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3 Dec, 2025
Eigenvalues and eigenvectors are fundamental concepts in linear algebra, used in various applications such as matrix diagonalization, stability analysis, and data analysis (e.g.,
Principal Component Analysis
). They are associated with a square matrix and provide insights into its properties.
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Eigenvalues
Eigenvalues are unique scalar values linked to a matrix or
linear transformation
. They indicate how much an eigenvector gets stretched or compressed during the transformation. The eigenvector's direction remains unchanged unless the eigenvalue is negative, in which case the direction is simply reversed.
The equation for eigenvalue is given by,
A
v
=
λ
v
Av = \lambda v
Where,
A is the matrix,
v is associated eigenvector and
λ
\lambda
is scalar eigenvalue.
Eigenvectors
Eigenvectors are non-zero vectors that, when multiplied by a matrix, only stretch or shrink without changing direction. The eigenvalue must be found first before the eigenvector. For any square matrix A of order n × n, the eigenvector is a column matrix of size n × 1. This is known as the right eigenvector, as matrix multiplication is not commutative.
Alternatively, the left eigenvector can be found using the equation
v
A
=
λ
v
vA = λv
, where v is a row matrix of size 1 × n.
Eigenvector Equation
The Eigenvector equation is the equation that is used to find the eigenvector of any square matrix. The eigenvector equation is
A
v
=
λ
v
Av = λv
Where,
A
is the given square matrix,
v
is the eigenvector of matrix A and
λ
\lambda
is any scaler multiple.
Here the left hand side is a matrix vector multiplication and right side is a scalar multiplication. So let's rewrite the right-hand side as matrix vector multiplication.
Right hand side represents scaling any vector by a factor of lambda which can be written as
λ
\lambda
I =
[
λ
0
0
0
λ
0
0
0
λ
]
\begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}
A
v
→
=
(
I
λ
)
v
→
A \overrightarrow{v} = (Iλ)\overrightarrow{v}
A
v
→
−
I
λ
v
→
=
0
A \overrightarrow{v} - Iλ\overrightarrow{v} = 0
(
A
−
I
λ
)
v
→
=
0
(A - Iλ)\overrightarrow{v} = 0
Product of a matrix to a non-zero vector is zero only if the transformation associated with that matrix squishes space into lower dimension. That implies the determinant of that matrix is zero.
i.e. det(A - ƛI) = 0
How to Find an Eigenvector?
The eigenvector of the following square matrix can be easily calculated using the steps below,
Step 1:
Find the eigenvalues of the matrix A, using the equation det |(A – λI| =0, where “I” is the identity matrix of order similar to matrix A
Step 2:
The value obtained in Step 2 are named as,
λ
1
,
λ
2
,
λ
3
…
.
λ _1, λ_2, λ_3….
Step 3:
Find the eigenvector (X) associated with the eigenvalue λ
1
using the equation,
(
A
–
λ
1
I
)
X
=
0
(A – λ_1I) X = 0
Step 4:
Repeat step 3 to find the eigenvector associated with other remaining eigenvalues
λ
2
,
λ
3
…
.
λ_2, λ_3….
Following these steps gives the eigenvector related to the given square matrix.
Types of Eigenvector
The eigenvectors calculated for the square matrix are of two types which are,
Right Eigenvector
Left Eigenvector
Right Eigenvector:
The eigenvector which is multiplied by the given square matrix from the right-hand side is called the right eigenvector. It is calculated by using the following equation,
A
V
R
=
λ
V
R
AV_R = \lambda V_R
Where,
A
is given square matrix of order n×n,
λ
is one of the eigenvalues and
V
R
is the column vector matrix
The value of V
R
is,
V
R
=
[
v
1
v
2
v
3
.
.
v
n
]
\bold{V_{R} = \begin{bmatrix} v_{1}\\ v_{2}\\ v_{3}\\ .\\ .\\ v_{n}\\ \end{bmatrix}}
Left Eigenvector:
The eigenvector which is multiplied by the given square matrix from the left-hand side is called the left eigenvector. It is calculated by using the following equation,
V
L
A
=
V
L
λ
V_LA = V_L\lambda
Where,
A
is given square matrix of order n×n,
λ
is one of the eigenvalues and
V
L
V_L
is the row vector matrix.
The value of V
L
is,
V
L
=
[
v
1
,
v
2
,
v
3
,
.
.
.
,
v
n
]
V_L = [v_1, v_2, v_3,..., v_n]
Eigenvectors of a Square Matrix
We can easily find the eigenvector of square matrices of order n × n. Now, let's find the following square matrices:
Eigenvectors of a 2 × 2 matrix
Eigenvectors of a 3 × 3 matrix.
Eigenvector of a 2 × 2 matrix
The Eigenvector of the 2 × 2 matrix can be calculated by,
Find the eigenvalues
of the matrix using the formula
d
e
t
(
A
−
λ
I
)
=
0
det(A - \lambda I) = 0
, where A is the matrix and
I
is the identity matrix.
Substitute the eigenvalue(s)
into the equation
(
A
−
λ
I
)
v
=
0
(A - \lambda I)v = 0
to find the corresponding eigenvector(s).
An example of the same is,
Example:
Find the eigenvalues and the eigenvector for the matrix A =
[
1
2
5
4
]
\begin{bmatrix} 1 & 2\\ 5& 4 \end{bmatrix}
Solution:
If eigenvalues are represented using λ and the eigenvector is represented as v =
[
a
b
]
\begin{bmatrix} a\\b \end{bmatrix}
Then the eigenvector is calculated by using the equation,
∣
A
−
λ
I
∣
=
0
|A- λI| = 0
[
1
2
5
4
]
−
λ
[
1
0
0
1
]
=
[
0
0
0
0
]
\begin{bmatrix}1 & 2\\ 5& 4\end{bmatrix} -λ\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} = \begin{bmatrix}0 & 0\\ 0& 0\end{bmatrix}
[
1
−
λ
2
5
4
−
λ
]
\begin{bmatrix} 1 - λ& 2\\ 5& 4 - λ \end{bmatrix}
= 0
(1-λ) ⨉ (4-λ) - (2 ⨉ 5) = 0
⇒ 4 - λ - 4λ + λ
2
- 10 = 0
⇒ λ
2
- 5λ - 6 = 0
⇒ λ
2
- (6λ - λ) - 6 = 0
⇒ λ
2
- 6λ + λ - 6 = 0
⇒ (λ-6)(λ+1) = 0
λ = 6 and λ = -1
Thus, the eigenvalues are 6 and -1.
For λ = 6
(A-λI)v = 0
⇒
[
1
−
6
2
5
4
−
6
]
.
[
a
b
]
\begin{bmatrix}1 - 6& 2\\ 5& 4 - 6\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix}
= 0
⇒
[
−
5
2
5
−
2
]
.
[
a
b
]
\begin{bmatrix}-5& 2\\ 5& -2\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix}
= 0
⇒ 5a - 2b = 0
Simplifying the above equation we get, 5a=2b
The required eigenvector is,
[
a
b
]
=
[
2
5
]
\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}2\\5\end{bmatrix}
For λ = -1
⇒
[
1
−
(
−
1
)
2
5
4
−
(
−
1
)
]
.
[
a
b
]
\begin{bmatrix}1 - (-1)& 2\\ 5& 4 - (-1)\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix}
= 0
⇒
[
2
2
5
5
]
.
[
a
b
]
\begin{bmatrix}2& 2\\ 5& 5\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix}
= 0
⇒ 2a + 2b = 0, ⇒ 5a + 5b = 0
simplifying the above equation we get, a = -b
The required eigenvector is,
[
a
b
]
=
[
1
−
1
]
\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix} 1\\-1\end{bmatrix}
Then the eigenvectors of the given 2 × 2 matrix are
[
a
b
]
=
[
2
5
]
,
[
a
b
]
=
[
1
−
1
]
\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}2\\5\end{bmatrix}, \begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}
These are two possible eigen vectors but many of the corresponding multiples of these eigen vectors can also be considered as other possible eigen vectors.
Eigenvector of a 3 × 3 Matrix
An example of eigenvector of 3x3 matrix is,
Example:
Find the eigenvalues and the eigenvector for the matrix A =
[
2
2
2
2
2
2
2
2
2
]
\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2\\2 & 2 & 2 \end{bmatrix}
Solution:
If eigenvalues are represented using λ and the eigenvector is represented as v =
[
a
b
c
]
\begin{bmatrix} a\\b\\c \end{bmatrix}
Then the eigenvector is calculated by using the equation,
|A- λI| = 0
[
2
2
2
2
2
2
2
2
2
]
−
λ
[
1
0
0
0
1
0
0
0
1
]
=
[
0
0
0
0
0
0
0
0
0
]
\begin{bmatrix}2 & 2 & 2\\ 2 & 2 & 2\\ 2 & 2 & 2\end{bmatrix} -λ\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}
[
2
−
λ
2
2
2
2
−
λ
2
2
2
2
−
λ
]
\begin{bmatrix} 2 - λ & 2 & 2 \\ 2 & 2 - λ & 2 \\ 2 & 2 & 2- λ\end{bmatrix}
= 0
Simplifying the above determinant we get
⇒ (2-λ)(λ
2
) + 2λ
2
+ 2λ
2
= 0 ⇒ (-λ
3
) + 6λ
2
= 0
⇒ λ
2
(6 - λ) = 0
⇒ λ = 0, λ = 6, λ = 0
For λ = 0
(A - λI) v = 0
⇒
[
2
−
0
2
2
2
2
−
0
2
2
2
2
−
0
]
.
[
a
b
c
]
\begin{bmatrix}2 - 0& 2& 2\\ 2& 2 - 0&2\\2 & 2 & 2-0\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix}
= 0
⇒
[
2
2
2
2
2
2
2
2
2
]
.
[
a
b
c
]
\begin{bmatrix}2& 2& 2\\ 2& 2 &2\\2 & 2 & 2\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix}
= 0
Simplifying the above equation we get
2a + 2b + 2c = 0
⇒ a + b + c = 0
Let b = k
1
and c = k
2
⇒ a + k
1
+ k
2
= 0
⇒ a = -(k
1
+ k
2
)
Thus, the eigenvector is,
[
a
b
c
]
=
[
−
(
k
1
+
k
2
)
k
1
k
2
]
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}-(k_{1}+k_{2})\\ k_{1}\\k_{2}\end{bmatrix}
taking k
1
= 1 and k
2
= 0, the eigenvector is,
[
a
b
c
]
=
[
−
1
1
0
]
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}-1\\ 1\\0\end{bmatrix}
taking k
1
= 0 and k
2
= 1, the eigenvector is,
[
a
b
c
]
=
[
−
1
0
1
]
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}-1\\ 0\\1\end{bmatrix}
For λ = 6
(A - λI) v = 0
⇒
[
2
−
6
2
2
2
2
−
6
2
2
2
2
−
6
]
.
[
a
b
c
]
\begin{bmatrix}2 - 6& 2& 2\\ 2& 2 -6&2\\2 & 2 & 2-6\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix}
= 0
⇒
[
−
4
2
2
2
−
4
2
2
2
−
4
]
.
[
a
b
c
]
\begin{bmatrix}-4& 2& 2\\ 2& -4 &2\\2 & 2 & -4\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix}
= 0
Simplifying the above equation we get,
-4a +2b +2c = 0 ⇒ 2 (-2a + b + c) = 0
⇒ 2a = b + c
Let b = k
1
and c = k
2
and taking k
1
= k
2
= 1 we get,
[
a
b
c
]
=
[
1
1
1
]
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}1\\ 1\\1\end{bmatrix}
Thus, the eigenvector is,
[
a
b
c
]
=
[
1
1
1
]
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}1\\ 1\\1\end{bmatrix}
Eigenspace
We define the eigenspace of a matrix as the set of all the eigenvectors of the matrix. All the vectors in the eigenspace are linearly independent of each other.
To find the Eigenspace of the matrix we have to follow the following steps
Step 1:
Find all the eigenvalues of the given square matrix.
Step 2:
For each eigenvalue find the corresponding eigenvector.
Step 3:
Take the set of all the eigenvectors (say A). The resultant set so formed is called the Eigenspace of the following vector.
From the above example of given 3 × 3 matrix A, the eigenspace so formed is
[
−
1
1
0
]
,
[
−
1
0
1
]
,
[
1
1
1
]
\begin{bmatrix}-1\\ 1\\0\end{bmatrix},\begin{bmatrix}-1\\ 0\\1\end{bmatrix},\begin{bmatrix}1\\ 1\\1\end{bmatrix}
Diagonalize Matrix Using Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors are used to find diagonal matrices. A
diagonal matrix
is a matrix which can be written as,
A = XDX
-1
Where,
A is the original matrix.
X is the matrix formed by eigenvectors.
D is a diagonal matrix where the diagonal elements are the eigenvalues of A.
X
-1
is the inverse of the matrix X.
We can understand the concept of a diagonal matrix by taking the following example.
Example:
Diagonalize the matrix A =
[
2
2
2
2
2
2
2
2
2
]
\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2\\2 & 2 & 2 \end{bmatrix}
Solution:
We have already solved for the eigenvalues and the eigenvectors of the A
=
[
2
2
2
2
2
2
2
2
2
]
\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2\\2 & 2 & 2 \end{bmatrix}
The eigenvalues of the A are λ = 6, λ = 0 and λ = 0
The eigenvectors of A are
[
1
1
1
]
,
[
−
1
1
0
]
,
[
−
1
0
1
]
\begin{bmatrix}1\\ 1\\1\end{bmatrix},\begin{bmatrix}-1\\ 1\\0\end{bmatrix},\begin{bmatrix}-1\\ 0\\1\end{bmatrix}
Thus,
D =
[
6
0
0
0
0
0
0
0
0
]
\begin{bmatrix}6 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}
X =
[
1
1
1
1
−
1
0
1
0
−
1
]
\begin{bmatrix} 1 & 1 & 1\\1 & -1 & 0\\1 & 0 & -1\end{bmatrix}
Applications of Eigen Values and Eigen Vectors
Some of the common applications of eigen values are:
Google PageRank Algorithm
Web pages form a graph; eigenvector corresponding to eigenvalue 1 of the link matrix gives each page’s importance score.
Markov Processes & Steady-State Analysis
Used in simulations, natural language processing, recommendation systems, and stochastic modeling.
The eigenvector for eigenvalue 1 gives long-term stable probabilities.
Principal Component Analysis (PCA)
Reduces dimensionality of data, removes noise, and extracts key patterns.
Common in image recognition, machine learning, and data compression.
Latent Semantic Analysis (LSA) in NLP
Eigen-decomposition (via SVD) finds semantic relationships between words and documents.
Improves search, document clustering, and recommendation engines.
Spectral Graph Theory in Network Analysis
Eigenvalues of the graph Laplacian help detect communities, find graph partitions, and measure connectivity in social networks or computer networks.
Computer Vision (Eigenfaces)
Eigenvectors from face image datasets form a “basis” for recognizing and comparing faces.
Control System Stability in Robotics & AI
Eigenvalues of system matrices determine stability and convergence of autonomous systems and learning algorithms.
Signal Processing & Data Transmission
Eigen-decomposition optimizes communication channels, feature extraction in audio/speech, and noise filtering.
Solved Examples on Eigenvectors
Example 1: Find the eigenvectors of the matrix A =
[
1
1
0
0
1
1
0
0
1
]
\begin{bmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 & 1\end{bmatrix}
Solution:
The eigen values of the matrix is found using,
|A - λI| = 0
[
1
−
λ
1
0
0
1
−
λ
1
0
0
1
−
λ
]
\begin{bmatrix}1-λ & 1 & 0\\0 & 1-λ & 1\\0 & 0 & 1-λ\end{bmatrix}
= 0
(1 - λ)
3
= 0
Thus, the eigen values are, λ = 1, 1, 1
As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 1, using (A - λI)v = O
[
1
−
1
1
0
0
1
−
1
1
0
0
1
−
1
]
.
[
a
b
c
]
=
[
0
0
0
]
\begin{bmatrix}1-1 & 1 & 0\\0 & 1-1 & 1\\0 & 0 & 1-1\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}0\\ 0\\0\end{bmatrix}
[
0
1
0
0
0
1
0
0
0
]
.
[
a
b
c
]
=
[
0
0
0
]
\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}0\\ 0\\0\end{bmatrix}
solving the above equation we get,
a = K
y = 0
z = 0
Then the eigenvector is,
[
a
b
c
]
=
[
k
0
0
]
=
k
[
1
0
0
]
\begin{bmatrix}a\\ b\\c\end{bmatrix}= \begin{bmatrix}k\\ 0\\0\end{bmatrix} = k\begin{bmatrix}1\\ 0\\0\end{bmatrix}
Example 2: Find the eigenvectors of the matrix A =
[
5
0
0
5
]
\begin{bmatrix}5 & 0\\0 & 5 \end{bmatrix}
Solution:
The eigen values of the matrix is found using,
|A - λI| = 0
[
5
−
λ
0
0
5
−
λ
]
\begin{bmatrix}5-λ & 0\\0 & 5-λ \end{bmatrix}
= 0
(5 - λ)
2
= 0
Thus, the eigen values are,
λ = 5, 5
As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 5, using
(A - λI)v = O
[
5
−
5
0
0
5
−
5
]
.
[
a
b
]
=
[
0
0
]
\begin{bmatrix}5-5 & 0 \\ 0 & 5-5\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix} = \begin{bmatrix}0\\ 0\end{bmatrix}
Simplifying the above we get,
a = 1, b = 0
a = 0, b = 1
Then the eigenvector is,
[
a
b
]
=
[
1
0
]
,
[
0
1
]
\begin{bmatrix}a\\ b\end{bmatrix}= \begin{bmatrix}1\\ 0\end{bmatrix} , \begin{bmatrix}0\\ 1\end{bmatrix} |
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# Eigenvalues and Eigenvectors
Last Updated : 3 Dec, 2025
Eigenvalues and eigenvectors are fundamental concepts in linear algebra, used in various applications such as matrix diagonalization, stability analysis, and data analysis (e.g., [Principal Component Analysis](https://www.geeksforgeeks.org/data-analysis/principal-component-analysis-pca/)). They are associated with a square matrix and provide insights into its properties.
2 / 2
## Eigenvalues
Eigenvalues are unique scalar values linked to a matrix or [linear transformation](https://www.geeksforgeeks.org/machine-learning/linear-mapping/). They indicate how much an eigenvector gets stretched or compressed during the transformation. The eigenvector's direction remains unchanged unless the eigenvalue is negative, in which case the direction is simply reversed.
The equation for eigenvalue is given by,
> A v \= λ v Av = \\lambda v Av\=λv
Where,
- A is the matrix,
- v is associated eigenvector and
- λ
\\lambda
λ
is scalar eigenvalue.
## Eigenvectors
Eigenvectors are non-zero vectors that, when multiplied by a matrix, only stretch or shrink without changing direction. The eigenvalue must be found first before the eigenvector. For any square matrix A of order n × n, the eigenvector is a column matrix of size n × 1. This is known as the right eigenvector, as matrix multiplication is not commutative.
Alternatively, the left eigenvector can be found using the equation v A \= λ v vA = λv vA\=λv, where v is a row matrix of size 1 × n.
### Eigenvector Equation
The Eigenvector equation is the equation that is used to find the eigenvector of any square matrix. The eigenvector equation is
> A v \= λ v Av = λv Av\=λv
Where,
- ****A**** is the given square matrix,
- ****v**** is the eigenvector of matrix A and
- λ
\\lambda
λ
is any scaler multiple.
Here the left hand side is a matrix vector multiplication and right side is a scalar multiplication. So let's rewrite the right-hand side as matrix vector multiplication.
Right hand side represents scaling any vector by a factor of lambda which can be written as
λ \\lambda λI = \[ λ 0 0 0 λ 0 0 0 λ \] \\begin{bmatrix} \\lambda & 0 & 0 \\\\ 0 & \\lambda & 0 \\\\ 0 & 0 & \\lambda \\end{bmatrix} ⎣ ⎡ λ000λ000λ ⎦ ⎤
A v → \= ( I λ ) v → A \\overrightarrow{v} = (Iλ)\\overrightarrow{v} A v \= (Iλ) v
A v → − I λ v → \= 0 A \\overrightarrow{v} - Iλ\\overrightarrow{v} = 0 A v − Iλ v \= 0
( A − I λ ) v → \= 0 (A - Iλ)\\overrightarrow{v} = 0 (A− Iλ) v \= 0
Product of a matrix to a non-zero vector is zero only if the transformation associated with that matrix squishes space into lower dimension. That implies the determinant of that matrix is zero.
i.e. det(A - ƛI) = 0
## How to Find an Eigenvector?
The eigenvector of the following square matrix can be easily calculated using the steps below,
****Step 1:**** Find the eigenvalues of the matrix A, using the equation det \|(A – λI\| =0, where “I” is the identity matrix of order similar to matrix A
****Step 2:**** The value obtained in Step 2 are named as,λ 1 , λ 2 , λ 3 … . λ \_1, λ\_2, λ\_3…. λ1,λ2,λ3….
****Step 3:**** Find the eigenvector (X) associated with the eigenvalue λ1 using the equation, ( A – λ 1 I ) X \= 0 (A – λ\_1I) X = 0 (A–λ1I)X\=0
****Step 4:**** Repeat step 3 to find the eigenvector associated with other remaining eigenvalues λ 2 , λ 3 … . λ\_2, λ\_3…. λ2,λ3….
Following these steps gives the eigenvector related to the given square matrix.
### Types of Eigenvector
The eigenvectors calculated for the square matrix are of two types which are,
- Right Eigenvector
- Left Eigenvector
****Right Eigenvector:**** The eigenvector which is multiplied by the given square matrix from the right-hand side is called the right eigenvector. It is calculated by using the following equation,
> A V R \= λ V R AV\_R = \\lambda V\_R AVR\=λVR
Where,
- ****A**** is given square matrix of order n×n,
- ****λ**** is one of the eigenvalues and
- ****V********R**** is the column vector matrix
The value of VR is, V R \= \[ v 1 v 2 v 3 . . v n \] \\bold{V\_{R} = \\begin{bmatrix} v\_{1}\\\\ v\_{2}\\\\ v\_{3}\\\\ .\\\\ .\\\\ v\_{n}\\\\ \\end{bmatrix}} VR\= ⎣ ⎡ v1v2v3..vn ⎦ ⎤
****Left Eigenvector:**** The eigenvector which is multiplied by the given square matrix from the left-hand side is called the left eigenvector. It is calculated by using the following equation,
> V L A \= V L λ V\_LA = V\_L\\lambda VLA\=VLλ
Where,
- ****A**** is given square matrix of order n×n,
- ****λ**** is one of the eigenvalues and
- V
L
V\_L
VL
is the row vector matrix.
The value of VL is,
V L \= \[ v 1 , v 2 , v 3 , . . . , v n \] V\_L = \[v\_1, v\_2, v\_3,..., v\_n\] VL\=\[v1,v2,v3,...,vn\]
## Eigenvectors of a Square Matrix
We can easily find the eigenvector of square matrices of order n × n. Now, let's find the following square matrices:
- Eigenvectors of a 2 × 2 matrix
- Eigenvectors of a 3 × 3 matrix.
### ****Eigenvector of a 2 × 2 matrix****
The Eigenvector of the 2 × 2 matrix can be calculated by,
- ****Find the eigenvalues**** of the matrix using the formula
d
e
t
(
A
−
λ
I
)
\=
0
det(A - \\lambda I) = 0
det(A−λI)\=0
, where A is the matrix and ****I**** is the identity matrix.
- ****Substitute the eigenvalue(s)**** into the equation
(
A
−
λ
I
)
v
\=
0
(A - \\lambda I)v = 0
(A−λI)v\=0
to find the corresponding eigenvector(s).
An example of the same is,
****Example:**** Find the eigenvalues and the eigenvector for the matrix A =\[ 1 2 5 4 \] \\begin{bmatrix} 1 & 2\\\\ 5& 4 \\end{bmatrix} \[1524\]
****Solution:****
If eigenvalues are represented using λ and the eigenvector is represented as v = \[ a b \] \\begin{bmatrix} a\\\\b \\end{bmatrix} \[ab\]
Then the eigenvector is calculated by using the equation,
∣ A − λ I ∣ \= 0 \|A- λI\| = 0 ∣A−λI∣\=0
> \[ 1 2 5 4 \] − λ \[ 1 0 0 1 \] \= \[ 0 0 0 0 \] \\begin{bmatrix}1 & 2\\\\ 5& 4\\end{bmatrix} -λ\\begin{bmatrix}1 & 0\\\\ 0 & 1\\end{bmatrix} = \\begin{bmatrix}0 & 0\\\\ 0& 0\\end{bmatrix} \[1524\]−λ\[1001\]\=\[0000\]
>
> \[ 1 − λ 2 5 4 − λ \] \\begin{bmatrix} 1 - λ& 2\\\\ 5& 4 - λ \\end{bmatrix} \[1−λ524−λ\] = 0
>
> (1-λ) ⨉ (4-λ) - (2 ⨉ 5) = 0
> ⇒ 4 - λ - 4λ + λ2 - 10 = 0
> ⇒ λ2 - 5λ - 6 = 0
> ⇒ λ2 - (6λ - λ) - 6 = 0
> ⇒ λ2 - 6λ + λ - 6 = 0
> ⇒ (λ-6)(λ+1) = 0
> λ = 6 and λ = -1
Thus, the eigenvalues are 6 and -1.
- ****For λ = 6****
> (A-λI)v = 0
>
> ⇒ \[ 1 − 6 2 5 4 − 6 \] . \[ a b \] \\begin{bmatrix}1 - 6& 2\\\\ 5& 4 - 6\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} \[1−6524−6\].\[ab\] = 0
>
> ⇒ \[ − 5 2 5 − 2 \] . \[ a b \] \\begin{bmatrix}-5& 2\\\\ 5& -2\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} \[−552−2\].\[ab\] = 0
> ⇒ 5a - 2b = 0
>
> Simplifying the above equation we get, 5a=2b
>
> The required eigenvector is,\[ a b \] \= \[ 2 5 \] \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix}2\\\\5\\end{bmatrix} \[ab\]\=\[25\]
- ****For λ = -1****
> ⇒ \[ 1 − ( − 1 ) 2 5 4 − ( − 1 ) \] . \[ a b \] \\begin{bmatrix}1 - (-1)& 2\\\\ 5& 4 - (-1)\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} \[1−(−1)524−(−1)\].\[ab\] = 0
>
> ⇒ \[ 2 2 5 5 \] . \[ a b \] \\begin{bmatrix}2& 2\\\\ 5& 5\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} \[2525\].\[ab\] = 0
> ⇒ 2a + 2b = 0, ⇒ 5a + 5b = 0
>
> simplifying the above equation we get, a = -b
>
> The required eigenvector is,\[ a b \] \= \[ 1 − 1 \] \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix} 1\\\\-1\\end{bmatrix} \[ab\]\=\[1−1\]
Then the eigenvectors of the given 2 × 2 matrix are \[ a b \] \= \[ 2 5 \] , \[ a b \] \= \[ 1 − 1 \] \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix}2\\\\5\\end{bmatrix}, \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix}1\\\\-1\\end{bmatrix} \[ab\]\=\[25\],\[ab\]\=\[1−1\]
These are two possible eigen vectors but many of the corresponding multiples of these eigen vectors can also be considered as other possible eigen vectors.
### Eigenvector of a 3 × 3 Matrix
An example of eigenvector of 3x3 matrix is,
****Example:**** Find the eigenvalues and the eigenvector for the matrix A =\[ 2 2 2 2 2 2 2 2 2 \] \\begin{bmatrix} 2 & 2 & 2 \\\\ 2 & 2 & 2\\\\2 & 2 & 2 \\end{bmatrix} ⎣ ⎡ 222222222 ⎦ ⎤
****Solution:****
If eigenvalues are represented using λ and the eigenvector is represented as v = \[ a b c \] \\begin{bmatrix} a\\\\b\\\\c \\end{bmatrix} ⎣ ⎡ abc ⎦ ⎤
Then the eigenvector is calculated by using the equation,
> \|A- λI\| = 0
> \[ 2 2 2 2 2 2 2 2 2 \] − λ \[ 1 0 0 0 1 0 0 0 1 \] \= \[ 0 0 0 0 0 0 0 0 0 \] \\begin{bmatrix}2 & 2 & 2\\\\ 2 & 2 & 2\\\\ 2 & 2 & 2\\end{bmatrix} -λ\\begin{bmatrix}1 & 0 & 0\\\\ 0 & 1 & 0\\\\0 & 0 & 1\\end{bmatrix} = \\begin{bmatrix}0 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0\\end{bmatrix} ⎣ ⎡ 222222222 ⎦ ⎤ − λ ⎣ ⎡ 100010001 ⎦ ⎤ \= ⎣ ⎡ 000000000 ⎦ ⎤
>
> \[ 2 − λ 2 2 2 2 − λ 2 2 2 2 − λ \] \\begin{bmatrix} 2 - λ & 2 & 2 \\\\ 2 & 2 - λ & 2 \\\\ 2 & 2 & 2- λ\\end{bmatrix} ⎣ ⎡ 2−λ2222−λ2222−λ ⎦ ⎤ = 0
>
> Simplifying the above determinant we get
> ⇒ (2-λ)(λ2) + 2λ2 + 2λ2 = 0 ⇒ (-λ3) + 6λ2 = 0
> ⇒ λ2(6 - λ) = 0
> ⇒ λ = 0, λ = 6, λ = 0
****For λ = 0****
> (A - λI) v = 0
> ⇒ \[ 2 − 0 2 2 2 2 − 0 2 2 2 2 − 0 \] . \[ a b c \] \\begin{bmatrix}2 - 0& 2& 2\\\\ 2& 2 - 0&2\\\\2 & 2 & 2-0\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} ⎣ ⎡ 2−02222−02222−0 ⎦ ⎤ . ⎣ ⎡ abc ⎦ ⎤ = 0
>
> ⇒ \[ 2 2 2 2 2 2 2 2 2 \] . \[ a b c \] \\begin{bmatrix}2& 2& 2\\\\ 2& 2 &2\\\\2 & 2 & 2\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} ⎣ ⎡ 222222222 ⎦ ⎤ . ⎣ ⎡ abc ⎦ ⎤ = 0
>
> Simplifying the above equation we get
> 2a + 2b + 2c = 0
> ⇒ a + b + c = 0
>
> Let b = k1 and c = k2
> ⇒ a + k1 + k2 = 0
> ⇒ a = -(k1 + k2)
>
> Thus, the eigenvector is,
> \[ a b c \] \= \[ − ( k 1 \+ k 2 ) k 1 k 2 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}-(k\_{1}+k\_{2})\\\\ k\_{1}\\\\k\_{2}\\end{bmatrix} ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ −(k1\+k2)k1k2 ⎦ ⎤
>
> taking k1 = 1 and k2 \= 0, the eigenvector is,
> \[ a b c \] \= \[ − 1 1 0 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}-1\\\\ 1\\\\0\\end{bmatrix} ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ −110 ⎦ ⎤
>
> taking k1 = 0 and k2\= 1, the eigenvector is,
> \[ a b c \] \= \[ − 1 0 1 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}-1\\\\ 0\\\\1\\end{bmatrix} ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ −101 ⎦ ⎤
****For λ = 6****
> (A - λI) v = 0
> ⇒ \[ 2 − 6 2 2 2 2 − 6 2 2 2 2 − 6 \] . \[ a b c \] \\begin{bmatrix}2 - 6& 2& 2\\\\ 2& 2 -6&2\\\\2 & 2 & 2-6\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} ⎣ ⎡ 2−62222−62222−6 ⎦ ⎤ . ⎣ ⎡ abc ⎦ ⎤ = 0
>
> ⇒ \[ − 4 2 2 2 − 4 2 2 2 − 4 \] . \[ a b c \] \\begin{bmatrix}-4& 2& 2\\\\ 2& -4 &2\\\\2 & 2 & -4\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} ⎣ ⎡ −4222−4222−4 ⎦ ⎤ . ⎣ ⎡ abc ⎦ ⎤ = 0
>
> Simplifying the above equation we get,
> \-4a +2b +2c = 0 ⇒ 2 (-2a + b + c) = 0
> ⇒ 2a = b + c
>
> Let b = k1 and c = k2 and taking k1 = k2 = 1 we get,
> \[ a b c \] \= \[ 1 1 1 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix} ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ 111 ⎦ ⎤
>
> Thus, the eigenvector is,
> \[ a b c \] \= \[ 1 1 1 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix} ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ 111 ⎦ ⎤
## Eigenspace
We define the eigenspace of a matrix as the set of all the eigenvectors of the matrix. All the vectors in the eigenspace are linearly independent of each other.
To find the Eigenspace of the matrix we have to follow the following steps
- ****Step 1:**** Find all the eigenvalues of the given square matrix.
- ****Step 2:**** For each eigenvalue find the corresponding eigenvector.
- ****Step 3:**** Take the set of all the eigenvectors (say A). The resultant set so formed is called the Eigenspace of the following vector.
From the above example of given 3 × 3 matrix A, the eigenspace so formed is \[ − 1 1 0 \] , \[ − 1 0 1 \] , \[ 1 1 1 \] \\begin{bmatrix}-1\\\\ 1\\\\0\\end{bmatrix},\\begin{bmatrix}-1\\\\ 0\\\\1\\end{bmatrix},\\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix} ⎣ ⎡ −110 ⎦ ⎤ , ⎣ ⎡ −101 ⎦ ⎤ , ⎣ ⎡ 111 ⎦ ⎤
## Diagonalize Matrix Using Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors are used to find diagonal matrices. A [diagonal matrix](https://www.geeksforgeeks.org/maths/diagonal-matrix/) is a matrix which can be written as,
> ****A = XDX********\-1****
Where,
- A is the original matrix.
- X is the matrix formed by eigenvectors.
- D is a diagonal matrix where the diagonal elements are the eigenvalues of A.
- X\-1 is the inverse of the matrix X.
We can understand the concept of a diagonal matrix by taking the following example.
****Example:**** Diagonalize the matrix A = \[ 2 2 2 2 2 2 2 2 2 \] \\begin{bmatrix} 2 & 2 & 2 \\\\ 2 & 2 & 2\\\\2 & 2 & 2 \\end{bmatrix} ⎣ ⎡ 222222222 ⎦ ⎤
****Solution:****
> We have already solved for the eigenvalues and the eigenvectors of the A ****\=**** \[ 2 2 2 2 2 2 2 2 2 \] \\begin{bmatrix} 2 & 2 & 2 \\\\ 2 & 2 & 2\\\\2 & 2 & 2 \\end{bmatrix} ⎣ ⎡ 222222222 ⎦ ⎤
>
> The eigenvalues of the A are λ = 6, λ = 0 and λ = 0
> The eigenvectors of A are \[ 1 1 1 \] , \[ − 1 1 0 \] , \[ − 1 0 1 \] \\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix},\\begin{bmatrix}-1\\\\ 1\\\\0\\end{bmatrix},\\begin{bmatrix}-1\\\\ 0\\\\1\\end{bmatrix} ⎣ ⎡ 111 ⎦ ⎤ , ⎣ ⎡ −110 ⎦ ⎤ , ⎣ ⎡ −101 ⎦ ⎤
> Thus,
> D = \[ 6 0 0 0 0 0 0 0 0 \] \\begin{bmatrix}6 & 0 & 0\\\\0 & 0 & 0\\\\0 & 0 & 0\\end{bmatrix} ⎣ ⎡ 600000000 ⎦ ⎤
>
> X = \[ 1 1 1 1 − 1 0 1 0 − 1 \] \\begin{bmatrix} 1 & 1 & 1\\\\1 & -1 & 0\\\\1 & 0 & -1\\end{bmatrix} ⎣ ⎡ 1111−1010−1 ⎦ ⎤
## Applications of Eigen Values and Eigen Vectors
Some of the common applications of eigen values are:
****Google PageRank Algorithm****
- Web pages form a graph; eigenvector corresponding to eigenvalue 1 of the link matrix gives each page’s importance score.
****Markov Processes & Steady-State Analysis****
- Used in simulations, natural language processing, recommendation systems, and stochastic modeling.
- The eigenvector for eigenvalue 1 gives long-term stable probabilities.
****Principal Component Analysis (PCA)****
- Reduces dimensionality of data, removes noise, and extracts key patterns.
- Common in image recognition, machine learning, and data compression.
****Latent Semantic Analysis (LSA) in NLP****
- Eigen-decomposition (via SVD) finds semantic relationships between words and documents.
- Improves search, document clustering, and recommendation engines.
****Spectral Graph Theory in Network Analysis****
- Eigenvalues of the graph Laplacian help detect communities, find graph partitions, and measure connectivity in social networks or computer networks.
****Computer Vision (Eigenfaces)****
- Eigenvectors from face image datasets form a “basis” for recognizing and comparing faces.
****Control System Stability in Robotics & AI****
- Eigenvalues of system matrices determine stability and convergence of autonomous systems and learning algorithms.
****Signal Processing & Data Transmission****
- Eigen-decomposition optimizes communication channels, feature extraction in audio/speech, and noise filtering.
## Solved Examples on Eigenvectors
****Example 1: Find the eigenvectors of the matrix A =**** \[ 1 1 0 0 1 1 0 0 1 \] \\begin{bmatrix}1 & 1 & 0\\\\0 & 1 & 1\\\\0 & 0 & 1\\end{bmatrix} ⎣ ⎡ 100110011 ⎦ ⎤
****Solution:****
> The eigen values of the matrix is found using,
> \|A - λI\| = 0
> \[ 1 − λ 1 0 0 1 − λ 1 0 0 1 − λ \] \\begin{bmatrix}1-λ & 1 & 0\\\\0 & 1-λ & 1\\\\0 & 0 & 1-λ\\end{bmatrix} ⎣ ⎡ 1−λ0011−λ0011−λ ⎦ ⎤ = 0
> (1 - λ)3 = 0
>
> Thus, the eigen values are, λ = 1, 1, 1
>
> As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 1, using (A - λI)v = O
>
> \[ 1 − 1 1 0 0 1 − 1 1 0 0 1 − 1 \] . \[ a b c \] \= \[ 0 0 0 \] \\begin{bmatrix}1-1 & 1 & 0\\\\0 & 1-1 & 1\\\\0 & 0 & 1-1\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}0\\\\ 0\\\\0\\end{bmatrix} ⎣ ⎡ 1−10011−10011−1 ⎦ ⎤ . ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ 000 ⎦ ⎤
>
> \[ 0 1 0 0 0 1 0 0 0 \] . \[ a b c \] \= \[ 0 0 0 \] \\begin{bmatrix}0 & 1 & 0\\\\0 & 0 & 1\\\\0 & 0 & 0\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}0\\\\ 0\\\\0\\end{bmatrix} ⎣ ⎡ 000100010 ⎦ ⎤ . ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ 000 ⎦ ⎤
> solving the above equation we get,
>
> - a = K
> - y = 0
> - z = 0
>
> Then the eigenvector is,
> \[ a b c \] \= \[ k 0 0 \] \= k \[ 1 0 0 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix}= \\begin{bmatrix}k\\\\ 0\\\\0\\end{bmatrix} = k\\begin{bmatrix}1\\\\ 0\\\\0\\end{bmatrix} ⎣ ⎡ abc ⎦ ⎤ \= ⎣ ⎡ k00 ⎦ ⎤ \= k ⎣ ⎡ 100 ⎦ ⎤
****Example 2: Find the eigenvectors of the matrix A =**** \[ 5 0 0 5 \] \\begin{bmatrix}5 & 0\\\\0 & 5 \\end{bmatrix} \[5005\]
****Solution:****
> The eigen values of the matrix is found using,
> \|A - λI\| = 0
> \[ 5 − λ 0 0 5 − λ \] \\begin{bmatrix}5-λ & 0\\\\0 & 5-λ \\end{bmatrix} \[5−λ005−λ\] = 0
> (5 - λ)2 = 0
>
> Thus, the eigen values are,
> λ = 5, 5
>
> As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 5, using
> (A - λI)v = O
> \[ 5 − 5 0 0 5 − 5 \] . \[ a b \] \= \[ 0 0 \] \\begin{bmatrix}5-5 & 0 \\\\ 0 & 5-5\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} = \\begin{bmatrix}0\\\\ 0\\end{bmatrix} \[5−5005−5\].\[ab\]\=\[00\]
>
> Simplifying the above we get,
> a = 1, b = 0
> a = 0, b = 1
>
> Then the eigenvector is,
> \[ a b \] \= \[ 1 0 \] , \[ 0 1 \] \\begin{bmatrix}a\\\\ b\\end{bmatrix}= \\begin{bmatrix}1\\\\ 0\\end{bmatrix} , \\begin{bmatrix}0\\\\ 1\\end{bmatrix} \[ab\]\=\[10\],\[01\]
## Eigenvalues and Eigenvectors
## 15\. Linear Algebra \| Eigen Values and Eigen Vectors
## Eigenvalues & Eigenvectors: Definition, Formula, Examples
Eigenvalues and Eigenvectors
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| Readable Markdown | Last Updated : 3 Dec, 2025
Eigenvalues and eigenvectors are fundamental concepts in linear algebra, used in various applications such as matrix diagonalization, stability analysis, and data analysis (e.g., [Principal Component Analysis](https://www.geeksforgeeks.org/data-analysis/principal-component-analysis-pca/)). They are associated with a square matrix and provide insights into its properties.
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## Eigenvalues
Eigenvalues are unique scalar values linked to a matrix or [linear transformation](https://www.geeksforgeeks.org/machine-learning/linear-mapping/). They indicate how much an eigenvector gets stretched or compressed during the transformation. The eigenvector's direction remains unchanged unless the eigenvalue is negative, in which case the direction is simply reversed.
The equation for eigenvalue is given by,
> A v \= λ v Av = \\lambda v
Where,
- A is the matrix,
- v is associated eigenvector and
- λ
\\lambda
is scalar eigenvalue.
## Eigenvectors
Eigenvectors are non-zero vectors that, when multiplied by a matrix, only stretch or shrink without changing direction. The eigenvalue must be found first before the eigenvector. For any square matrix A of order n × n, the eigenvector is a column matrix of size n × 1. This is known as the right eigenvector, as matrix multiplication is not commutative.
Alternatively, the left eigenvector can be found using the equation v A \= λ v vA = λv, where v is a row matrix of size 1 × n.
### Eigenvector Equation
The Eigenvector equation is the equation that is used to find the eigenvector of any square matrix. The eigenvector equation is
> A v \= λ v Av = λv
Where,
- ****A**** is the given square matrix,
- ****v**** is the eigenvector of matrix A and
- λ
\\lambda
is any scaler multiple.
Here the left hand side is a matrix vector multiplication and right side is a scalar multiplication. So let's rewrite the right-hand side as matrix vector multiplication.
Right hand side represents scaling any vector by a factor of lambda which can be written as
λ \\lambdaI = \[ λ 0 0 0 λ 0 0 0 λ \] \\begin{bmatrix} \\lambda & 0 & 0 \\\\ 0 & \\lambda & 0 \\\\ 0 & 0 & \\lambda \\end{bmatrix}
A v → \= ( I λ ) v → A \\overrightarrow{v} = (Iλ)\\overrightarrow{v}
A v → − I λ v → \= 0 A \\overrightarrow{v} - Iλ\\overrightarrow{v} = 0
( A − I λ ) v → \= 0 (A - Iλ)\\overrightarrow{v} = 0
Product of a matrix to a non-zero vector is zero only if the transformation associated with that matrix squishes space into lower dimension. That implies the determinant of that matrix is zero.
i.e. det(A - ƛI) = 0
## How to Find an Eigenvector?
The eigenvector of the following square matrix can be easily calculated using the steps below,
****Step 1:**** Find the eigenvalues of the matrix A, using the equation det \|(A – λI\| =0, where “I” is the identity matrix of order similar to matrix A
****Step 2:**** The value obtained in Step 2 are named as,λ 1 , λ 2 , λ 3 … . λ \_1, λ\_2, λ\_3….
****Step 3:**** Find the eigenvector (X) associated with the eigenvalue λ1 using the equation, ( A – λ 1 I ) X \= 0 (A – λ\_1I) X = 0
****Step 4:**** Repeat step 3 to find the eigenvector associated with other remaining eigenvalues λ 2 , λ 3 … . λ\_2, λ\_3….
Following these steps gives the eigenvector related to the given square matrix.
### Types of Eigenvector
The eigenvectors calculated for the square matrix are of two types which are,
- Right Eigenvector
- Left Eigenvector
****Right Eigenvector:**** The eigenvector which is multiplied by the given square matrix from the right-hand side is called the right eigenvector. It is calculated by using the following equation,
> A V R \= λ V R AV\_R = \\lambda V\_R
Where,
- ****A**** is given square matrix of order n×n,
- ****λ**** is one of the eigenvalues and
- ****V********R**** is the column vector matrix
The value of VR is, V R \= \[ v 1 v 2 v 3 . . v n \] \\bold{V\_{R} = \\begin{bmatrix} v\_{1}\\\\ v\_{2}\\\\ v\_{3}\\\\ .\\\\ .\\\\ v\_{n}\\\\ \\end{bmatrix}}
****Left Eigenvector:**** The eigenvector which is multiplied by the given square matrix from the left-hand side is called the left eigenvector. It is calculated by using the following equation,
> V L A \= V L λ V\_LA = V\_L\\lambda
Where,
- ****A**** is given square matrix of order n×n,
- ****λ**** is one of the eigenvalues and
- V
L
V\_L
is the row vector matrix.
The value of VL is,
V L \= \[ v 1 , v 2 , v 3 , . . . , v n \] V\_L = \[v\_1, v\_2, v\_3,..., v\_n\]
## Eigenvectors of a Square Matrix
We can easily find the eigenvector of square matrices of order n × n. Now, let's find the following square matrices:
- Eigenvectors of a 2 × 2 matrix
- Eigenvectors of a 3 × 3 matrix.
### ****Eigenvector of a 2 × 2 matrix****
The Eigenvector of the 2 × 2 matrix can be calculated by,
- ****Find the eigenvalues**** of the matrix using the formula
d
e
t
(
A
−
λ
I
)
\=
0
det(A - \\lambda I) = 0
, where A is the matrix and ****I**** is the identity matrix.
- ****Substitute the eigenvalue(s)**** into the equation
(
A
−
λ
I
)
v
\=
0
(A - \\lambda I)v = 0
to find the corresponding eigenvector(s).
An example of the same is,
****Example:**** Find the eigenvalues and the eigenvector for the matrix A =\[ 1 2 5 4 \] \\begin{bmatrix} 1 & 2\\\\ 5& 4 \\end{bmatrix}
****Solution:****
If eigenvalues are represented using λ and the eigenvector is represented as v = \[ a b \] \\begin{bmatrix} a\\\\b \\end{bmatrix}
Then the eigenvector is calculated by using the equation,
∣ A − λ I ∣ \= 0 \|A- λI\| = 0
> \[ 1 2 5 4 \] − λ \[ 1 0 0 1 \] \= \[ 0 0 0 0 \] \\begin{bmatrix}1 & 2\\\\ 5& 4\\end{bmatrix} -λ\\begin{bmatrix}1 & 0\\\\ 0 & 1\\end{bmatrix} = \\begin{bmatrix}0 & 0\\\\ 0& 0\\end{bmatrix}
>
> \[ 1 − λ 2 5 4 − λ \] \\begin{bmatrix} 1 - λ& 2\\\\ 5& 4 - λ \\end{bmatrix} = 0
>
> (1-λ) ⨉ (4-λ) - (2 ⨉ 5) = 0
> ⇒ 4 - λ - 4λ + λ2 - 10 = 0
> ⇒ λ2 - 5λ - 6 = 0
> ⇒ λ2 - (6λ - λ) - 6 = 0
> ⇒ λ2 - 6λ + λ - 6 = 0
> ⇒ (λ-6)(λ+1) = 0
> λ = 6 and λ = -1
Thus, the eigenvalues are 6 and -1.
- ****For λ = 6****
> (A-λI)v = 0
>
> ⇒ \[ 1 − 6 2 5 4 − 6 \] . \[ a b \] \\begin{bmatrix}1 - 6& 2\\\\ 5& 4 - 6\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} = 0
>
> ⇒ \[ − 5 2 5 − 2 \] . \[ a b \] \\begin{bmatrix}-5& 2\\\\ 5& -2\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} = 0
> ⇒ 5a - 2b = 0
>
> Simplifying the above equation we get, 5a=2b
>
> The required eigenvector is,\[ a b \] \= \[ 2 5 \] \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix}2\\\\5\\end{bmatrix}
- ****For λ = -1****
> ⇒ \[ 1 − ( − 1 ) 2 5 4 − ( − 1 ) \] . \[ a b \] \\begin{bmatrix}1 - (-1)& 2\\\\ 5& 4 - (-1)\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} = 0
>
> ⇒ \[ 2 2 5 5 \] . \[ a b \] \\begin{bmatrix}2& 2\\\\ 5& 5\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} = 0
> ⇒ 2a + 2b = 0, ⇒ 5a + 5b = 0
>
> simplifying the above equation we get, a = -b
>
> The required eigenvector is,\[ a b \] \= \[ 1 − 1 \] \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix} 1\\\\-1\\end{bmatrix}
Then the eigenvectors of the given 2 × 2 matrix are \[ a b \] \= \[ 2 5 \] , \[ a b \] \= \[ 1 − 1 \] \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix}2\\\\5\\end{bmatrix}, \\begin{bmatrix}a\\\\b\\end{bmatrix} = \\begin{bmatrix}1\\\\-1\\end{bmatrix}
These are two possible eigen vectors but many of the corresponding multiples of these eigen vectors can also be considered as other possible eigen vectors.
### Eigenvector of a 3 × 3 Matrix
An example of eigenvector of 3x3 matrix is,
****Example:**** Find the eigenvalues and the eigenvector for the matrix A =\[ 2 2 2 2 2 2 2 2 2 \] \\begin{bmatrix} 2 & 2 & 2 \\\\ 2 & 2 & 2\\\\2 & 2 & 2 \\end{bmatrix}
****Solution:****
If eigenvalues are represented using λ and the eigenvector is represented as v = \[ a b c \] \\begin{bmatrix} a\\\\b\\\\c \\end{bmatrix}
Then the eigenvector is calculated by using the equation,
> \|A- λI\| = 0
> \[ 2 2 2 2 2 2 2 2 2 \] − λ \[ 1 0 0 0 1 0 0 0 1 \] \= \[ 0 0 0 0 0 0 0 0 0 \] \\begin{bmatrix}2 & 2 & 2\\\\ 2 & 2 & 2\\\\ 2 & 2 & 2\\end{bmatrix} -λ\\begin{bmatrix}1 & 0 & 0\\\\ 0 & 1 & 0\\\\0 & 0 & 1\\end{bmatrix} = \\begin{bmatrix}0 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0\\end{bmatrix}
>
> \[ 2 − λ 2 2 2 2 − λ 2 2 2 2 − λ \] \\begin{bmatrix} 2 - λ & 2 & 2 \\\\ 2 & 2 - λ & 2 \\\\ 2 & 2 & 2- λ\\end{bmatrix} = 0
>
> Simplifying the above determinant we get
> ⇒ (2-λ)(λ2) + 2λ2 + 2λ2 = 0 ⇒ (-λ3) + 6λ2 = 0
> ⇒ λ2(6 - λ) = 0
> ⇒ λ = 0, λ = 6, λ = 0
****For λ = 0****
> (A - λI) v = 0
> ⇒ \[ 2 − 0 2 2 2 2 − 0 2 2 2 2 − 0 \] . \[ a b c \] \\begin{bmatrix}2 - 0& 2& 2\\\\ 2& 2 - 0&2\\\\2 & 2 & 2-0\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = 0
>
> ⇒ \[ 2 2 2 2 2 2 2 2 2 \] . \[ a b c \] \\begin{bmatrix}2& 2& 2\\\\ 2& 2 &2\\\\2 & 2 & 2\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = 0
>
> Simplifying the above equation we get
> 2a + 2b + 2c = 0
> ⇒ a + b + c = 0
>
> Let b = k1 and c = k2
> ⇒ a + k1 + k2 = 0
> ⇒ a = -(k1 + k2)
>
> Thus, the eigenvector is,
> \[ a b c \] \= \[ − ( k 1 \+ k 2 ) k 1 k 2 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}-(k\_{1}+k\_{2})\\\\ k\_{1}\\\\k\_{2}\\end{bmatrix}
>
> taking k1 = 1 and k2 \= 0, the eigenvector is,
> \[ a b c \] \= \[ − 1 1 0 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}-1\\\\ 1\\\\0\\end{bmatrix}
>
> taking k1 = 0 and k2\= 1, the eigenvector is,
> \[ a b c \] \= \[ − 1 0 1 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}-1\\\\ 0\\\\1\\end{bmatrix}
****For λ = 6****
> (A - λI) v = 0
> ⇒ \[ 2 − 6 2 2 2 2 − 6 2 2 2 2 − 6 \] . \[ a b c \] \\begin{bmatrix}2 - 6& 2& 2\\\\ 2& 2 -6&2\\\\2 & 2 & 2-6\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = 0
>
> ⇒ \[ − 4 2 2 2 − 4 2 2 2 − 4 \] . \[ a b c \] \\begin{bmatrix}-4& 2& 2\\\\ 2& -4 &2\\\\2 & 2 & -4\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = 0
>
> Simplifying the above equation we get,
> \-4a +2b +2c = 0 ⇒ 2 (-2a + b + c) = 0
> ⇒ 2a = b + c
>
> Let b = k1 and c = k2 and taking k1 = k2 = 1 we get,
> \[ a b c \] \= \[ 1 1 1 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix}
>
> Thus, the eigenvector is,
> \[ a b c \] \= \[ 1 1 1 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix}
## Eigenspace
We define the eigenspace of a matrix as the set of all the eigenvectors of the matrix. All the vectors in the eigenspace are linearly independent of each other.
To find the Eigenspace of the matrix we have to follow the following steps
- ****Step 1:**** Find all the eigenvalues of the given square matrix.
- ****Step 2:**** For each eigenvalue find the corresponding eigenvector.
- ****Step 3:**** Take the set of all the eigenvectors (say A). The resultant set so formed is called the Eigenspace of the following vector.
From the above example of given 3 × 3 matrix A, the eigenspace so formed is \[ − 1 1 0 \] , \[ − 1 0 1 \] , \[ 1 1 1 \] \\begin{bmatrix}-1\\\\ 1\\\\0\\end{bmatrix},\\begin{bmatrix}-1\\\\ 0\\\\1\\end{bmatrix},\\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix}
## Diagonalize Matrix Using Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors are used to find diagonal matrices. A [diagonal matrix](https://www.geeksforgeeks.org/maths/diagonal-matrix/) is a matrix which can be written as,
> ****A = XDX********\-1****
Where,
- A is the original matrix.
- X is the matrix formed by eigenvectors.
- D is a diagonal matrix where the diagonal elements are the eigenvalues of A.
- X\-1 is the inverse of the matrix X.
We can understand the concept of a diagonal matrix by taking the following example.
****Example:**** Diagonalize the matrix A = \[ 2 2 2 2 2 2 2 2 2 \] \\begin{bmatrix} 2 & 2 & 2 \\\\ 2 & 2 & 2\\\\2 & 2 & 2 \\end{bmatrix}
****Solution:****
> We have already solved for the eigenvalues and the eigenvectors of the A ****\=**** \[ 2 2 2 2 2 2 2 2 2 \] \\begin{bmatrix} 2 & 2 & 2 \\\\ 2 & 2 & 2\\\\2 & 2 & 2 \\end{bmatrix}
>
> The eigenvalues of the A are λ = 6, λ = 0 and λ = 0
> The eigenvectors of A are \[ 1 1 1 \] , \[ − 1 1 0 \] , \[ − 1 0 1 \] \\begin{bmatrix}1\\\\ 1\\\\1\\end{bmatrix},\\begin{bmatrix}-1\\\\ 1\\\\0\\end{bmatrix},\\begin{bmatrix}-1\\\\ 0\\\\1\\end{bmatrix}
> Thus,
> D = \[ 6 0 0 0 0 0 0 0 0 \] \\begin{bmatrix}6 & 0 & 0\\\\0 & 0 & 0\\\\0 & 0 & 0\\end{bmatrix}
>
> X = \[ 1 1 1 1 − 1 0 1 0 − 1 \] \\begin{bmatrix} 1 & 1 & 1\\\\1 & -1 & 0\\\\1 & 0 & -1\\end{bmatrix}
## Applications of Eigen Values and Eigen Vectors
Some of the common applications of eigen values are:
****Google PageRank Algorithm****
- Web pages form a graph; eigenvector corresponding to eigenvalue 1 of the link matrix gives each page’s importance score.
****Markov Processes & Steady-State Analysis****
- Used in simulations, natural language processing, recommendation systems, and stochastic modeling.
- The eigenvector for eigenvalue 1 gives long-term stable probabilities.
****Principal Component Analysis (PCA)****
- Reduces dimensionality of data, removes noise, and extracts key patterns.
- Common in image recognition, machine learning, and data compression.
****Latent Semantic Analysis (LSA) in NLP****
- Eigen-decomposition (via SVD) finds semantic relationships between words and documents.
- Improves search, document clustering, and recommendation engines.
****Spectral Graph Theory in Network Analysis****
- Eigenvalues of the graph Laplacian help detect communities, find graph partitions, and measure connectivity in social networks or computer networks.
****Computer Vision (Eigenfaces)****
- Eigenvectors from face image datasets form a “basis” for recognizing and comparing faces.
****Control System Stability in Robotics & AI****
- Eigenvalues of system matrices determine stability and convergence of autonomous systems and learning algorithms.
****Signal Processing & Data Transmission****
- Eigen-decomposition optimizes communication channels, feature extraction in audio/speech, and noise filtering.
## Solved Examples on Eigenvectors
****Example 1: Find the eigenvectors of the matrix A =**** \[ 1 1 0 0 1 1 0 0 1 \] \\begin{bmatrix}1 & 1 & 0\\\\0 & 1 & 1\\\\0 & 0 & 1\\end{bmatrix}
****Solution:****
> The eigen values of the matrix is found using,
> \|A - λI\| = 0
> \[ 1 − λ 1 0 0 1 − λ 1 0 0 1 − λ \] \\begin{bmatrix}1-λ & 1 & 0\\\\0 & 1-λ & 1\\\\0 & 0 & 1-λ\\end{bmatrix} = 0
> (1 - λ)3 = 0
>
> Thus, the eigen values are, λ = 1, 1, 1
>
> As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 1, using (A - λI)v = O
>
> \[ 1 − 1 1 0 0 1 − 1 1 0 0 1 − 1 \] . \[ a b c \] \= \[ 0 0 0 \] \\begin{bmatrix}1-1 & 1 & 0\\\\0 & 1-1 & 1\\\\0 & 0 & 1-1\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}0\\\\ 0\\\\0\\end{bmatrix}
>
> \[ 0 1 0 0 0 1 0 0 0 \] . \[ a b c \] \= \[ 0 0 0 \] \\begin{bmatrix}0 & 1 & 0\\\\0 & 0 & 1\\\\0 & 0 & 0\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix} = \\begin{bmatrix}0\\\\ 0\\\\0\\end{bmatrix}
> solving the above equation we get,
>
> - a = K
> - y = 0
> - z = 0
>
> Then the eigenvector is,
> \[ a b c \] \= \[ k 0 0 \] \= k \[ 1 0 0 \] \\begin{bmatrix}a\\\\ b\\\\c\\end{bmatrix}= \\begin{bmatrix}k\\\\ 0\\\\0\\end{bmatrix} = k\\begin{bmatrix}1\\\\ 0\\\\0\\end{bmatrix}
****Example 2: Find the eigenvectors of the matrix A =**** \[ 5 0 0 5 \] \\begin{bmatrix}5 & 0\\\\0 & 5 \\end{bmatrix}
****Solution:****
> The eigen values of the matrix is found using,
> \|A - λI\| = 0
> \[ 5 − λ 0 0 5 − λ \] \\begin{bmatrix}5-λ & 0\\\\0 & 5-λ \\end{bmatrix} = 0
> (5 - λ)2 = 0
>
> Thus, the eigen values are,
> λ = 5, 5
>
> As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 5, using
> (A - λI)v = O
> \[ 5 − 5 0 0 5 − 5 \] . \[ a b \] \= \[ 0 0 \] \\begin{bmatrix}5-5 & 0 \\\\ 0 & 5-5\\end{bmatrix}.\\begin{bmatrix}a\\\\ b\\end{bmatrix} = \\begin{bmatrix}0\\\\ 0\\end{bmatrix}
>
> Simplifying the above we get,
> a = 1, b = 0
> a = 0, b = 1
>
> Then the eigenvector is,
> \[ a b \] \= \[ 1 0 \] , \[ 0 1 \] \\begin{bmatrix}a\\\\ b\\end{bmatrix}= \\begin{bmatrix}1\\\\ 0\\end{bmatrix} , \\begin{bmatrix}0\\\\ 1\\end{bmatrix} |
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