Eigenvalues and eigenvectors are fundamental concepts in linear algebra, used in various applications such as matrix diagonalization, stability analysis and data analysis (e.g., PCA). They are associated with a square matrix and provide insights into its properties.
Eigen value and Eigen vector1. 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,
Av = \lambda v
Where,
- A is the matrix,
- v is associated eigenvector and
- \lambda is scalar eigenvalue.
2. 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 vA = λv, where v is a row matrix of size 1 × n.
2.1 Eigenvector Equation
The Eigenvector equation is the equation that is used to find the eigenvector of any square matrix. The eigenvector equation is
Av = λv
Where,
- A is the given square matrix,
- v is the eigenvector of matrix A and
- \lambda is any scaler multiple.
2.2 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….
Step 3: Find the eigenvector (X) associated with the eigenvalue λ1 using the equation, (A – λ_1I) X = 0
Step 4: Repeat step 3 to find the eigenvector associated with other remaining eigenvalues λ_2, λ_3….
Following these steps gives the eigenvector related to the given square matrix.
2.3 Types of Eigenvector
The eigenvectors calculated for the square matrix are of two types which are,
- Right Eigenvector
- Left Eigenvector
2.3.1 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,
AV_R = \lambda V_R
Where,
- A is given square matrix of order n×n,
- λ is one of the eigenvalues and
- VR is the column vector matrix
The value of VR is, \bold{V_{R} = \begin{bmatrix} v_{1}\\ v_{2}\\ v_{3}\\ .\\ .\\ v_{n}\\ \end{bmatrix}}
2.3.2 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_LA = V_L\lambda
Where,
- A is given square matrix of order n×n,
- λ is one of the eigenvalues and
- V_L is the row vector matrix.
The value of VL is,
V_L = [v_1, v_2, v_3,..., v_n]
3. 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.
3.1 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 det(A - \lambda I) = 0 , where A is the matrix and I is the identity matrix.
- Substitute the eigenvalue(s) into the equation (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 = \begin{bmatrix} 1 & 2\\ 5& 4 \end{bmatrix}
Solution:
If eigenvalues are represented using λ and the eigenvector is represented as v = \begin{bmatrix} a\\b \end{bmatrix}
Then the eigenvector is calculated by using the equation,
|A- λI| = 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}
\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 -6λ + λ - 6 = 0
⇒ (λ-6)(λ+1) = 0
λ = 6 and λ = -1
Thus, the eigenvalues are 6 and -1.
(A-λI)v = 0
⇒ \begin{bmatrix}1 - 6& 2\\ 5& 4 - 6\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix} = 0
⇒ \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, \begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}2\\5\end{bmatrix}
⇒ \begin{bmatrix}1 - (-1)& 2\\ 5& 4 - (-1)\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix} = 0
⇒ \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, \begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix} 1\\-1\end{bmatrix}
Then the eigenvectors of the given 2 × 2 matrix are \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.
3.2 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 = \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 = \begin{bmatrix} a\\b\\c \end{bmatrix}
Then the eigenvector is calculated by using the equation,
|A- λI| = 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}
\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
⇒ \begin{bmatrix}2 - 0& 2& 2\\ 2& 2 - 0&2\\2 & 2 & 2-0\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix} = 0
⇒ \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,
\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,
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}-1\\ 1\\0\end{bmatrix}
taking k1 = 0 and k2= 1, the eigenvector is,
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}-1\\ 0\\1\end{bmatrix}
For λ = 6
(A - λI) v = 0
⇒ \begin{bmatrix}2 - 6& 2& 2\\ 2& 2 -6&2\\2 & 2 & 2-6\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix} = 0
⇒ \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,
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}1\\ 1\\1\end{bmatrix}
Thus, the eigenvector is,
\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}1\\ 1\\1\end{bmatrix}
4. 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 \begin{bmatrix}-1\\ 1\\0\end{bmatrix},\begin{bmatrix}-1\\ 0\\1\end{bmatrix},\begin{bmatrix}1\\ 1\\1\end{bmatrix}
5. 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 = \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 = \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 \begin{bmatrix}1\\ 1\\1\end{bmatrix},\begin{bmatrix}-1\\ 1\\0\end{bmatrix},\begin{bmatrix}-1\\ 0\\1\end{bmatrix}
Thus,
D = \begin{bmatrix}6 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}
X = \begin{bmatrix} 1 & 1 & 1\\1 & -1 & 0\\1 & 0 & -1\end{bmatrix}
6. Applications of Eigen Values in Engineering
Some of the common applications of eigen values are:
1. Linear Algebra:
- Diagonalization: Eigenvalues help diagonalize matrices, simplifying computations and enabling more efficient solutions to linear systems.
- Matrix Exponentiation: Eigenvalues are essential for matrix exponentiation, particularly in solving differential equations.
2. Quantum Mechanics: Eigenvalues of the Hamiltonian operator define the energy levels of quantum systems, giving insight into possible states.
3. Vibrations and Structural Analysis:
- Mechanical Vibrations: Eigenvalues represent the natural frequencies of vibrating systems, critical for analyzing resonances and stability.
- Structural Stability: In structural analysis, eigenvalues help assess the stability and behavior of structures under various forces.
4. Statistics: Eigenvalues play a role in analyzing covariance matrices, offering insight into the data's spread and orientation.
5. Computer Graphics: Eigenvalues are used in PCA to identify the principal components of a dataset, reducing dimensionality while preserving key information.
6. Control Systems: Eigenvalues of the system matrix are crucial for evaluating the stability of control systems, ensuring that the system's response remains bounded.
7. Solved Examples on Eigenvectors
Example 1: Find the eigenvectors of the matrix A = \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
\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
\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}
\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,
Then the eigenvector is,
\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 = \begin{bmatrix}5 & 0\\0 & 5 \end{bmatrix}
Solution:
The eigen values of the matrix is found using,
|A - λI| = 0
\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
\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,
\begin{bmatrix}a\\ b\end{bmatrix}= \begin{bmatrix}1\\ 0\end{bmatrix} , \begin{bmatrix}0\\ 1\end{bmatrix}
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