# Eigenvectors of a Matrix

Eigenvector of a matrix is also known as latent vector, proper vector or characteristic vector. These are defined in the reference of a square matrix.Matrix is an important branch that is studied under linear algebra. 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 vector that is associated with a set of linear equations. These are also useful in solving differential equations and many other applications related to them. Let us go ahead and understand eigenvector, how to find eigenvalue of a 2×2 matrix, its technique and various other concepts related to it.

## Eigenvector Method

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

If A be an $n \times n$ matrix and $\lambda$ be the eigenvalues associated with it. Then, eigenvector $v$ can be defined by the following relation:

$A v$ = $\lambda v$

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

($(A- \lambda I) v= 0$

Eigenvector associated with matrix $A$ can be determined using above method.

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

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

Eigenvector Equation

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

$A X$ = $\lambda$ $X$

It is formally known as eigenvector equation.

In place of $\lambda$, we one by one put each eigenvalue and get the eigenvector equation which enables us to solve for eigen vector belonging to each eigenvalue.

For example: Suppose that there are two eigenvalues $\lambda_{1}$ = 0 and $\lambda_{2}$ = 1 of any 2 $\times$ 2 matrix.

Then,

$A X$ = $\lambda_{1}$ $X$ $A X$ = O …..(1)

and

$A X$ = $\lambda_{2}$ $X$ $A$ = 1

$(A – I) X$ = O …. (2)

Equations (1) and (2) are eigen vector equations for given matrix.

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

O = zero matrix of same order as $A$ $X$ = Eigen vector which is equal to $\begin{bmatrix} x\\ y \end{bmatrix}$

(as A is of order 2)

## How to Find Eigenvector

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

Step 1: Determine the eigenvalues of given matrix $A$ using the equation det ($A$$\lambda$ $I$) = 0, where I is equivalent order identity matrix as $A$. Denote each eigenvalue of $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, …

Step 2: Substitute the value of $\lambda_{1}$ in equation $A X$ = $\lambda_{1}$ $X$ or ($A$$\lambda_{1}$ $I$) $X$ = O.

Step 3: Calculate the value of eigenvector $X$ which is associated with eigenvalue $\lambda_{1}$.

Step 4: Repeat steps 3 and 4 for other eigenvalues $\lambda_{2}$, $\lambda_{3}$, … as well.

## Generalized Eigenvector

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

A generalized eigenvector associated with an eigenvalue $\lambda$ of an $n \times n$ matrix is denoted by a nonzero vector $X$ and is defined as:

$(A-\lambda I)^{k}$ = 0

Where, $k$ is some positive integer.

For $k$ = 1 $\Rightarrow$ $(A-\lambda I)$ = 0

Therefore, if $k$ = 1, then 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, these are said to be orthogonal eigenvectors.

A symmetric matrix (in which $a_{ij}=a_{ji}$) does necessarily have orthogonal eigenvectors.

Left Eigenvector

There are basically two types of eigenvectors:

1) Left Eigenvector

2) Right Eigenvector

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

$A X_{L} = \lambda X_{L}$

Where, A is given matrix of order n and $\lambda$ be one of its eigenvalue. $X_{L}$ is denoted by row vector $\begin{bmatrix} x_{1} & x_{2} & … & x_{n} \end{bmatrix}$

Right Eigenvector

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

$A X_{R} = \lambda X_{R}$

In which, $A$ denotes an $n \times n$ square matrix and represents it eigenvalue.

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

## Power Method for Eigenvectors

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

Let us assume that $A$ be a matrix of order n x n and $\lambda_{1}, \lambda_{2}, …, \lambda_{n}$ be its eigenvalues, such that $\lambda_{1}$ be the dominant eigenvalue. We are to select an initial approximate value $x_{0}$ for a dominant eigenvector of $A$.

Then

$X_{1}=AX_{0}$ ….(1)

$X_{2}=AX_{1}=AA(X_{0})=A^{2}X_{0}$ …(using equation 1)

Similarly, we have

$X_{3}=A^{3}X_{0}$

….

$X_{k}=A^{k}X_{0}$

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

Determinant of a matrix

Rank of a matrix and special matrices

## Eigenvector Examples

Example 1: Find the eigenvalues for the following matrix?

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

Solution:

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

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

$\left | A-\lambda I \right |$ = 0

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

(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?

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

Solution : $A=\begin{bmatrix}1 & 4\\-4 & -7\end{bmatrix}$ $A-\lambda I=\begin{bmatrix}1-\lambda & 4\\-4 & -7-\lambda\end{bmatrix}$ $\left | A-\lambda I \right |=\begin{vmatrix}1-\lambda & 4\\-4 & -7-\lambda \end{vmatrix}$ $(1 – \lambda) (- 7 – \lambda) – 4 (-4)$ = 0

$(\lambda+3)^{2}$ = 0

$\lambda$ = -3, -3

Using eigenvector equation –

A X = – 3 X

A + 3 I = O

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

Which gives :

4x + 4y = 0

or

x + y = 0

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

Therefore, required eigenvector is:

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

Example 3: Find the eigenvectors of the matrix A = $\begin{pmatrix}1&2&1\\ \:6&-1&0\\ \:-1&-2&-1\end{pmatrix}$

Solution:

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

The system associated with the eigenvalue λ = 0 is $\left(A-0I\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1&0&\frac{1}{13}\\ 0&1&\frac{6}{13}\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\\$

This reduces to the following system of equations

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

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

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

## Eigenvector Applications

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

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

2) Eigenvectors are used in physics in simple mode of oscillation.

3) This concept is widely used in quantum mechanics.

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