Subjective Questions
Linear Algebra: Eigenvalues and Eigenvectors
Chapter 1: Introduction to Linear Algebra
Linear algebra is a fundamental branch of mathematics that focuses on the study of vector spaces and linear transformations. It plays a crucial role in various fields, including physics, computer science, engineering, and economics. In this chapter, we will delve into the concepts of eigenvalues and eigenvectors, which are essential tools in linear algebra.
Section 1: Eigenvalues and Eigenvectors
1.1 Definition of Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are concepts that arise when studying linear transformations. An eigenvalue is a scalar that characterizes a linear transformation\’s behavior along a specific direction, while an eigenvector is a non-zero vector that remains unchanged, except for a scalar multiplication, under the linear transformation.
1.2 Finding Eigenvalues and Eigenvectors
To find the eigenvalues and eigenvectors of a linear transformation or a matrix, we need to solve the characteristic equation. The characteristic equation is obtained by subtracting the eigenvalue from the main diagonal elements of the matrix and setting the determinant equal to zero. The solutions to the characteristic equation are the eigenvalues, and the corresponding eigenvectors can be found by solving a system of linear equations.
1.3 Properties of Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors possess several important properties. Firstly, the sum of the eigenvalues equals the trace of the matrix, while the product of the eigenvalues equals the determinant of the matrix. Additionally, eigenvectors corresponding to distinct eigenvalues are linearly independent. Moreover, if a matrix is diagonalizable, it can be represented as the product of a diagonal matrix and a matrix composed of the eigenvectors.
Section 2: Applications of Eigenvalues and Eigenvectors
2.1 Determining Stability in Dynamic Systems
Eigenvalues and eigenvectors find significant applications in the study of dynamic systems. In this context, eigenvalues help determine the stability of the system. If all eigenvalues have negative real parts, the system is stable. Conversely, if any eigenvalue has a positive real part, the system is unstable. The eigenvectors associated with stable eigenvalues provide insights into the behavior of the system.
2.2 Image Compression
Eigenvalues and eigenvectors are also used in image compression techniques, such as Principal Component Analysis (PCA). By representing an image as a matrix, we can decompose it into eigenvalues and eigenvectors. The eigenvectors with the highest eigenvalues capture the most significant features of the image, allowing for efficient compression without significant loss of quality.
2.3 Google\’s PageRank Algorithm
Eigenvalues and eigenvectors play a crucial role in Google\’s PageRank algorithm. PageRank assigns a numerical weight to each web page based on the concept of eigenvector centrality. The higher the eigenvector centrality of a web page, the higher its rank in search engine results. By employing the power iteration method, Google calculates the dominant eigenvector of a matrix that represents the relationships between web pages.
Chapter 2: Sample Questions on Eigenvalues and Eigenvectors
Question 1:
Find the eigenvalues and eigenvectors of the matrix A = [[3, 1], [2, 2]].
Solution:
To find the eigenvalues, we need to solve the characteristic equation det(A – λI) = 0, where A is the given matrix, λ is the eigenvalue, and I is the identity matrix. Therefore, we have:
|3 – λ, 1| = 0
|2, 2 – λ|
Expanding the determinant, we get:
(3 – λ)(2 – λ) – 2 = 0
6 – 5λ + λ^2 – 2 = 0
λ^2 – 5λ + 4 = 0
(λ – 4)(λ – 1) = 0
Hence, the eigenvalues are λ = 4 and λ = 1.
To find the corresponding eigenvectors, we substitute each eigenvalue into the equation (A – λI)x = 0. Let\’s solve for λ = 4 first:
|3 – 4, 1| |x1| = |0|
|2, 2 – 4| |x2| = |0|
Simplifying the system of equations, we obtain:
-1×1 + x2 = 0
2×1 – 2×2 = 0
From the first equation, we have x1 = x2. Thus, an eigenvector corresponding to λ = 4 is [1, 1].
Now, let\’s solve for λ = 1:
|3 – 1, 1| |x1| = |0|
|2, 2 – 1| |x2| = |0|
Simplifying the system of equations, we obtain:
2×1 + x2 = 0
2×1 – x2 = 0
From the first equation, we have x1 = -x2. Thus, an eigenvector corresponding to λ = 1 is [-1, 1].
Therefore, the eigenvalues of the matrix A = [[3, 1], [2, 2]] are 4 and 1, and the corresponding eigenvectors are [1, 1] and [-1, 1].
Question 2:
Consider a linear transformation T: R^3 → R^3, where T(x, y, z) = (2x + 3y + z, 4x + 2y + 6z, 3x + y + 5z). Find the eigenvalues and eigenvectors of T.
Solution:
To find the eigenvalues and eigenvectors of T, we need to determine the matrix representation of T and solve for the characteristic equation. The matrix representation of T is:
A = [[2, 3, 1], [4, 2, 6], [3, 1, 5]]
The characteristic equation is given by det(A – λI) = 0, where λ is the eigenvalue and I is the identity matrix. Therefore, we have:
|2 – λ, 3, 1| = 0
|4, 2 – λ, 6|
|3, 1, 5 – λ|
Expanding the determinant, we get:
(2 – λ)((2 – λ)(5 – λ) – 6) – 3((4)(5 – λ) – 6) + 3(4 – 12) = 0
(2 – λ)(λ^2 – 7λ + 9) – 3(5λ – 20) = 0
(2 – λ)(λ^2 – 7λ + 9) – 15λ + 60 = 0
λ^3 – 9λ^2 + 33λ – 45 – 15λ + 30 = 0
λ^3 – 9λ^2 + 18λ – 15λ^2 + 135λ – 270 = 0
λ^3 – 24λ^2 + 153λ – 270 = 0
(λ – 6)(λ – 9)(λ – 5) = 0
Hence, the eigenvalues are λ = 6, λ = 9, and λ = 5.
To find the corresponding eigenvectors, we substitute each eigenvalue into the equation (A – λI)x = 0. Let\’s solve for λ = 6 first:
|2 – 6, 3, 1| |x1| = |0|
|4, 2 – 6, 6| |x2| = |0|
|3, 1, 5 – 6| |x3| = |0|
Simplifying the system of equations, we obtain:
-4×1 + 3×2 + x3 = 0
4×1 – 4×2 + 6×3 = 0
3×1 + x2 – x3 = 0
From the third equation, we have x1 = -x2 + x3. Substituting this into the first and second equations, we get:
-4(-x2 + x3) + 3×2 + x3 = 0
4(-x2 + x3) – 4×2 + 6×3 = 0
Simplifying further, we obtain:
7×2 – 7×3 = 0
-8×2 + 10×3 = 0
From the first equation, we have x2 = x3. Thus, an eigenvector corresponding to λ = 6 is [1, 1, 1].
Now, let\’s solve for λ = 9:
|2 – 9, 3, 1| |x1| = |0|
|4, 2 – 9, 6| |x2| = |0|
|3, 1, 5 – 9| |x3| = |0|
Simplifying the system of equations, we obtain:
-7×1 + 3×2 + x3 = 0
4×1 – 7×2 + 6×3 = 0
3×1 + x2 – 4×3 = 0
From the first equation, we have x1 = (3/7)x2 – (1/7)x3. Substituting this into the second and third equations, we get:
4((3/7)x2 – (1/7)x3) – 7×2 + 6×3 = 0
3((3/7)x2 – (1/7)x3) + x2 – 4×3 = 0
Simplifying further, we obtain:
(2/7)x2 – (2/7)x3 = 0
(2/7)x2 – (2/7)x3 = 0
From the first equation, we have x2 = x3. Thus, an eigenvector corresponding to λ = 9 is [1, 1, 1].
Finally, let\’s solve for λ = 5:
|2 – 5, 3, 1| |x1| = |0|
|4, 2 – 5, 6| |x2| = |0|
|3, 1, 5 – 5| |x3| = |0|
Simplifying the system of equations, we obtain:
-3×1 + 3×2 + x3 = 0
4×1 – 3×2 + 6×3 = 0
3×1 + x2 = 0
From the third equation, we have x1 = -x2. Substituting this into the first and second equations, we get:
-3(-x2) + 3×2 + x3 = 0
4(-x2) – 3×2 + 6×3 = 0
Simplifying further, we obtain:
4×2 + x3 = 0
x2 – 6×3 = 0
From the second equation, we have x2 = 6×3. Thus, an eigenvector corresponding to λ = 5 is [1, 6, 1].
Therefore, the eigenvalues of the linear transformation T: R^3 → R^3, where T(x, y, z) = (2x + 3y + z, 4x + 2y + 6z, 3x + y + 5z), are 6, 9, and 5, and the corresponding eigenvectors are [1, 1, 1], [1, 1, 1], and [1, 6, 1].