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×nn \times n matrix and λ\lambda be the eigenvalues associated with it. Then, eigenvector vv can be defined by the following relation:

AvA v = λv\lambda v

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

(AλI)v=0(A- \lambda I) v= 0

Eigenvector associated with matrix AA can be determined using above method.

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

v=[v1v2...vn]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:

AXA X = λ\lambda XX

It is formally known as eigenvector equation.

In place of λ\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 λ1\lambda_{1} = 0 and λ2\lambda_{2} = 1 of any 2 ×\times 2 matrix.


AXA X = λ1X\lambda_{1}X,

AXA X = O …..(1)


AXA X = λ2X\lambda_{2}X,

AA = 1,

(AI)X(A – I) X = O …. (2)

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

Where, II = Identity matrix of same order as AA

O = zero matrix of same order as AA XX = Eigen vector which is equal to [xy]\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 AA using the equation det (AAλ\lambda II) = 0, where I is equivalent order identity matrix as AA. Denote each eigenvalue of λ1\lambda_{1}, λ2\lambda_{2}, λ3\lambda_{3}, …

Step 2: Substitute the value of λ1\lambda_{1} in equation AXA X = λ1\lambda_{1} XX or (AAλ1\lambda_{1} II) XX = O.

Step 3: Calculate the value of eigenvector XX which is associated with eigenvalue λ1\lambda_{1}.

Step 4: Repeat steps 3 and 4 for other eigenvalues λ2\lambda_{2}, λ3\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 λ\lambda of an n×nn \times n matrix is denoted by a nonzero vector XX and is defined as:

(AλI)k(A-\lambda I)^{k} = 0

Where, kk is some positive integer.

For kk = 1 \Rightarrow (AλI)(A-\lambda I) = 0

Therefore, if kk = 1, then eigenvector of matrix AA 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 aij=ajia_{ij}=a_{ji}) does necessarily have orthogonal eigenvectors.

There are basically two types of eigenvectors:

1) Left Eigenvector

2) Right Eigenvector

Left Eigenvector

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

AXL=λXLA X_{L} = \lambda X_{L}

Where, A is given matrix of order n and λ\lambda be one of its eigenvalue. XLX_{L} is denoted by row vector [x1x2xn]\begin{bmatrix} x_{1} & x_{2} & … & x_{n} \end{bmatrix}

Right Eigenvector

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

AXR=λXRA X_{R} = \lambda X_{R}

In which, AA denotes an n×nn \times n square matrix and represents it eigenvalue.

XR=[x1x2...xn]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 AA be a matrix of order n x n and λ1,λ2,,λn\lambda_{1}, \lambda_{2}, …, \lambda_{n} be its eigenvalues, such that λ1\lambda_{1} be the dominant eigenvalue. We are to select an initial approximate value x0x_{0} for a dominant eigenvector of AA.


X1=AX0X_{1}=AX_{0} ….(1)

X2=AX1=AA(X0)=A2X0X_{2}=AX_{1}=AA(X_{0})=A^{2}X_{0} …(using equation 1)

Similarly, we have




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

Also read

Determinant of a matrix

Rank of a matrix and special matrices

Eigenvector Solved Examples 

Example 1: Find the eigenvalues for the following matrix?

A = [4615]\begin{bmatrix} 4 & 6\\ 1 & 5 \end{bmatrix}


Given A = [4615]\begin{bmatrix} 4 & 6\\ 1 & 5 \end{bmatrix}

A-λI = [4λ615λ]\begin{bmatrix} 4-\lambda & 6\\ 1 & 5-\lambda \end{bmatrix}

AλI\left | A-\lambda I \right | = 0

 ⇒4λ615λ=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=[1447]A= \begin{bmatrix} 1 & 4\\ -4 & -7 \end{bmatrix}

Solution : A=[1447]A=\begin{bmatrix}1 & 4\\-4 & -7\end{bmatrix},

AλI=[1λ447λ]A-\lambda I=\begin{bmatrix}1-\lambda & 4\\-4 & -7-\lambda\end{bmatrix},

AλI=1λ447λ\left | A-\lambda I \right |=\begin{vmatrix}1-\lambda & 4\\-4 & -7-\lambda \end{vmatrix},

(1λ)(7λ)4(4)(1 – \lambda) (- 7 – \lambda) – 4 (-4) = 0,

(λ+3)2(\lambda+3)^{2} = 0,

λ\lambda = -3, -3

Using eigenvector equation –

A X = – 3 X

A + 3 I = O

([1447]+[3003])[xy]=[00]\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


x + y = 0

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

Therefore, required eigenvector is:

X=[xy]=k[11]X = \begin{bmatrix}x\\ y\end{bmatrix}=k\begin{bmatrix}1\\ -1\end{bmatrix}

Example 3: Find the eigenvectors of the matrix A = (121610121)\begin{pmatrix}1&2&1\\ \:6&-1&0\\ \:-1&-2&-1\end{pmatrix}


TheeigenvaluesofAaretherootsofthecharacteristicequationdet(AλI)=0\mathrm{The\:eigenvalues\:of}\:A\:\mathrm{are\:the\:roots\:of\:the\:characteristic\:equation}\:\det \left(A-λ\:I\right)=0\\

so, det((121610121)λ(100010001))=(121610121)λ(100010001)=(1λ2161λ0121λ)=det(1λ2161λ0121λ)=(1λ)det(1λ021λ)2det(6011λ)+1det(61λ12)=(1λ)(λ1)226(λ1)+1(λ13)=λ3λ2+12λ\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\\ SolvebyfactoringFactorλ3λ2+12λλ(λ3)(λ+4)=0Thesolutionsareλ=0,λ=3,λ=4Eigenvectorsforλ=0Solve(AλI):(121610121)0(100010001)=(121610121)Reduce(121610121):(1011301613000)\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 (A0I)(xyz)=(1011301613000)(xyz)=(000)\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

{x+113z=0y+613z=0}Isolate{x=113zy=613z}Pluginto(xyz)η=(113z613zz)  z0Letz=13(1613)\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.

Eigenvectorsforλ=3Solve(AλI):(121610121)3(100010001)(121610121)3(100010001)=(221640124)Tosolve(221640124)(xyz)=(000),reducethematrixReduce(221640124):(1010132000)Thesystemassociatedwiththeeigenvalueλ=3(A3I)(xyz)=(1010132000)(xyz)=(000)Thisreducestothefollowingsystemofequations{x+z=0y+32z=0}Letz=2(232)\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}\\ Eigenvectorsforλ=4:(121)Solve(AλI):(121610121)(4)(100010001)Tosolve(521630123)(xyz)=(000),reducethematrixThesystemassociatedwiththeeigenvalueλ=4(A+4I)(xyz)=(101012000)(xyz)=(000)Thisreducestothefollowingsystemofequations{x+z=0y2z=0}Letz=1(121)Theeigenvectorsfor(121610121)=(1613),(232),(121)\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.

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 measures the distortion induced by the transformation and eigenvectors explains about the orientation of the distortion. It gives a rough picture of the Principal Component Analysis = A statistical procedure.

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