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# Eigenvectors of a Matrix

The eigenvector of a matrix is also known as a latent vector, proper vector or characteristic vector. They are defined in the reference of a square matrix. Matrix is an important branch that is studied under linear algebra. A matrix is a rectangular array of numbers or other elements of the same kind. It generally represents a system of linear equations.

A very useful concept related to matrices is eigenvectors. It is a vector that is associated with a set of linear equations. They are also useful in solving differential equations and many other applications related to them. Let us go ahead and understand the eigenvector, how to find the eigenvalue of a 2×2 matrix, its technique and various other concepts related to it.

## Eigenvector Method

The method of determining the eigenvector of a matrix is given below:

If A be an n × n matrix and λ be the eigenvalues associated with it. Then, eigenvector v can be defined by the following relation:

Av = λv

If I is the identity matrix of the same order as A, then

(A – λI)v = 0

The eigenvector associated with matrix A can be determined using the above method.

Here, v is known as the eigenvector belonging to each eigenvalue and is written as:

$$\begin{array}{l}v=\begin{bmatrix} v_{1}\\ v_{2}\\ .\\ .\\ .\\ v_{n} \end{bmatrix}\end{array}$$

Eigenvector Equation

The equation corresponding to each eigenvalue of a matrix is given by:

AX = λX

It is formally known as the eigenvector equation.

In place of λ, put each eigenvalue one by one and get the eigenvector equation which enables us to solve for the eigenvector belonging to each eigenvalue.

For example: Suppose that there are two eigenvalues λ1 = 0 and λ2 = 1 of any 2 × 2 matrix.

Then,

AX = λ1X

AX = 0…..(1)

and

AX = λ2X

A = 1

(A – I) X = O ….. (2)

Equations (1) and (2) are eigenvector equations for the given matrix.

Where, I = Identity matrix of the same order as A

O = zero matrix of the same order as A

$$\begin{array}{l}X = \text{Eigen vector which is equal to}\ \begin{bmatrix} x\\ y \end{bmatrix}\ \text{(as A is of order 2)}\end{array}$$

## How to Find Eigenvector?

In order to find the eigenvectors of a matrix, one needs to follow the steps, given below:

Step 1: Determine the eigenvalues of given matrix A using the equation det (A – λI) = 0, where I is the equivalent order identity matrix as A. Denote each eigenvalue of λ1, λ2, λ3,…

Step 2: Substitute the value of λ1 in equation AX = λ1X or (A – λ1I)X = 0.

Step 3: Calculate the value of eigenvector X, which is associated with eigenvalue λ1.

Step 4: Repeat steps 3 and 4 for other eigenvalues λ2, λ3, … as well.

## Generalized Eigenvector

There is little difference between an eigenvector and a generalized eigenvector. It is defined in the following way:

A generalized eigenvector associated with an eigenvalue λ of an n × n matrix is denoted by a nonzero vector X and is defined as:

(A – λI)k = 0

Where, k is some positive integer.

For k = 1 (A – λI) = 0

Therefore, if k = 1, then the eigenvector of matrix A is its generalized eigenvector.

## Eigenvector Orthogonality

We know that a vector quantity possesses magnitude as well as direction. So, an eigenvector has some magnitude in a particular direction. Orthogonality is a concept of two eigenvectors of a matrix being perpendicular to each other. We can say that when two eigenvectors make a right angle between each other, they are said to be orthogonal eigenvectors.

A symmetric matrix (in which aij = aji) does necessarily have orthogonal eigenvectors.

There are basically two types of eigenvectors:

1) Left Eigenvector

2) Right Eigenvector

Left Eigenvector

The left eigenvector is a type of eigenvector that is represented in the form of a row vector which satisfies the following condition:

$$\begin{array}{l}A X_{L} = \lambda X_{L}\end{array}$$

Where, A is the given matrix of order n and λ is one of its eigenvalues. XL is denoted by row vector,

$$\begin{array}{l}\begin{bmatrix} x_{1} & x_{2} & … & x_{n} \end{bmatrix}\end{array}$$

Right Eigenvector

In the same way as the left eigenvector, the right eigenvector is denoted by XR. It is defined as an eigenvector that is written in the form of a column vector, satisfying the condition given below:

$$\begin{array}{l}A X_{R} = \lambda X_{R}\end{array}$$

In which, A denotes an n × n square matrix and represents its eigenvalue.

$$\begin{array}{l}X_{R} = \begin{bmatrix} x_{1}\\ x_{2}\\ .\\ .\\ .\\ x_{n} \end{bmatrix}\end{array}$$

## Power Method for Eigenvectors

The power method is an important method for computing the eigenvectors of a matrix. It is an iterative method used in numerical analysis. The power method works in the following way:

Let us assume that A be a matrix of order n x n and λ1, λ2,…, λn be its eigenvalues, such that λ1 be the dominant eigenvalue. We need to select an initial approximate value of x0 for a dominant eigenvector of A.

Then

$$\begin{array}{l}X_{1}=AX_{0}….(1)\end{array}$$
$$\begin{array}{l}X_{2}=AX_{1}=AA(X_{0})=A^{2}X_{0}….\text{(using equation 1)}\end{array}$$

Similarly, we have

$$\begin{array}{l}X_{3}=A^{3}X_{0}\end{array}$$

….

$$\begin{array}{l}X_{k}=A^{k}X_{0}\end{array}$$

As we proceed towards large powers of k, we get a better approximation of the eigenvector.

Determinant of a Matrix

Rank of a Matrix and Special Matrices

## Eigenvector Solved Examples

Example 1: Find the eigenvalues for the following matrix.

$$\begin{array}{l}A = \begin{bmatrix} 4 & 6\\ 1 & 5 \end{bmatrix}\end{array}$$

Solution:

Given,

$$\begin{array}{l}A = \begin{bmatrix} 4 & 6\\ 1 & 5 \end{bmatrix}\end{array}$$

$$\begin{array}{l}A-\lambda I = \begin{bmatrix} 4-\lambda & 6\\ 1 & 5-\lambda \end{bmatrix}\end{array}$$

$$\begin{array}{l}\left | A-\lambda I \right |=0\end{array}$$

$$\begin{array}{l}\Rightarrow \begin{vmatrix} 4-\lambda & 6\\ 1& 5-\lambda \end{vmatrix} = 0\end{array}$$

(4 – λ)(5 – λ) – 6 = 0

⇒ 20 – 5λ – 4λ + λ2-6 = 0

⇒ λ2 – 9λ +14 = 0

⇒ (λ – 7)(λ – 2) = 0

⇒ λ = 7 or λ= 2

Example 2: Find the eigenvectors for the following matrix.

$$\begin{array}{l}A= \begin{bmatrix} 1 & 4\\ -4 & -7 \end{bmatrix}\end{array}$$

Solution :

$$\begin{array}{l}A=\begin{bmatrix}1 & 4\\-4 & -7\end{bmatrix}\end{array}$$
$$\begin{array}{l}A-\lambda I=\begin{bmatrix}1-\lambda & 4\\-4 & -7-\lambda\end{bmatrix}\end{array}$$
$$\begin{array}{l}\left | A-\lambda I \right |=\begin{vmatrix}1-\lambda & 4\\-4 & -7-\lambda \end{vmatrix}\end{array}$$
$$\begin{array}{l}(1 – \lambda) (- 7 – \lambda) – 4 (-4)=0\end{array}$$
$$\begin{array}{l}(\lambda+3)^{2}=0\end{array}$$

λ = -3, -3

Using eigenvector equation,

A X = – 3 X

A + 3 I = O

$$\begin{array}{l}\left ( \begin{bmatrix} 1 & 4\\ -4 & -7 \end{bmatrix}+\begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix} \right )\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}\end{array}$$

Which gives,

4x + 4y = 0

or

x + y = 0

Let us set x = k, then y = -k

Therefore, the required eigenvector is:

$$\begin{array}{l}X = \begin{bmatrix}x\\ y\end{bmatrix}=k\begin{bmatrix}1\\ -1\end{bmatrix}\end{array}$$

Example 3:

$$\begin{array}{l}\text{Find the eigenvectors of the matrix A}=\begin{pmatrix}1&2&1\\ \:6&-1&0\\ \:-1&-2&-1\end{pmatrix}\end{array}$$

Solution:

$$\begin{array}{l}\mathrm{The\:eigenvalues\:of}\:A\:\mathrm{are\:the\:roots\:of\:the\:characteristic\:equation}\:\det \left(A-λ\:I\right)=0\\\end{array}$$

so,

$$\begin{array}{l}\det \left(\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-\lambda\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)\\ =\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-\lambda\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}1-\lambda&2&1\\ 6&-1-\lambda&0\\ -1&-2&-1-\lambda\end{pmatrix}\\ =\det \begin{pmatrix}1-\lambda&2&1\\ 6&-1-\lambda&0\\ -1&-2&-1-\lambda\end{pmatrix}\\ =\left(1-\lambda\right)\det \begin{pmatrix}-1-\lambda&0\\ -2&-1-\lambda\end{pmatrix}-2\cdot \det \begin{pmatrix}6&0\\ -1&-1-\lambda\end{pmatrix}+1\cdot \det \begin{pmatrix}6&-1-\lambda\\ -1&-2\end{pmatrix}\\ =\left(1-\lambda\right)\left(-\lambda-1\right)^2-2\cdot \:6\left(-\lambda-1\right)+1\cdot \left(-\lambda-13\right)\\ =-\lambda^3-\lambda^2+12\lambda\\\end{array}$$
$$\begin{array}{l}\mathrm{Solve\:by\:factoring}\\ \mathrm{Factor\:}-λ^3-λ^2+12λ\\ -λ\left(λ-3\right)\left(λ+4\right)=0\\ \mathrm{The\:solutions\:are}\\ λ=0,\:λ=3,\:λ=-4\\\mathrm{Eigenvectors\:for\:}λ=0\\\mathrm{Solve\:}\:\left(A-λ\:I\right):\:\begin{pmatrix}1&2&1\\ 6&-1&0\\-1&-2&-1\end{pmatrix}-0\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\\ =\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}\\\mathrm{Reduce\:}\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}:\quad \begin{pmatrix}1&0&\frac{1}{13}\\ 0&1&\frac{6}{13}\\ 0&0&0\end{pmatrix}\\\end{array}$$
,

The system associated with the eigenvalue λ = 0 is

$$\begin{array}{l}\left(A-0I\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1&0&\frac{1}{13}\\ 0&1&\frac{6}{13}\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\\\end{array}$$

This reduces to the following system of equations.

$$\begin{array}{l}\begin{Bmatrix}x+\frac{1}{13}z=0\\ y+\frac{6}{13}z=0\end{Bmatrix}\\\mathrm{Isolate}\\\begin{Bmatrix}x=-\frac{1}{13}z\\ y=-\frac{6}{13}z\end{Bmatrix}\\\mathrm{Plug\:into\:}\begin{pmatrix}x\\ y\\ z\end{pmatrix}\\η=\begin{pmatrix}-\frac{1}{13}z\\ -\frac{6}{13}z\\ z\end{pmatrix}\space\space\:z\ne \:0\\\mathrm{Let\:}z=13\\\begin{pmatrix}-1\\ -6\\ 13\end{pmatrix}\\\end{array}$$

Similarly, find the eigenvectors for λ = 3 and λ = -4.

$$\begin{array}{l}\mathrm{Eigenvectors\:for\:}λ=3\\\mathrm{Solve\:}\:\left(A-λ\:I\right):\:\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-3\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\\\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-3\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}-2&2&1\\ 6&-4&0\\ -1&-2&-4\end{pmatrix}\\\mathrm{To\:solve}\:\begin{pmatrix}-2&2&1\\ 6&-4&0\\ -1&-2&-4\end{pmatrix}\begin{pmatrix}\text{x}\\ \text{y}\\ \text{z}\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix},\:\mathrm{reduce\:the\:matrix}\\\mathrm{Reduce\:}\begin{pmatrix}-2&2&1\\ 6&-4&0\\ -1&-2&-4\end{pmatrix}:\quad \begin{pmatrix}1&0&1\\ 0&1&\frac{3}{2}\\ 0&0&0\end{pmatrix}\\\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}λ=3\\\left(A-3I\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1&0&1\\ 0&1&\frac{3}{2}\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\\\mathrm{This\:reduces\:to\:the\:following\:system\:of\:equations}\\\begin{Bmatrix}x+z=0\\ y+\frac{3}{2}z=0\end{Bmatrix}\\\mathrm{Let\:}z=2\\\begin{pmatrix}-2\\ -3\\ 2\end{pmatrix}\\\end{array}$$
$$\begin{array}{l}\mathrm{Eigenvectors\:for\:}λ=-4:\quad \begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}\\\mathrm{Solve\:}\:\left(A-λ\:I\right):\:\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-\left(-4\right)\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\\\mathrm{To\:solve}\:\begin{pmatrix}5&2&1\\ 6&3&0\\ -1&-2&3\end{pmatrix}\begin{pmatrix}\text{x}\\ \text{y}\\ \text{z}\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix},\:\mathrm{reduce\:the\:matrix}\\\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}λ=-4\\\left(A+4I\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1&0&1\\ 0&1&-2\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\\\mathrm{This\:reduces\:to\:the\:following\:system\:of\:equations}\\\begin{Bmatrix}x+z=0\\ y-2z=0\end{Bmatrix}\\\mathrm{Let\:}z=1\\\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}\\ \mathrm{The\:eigenvectors\:for\:}\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}\\=\begin{pmatrix}-1\\ -6\\ 13\end{pmatrix},\:\begin{pmatrix}-2\\ -3\\ 2\end{pmatrix},\:\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}\\\end{array}$$

## Eigenvector Applications

There are different applications of eigenvectors in real life. Some of the important ones are provided below:

1) In Mathematics, eigenvector decomposition is widely used in order to solve linear equations of the first order, in ranking matrices, in differential calculus, etc.

2) Eigenvectors are used in Physics in the simple mode of oscillation.

3) This concept is widely used in quantum mechanics.

4) Eigenvectors are widely applicable in almost all branches of engineering.

5) Eigenvalues can be used to calculate the theoretical limit of the transmission of information through a medium of communication like a telephone line or through the air. It is done by finding the eigenvalues and eigenvectors of the channel that is used for communication and then water-filling on the eigenvalues.

6) Eigenvalue is the natural frequency that is the smallest magnitude of a system that builds the bridge. The stability of the construction can be monitored by using this.

7) The concept of eigenvalues is used in the construction of a car stereo system. It helps in the reproduction of the vibration of the car due to the music.

8) It is used in decoupling three-phase systems along the symmetrical component transformation.

9) It helps in the reduction of a linear operation to separate simpler problems.

10) Many oil companies generally use eigenvalue analysis to survey land for oil.

11) It is used to explain natural occurrences.

12) The linear mapping eigenvalues measure the distortion induced by the transformation, and eigenvectors explain the orientation of the distortion. It gives a rough picture of the principal component analysis = A statistical procedure.

## Matrices and Determinants – Important Topics ## Matrices and Determinants – Important Questions Q1

### What are eigenvectors?

Eigenvectors of a square matrix are non-zero vectors, which, when multiplied by the square matrix, result in the scalar multiple of the vectors.

Q2

### Give the difference between eigenvector and eigenvalue.

Consider a square matrix A. If Av = λv, then v is the eigenvector and λ is the eigenvalue. A vector that transforms at most by a scale factor when a transformation is applied is called an eigenvector, and the corresponding scale factor is called an eigenvalue.

Q3

### Can an eigenvector be a zero vector?

No, an eigenvector cannot be a zero vector.

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