repeat them. In this section, we will learn some common mistakes and the ways to avoid them.
Find the eigenvalues for the matrix A, A = 24 31
The eigenvalues for matrix A are 5 and 2.
The eigenvalue can be calculated using the characteristic polynomial:
det(A - λI)
det24 - λ 3 - λ 1
= (4 - λ)(3 - λ) - 2 = λ2 - 7λ + 10
Solving λ2 - 7λ + 10 = 0
λ = -b ± b2 - 4ac2a
= -(-7) ± (-7)2 - 4 × 1 × 102 × 1
= 7 ± 49 - 402
= 7 ± 92 → 5, 2
The eigenvalues of the matrix are 5 and 2
Find the eigenvectors for the matrix A, A = 06 30
For λ = 6, the eigenvector is 01
For λ = 0 the eigenvector is -20
The matrix A is diagonal, the value of the eigenvalue is 6 and 3
Finding the eigenvector using the equation (A - λI)v
Finding the eigenvalue λ: det(A - λI) = 0
A - λI = 06 - λ -λ 3
det(A - λI) = (6 - λ)(-λ) - (0)(3)
= -λ(6 - λ)
So, λ = 0 or λ = 6
For λ = 6: (A - 6I) = 00 -6 3
00 -6 3 yx = 00
So, 3y = 0 → y = 0
So, v1 = 01
For λ = 0: (A - 0I) = 06 03
06 03 yx = 00
6x + 3y = 0 → 2x + y = 0
y = -2x
v2 = -2 0
Find the eigenvalue of the matrix B =
The eigenvalues of the matrix B are 1, 2, and 3
The eigenvalue of a diagonal matrix is its diagonal entries. Since matrix B is diagonal, its eigenvalues are the diagonal values that is 1, 2, and 3.
Find the eigenvectors for the symmetric matrix M = 12 21
For λ = 3, the eigenvector is 11
For λ = 1 the eigenvector is -1 1
Solving the characteristic equation to find eigenvalues:
det(M - λI) = 0
(M - λI) = 12 - λ 2 - λ 1
det(M - λI) = (2 - λ)2 - (1)(1)
= (2 - λ)2 - 1
= λ2 - 4λ + 3
Solving λ2 - 4λ + 3
λ = -b ± b2 - 4ac2a
= -(-4) ± (-4)2 - 4 × 1 × 32 × 1
= 4 ± 16 - 122
= 4 ± 22 → 3, 1
So, λ = 3 and λ = 1
Finding the eigenvector using M - λI
For λ = 3: M- 3I = 12 21 - 03 30
= 1-1 -1 1
(M - 3I)v = 0 → 1-1 -1 1yx = 00
So, -x + y = 0
x = y
Thus, v1 = 11
For λ = 1: M- 1I
= 12 21 - 01 10
= 11 11
(M - 1I)v = 0 → 11 11yx = 00
So, x + y = 0
x = -y
Thus, v2 = -1 1
Find the eigenvalue for the symmetric matrix M = 10 0-1
The eigenvalue for the matrix M is i and -i
To find the eigenvalue of the matrix, solve the characteristic equation:
det(M - λI) = 0
M - λI = 10 - λ 0 - λ -1
= 1- λ - λ -1
det(M - λI) = (-λ)(-λ) - (-1)(1)
= λ2 + 1
λ2 + 1 = 0
So, λ2 = -1
λ = ±i
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