TLDR;
This video explains eigenvalues and eigenvectors, fundamental concepts in linear algebra with applications in physics and other fields. It covers their definition, how to calculate them, and provides examples to illustrate the process. Key points include understanding the relationship between a matrix and its eigenvectors, solving the characteristic equation to find eigenvalues, and using row operations to determine eigenvectors.
- Eigenvalues and eigenvectors are useful in solving differential equations, describing vibrations, and distinguishing energy states.
- Eigenvectors remain in the same direction when a linear transformation is applied, only scaled by the eigenvalue.
- The characteristic equation, derived from the determinant of (A - λI), is crucial for finding eigenvalues.
- Row operations are used to simplify matrices and solve for eigenvectors after eigenvalues are known.
Introduction to Eigenvalues and Eigenvectors [0:00]
Eigenvalues and eigenvectors are important concepts in linear algebra with applications in math, physics (especially quantum physics). They can be used to solve systems of linear differential equations, describe natural frequencies of vibrations, separate modes of motion, and distinguish states of energy. Eigenvectors of a matrix A, when multiplied by A, result in the same vector scaled by a scalar value (lambda), which is called the eigenvalue.
Definition and Properties [0:44]
Given a square matrix A, an eigenvector x maintains its direction when multiplied by A, only scaled by a factor λ (lambda), the eigenvalue (Ax = λx). Eigenvectors must be nontrivial (not the zero vector). A matrix A can have multiple eigenvalues, up to the number of rows/columns in the matrix, each with its associated eigenvector.
Example Verification [1:45]
To verify if a vector x is an eigenvector of matrix A, multiply A by x. If the result is a scalar multiple of x, then x is an eigenvector, and the scalar is the eigenvalue. For example, given matrix A = [[-3, 1], [-2, 0]] and vector x = [1, 1], multiplying A by x yields [-2, -2], which is -2 times [1, 1]. Thus, [1, 1] is an eigenvector of A, and -2 is its eigenvalue.
Solving for Eigenvalues [2:51]
To solve for eigenvalues, start with the equation Ax = λx, subtract λx from both sides, and insert the identity matrix I to get Ax - λIx = 0. Factor out x to get (A - λI)x = 0. For nontrivial solutions (x ≠ 0), the matrix (A - λI) must not be invertible, meaning its determinant must be zero. The determinant of (A - λI) forms a polynomial in λ, called the characteristic polynomial, and setting it to zero gives the characteristic equation. Solving this equation yields the eigenvalues.
Concrete Example: Finding Eigenvalues [4:42]
To find the eigenvalues for the matrix A = [[1, 1], [4, 1]], subtract λI from A, resulting in the matrix [[1-λ, 1], [4, 1-λ]]. Calculate the determinant of this matrix: (1-λ)(1-λ) - (4*1). Simplify to get the characteristic equation λ² - 2λ - 3 = 0. Factor this equation into (λ - 3)(λ + 1) = 0. The solutions, λ = 3 and λ = -1, are the eigenvalues of matrix A.
Solving for Eigenvectors [6:31]
To solve for eigenvectors, plug each eigenvalue into the equation (A - λI)x = 0 and solve for x. Use row operations to get the matrix into row echelon form, which simplifies the system of equations. The solutions represent the form of the eigenvectors, and any scalar multiple of these solutions will also be an eigenvector.
Scalar Multiples of Eigenvectors [7:44]
If x is an eigenvector of A (Ax = λx), then any scalar multiple of x, denoted as cx, is also an eigenvector. This is because A(cx) = c(Ax) = c(λx) = λ(cx), satisfying the definition of an eigenvector.
Concrete Example: Finding Eigenvectors [8:25]
Using the matrix A = [[1, 1], [4, 1]] and its eigenvalues λ = 3 and λ = -1, solve for the corresponding eigenvectors. For λ = 3, A - λI = [[-2, 1], [4, -2]]. Apply row operations to get [[-2, 1], [0, 0]]. This gives the equation -2x1 + x2 = 0, so x2 = 2x1. Choosing x1 = 1, we get x2 = 2, resulting in the eigenvector [1, 2]. For λ = -1, A - λI = [[2, 1], [4, 2]]. Apply row operations to get [[2, 1], [0, 0]]. This gives the equation 2x1 + x2 = 0, so x2 = -2x1. Choosing x1 = 1, we get x2 = -2, resulting in the eigenvector [1, -2].
Example with a 3x3 Matrix: Finding Eigenvalues and Eigenvectors [11:30]
Given the matrix A = [[1, 0, 0], [3, -2, 0], [2, 3, 4]], find the eigenvalues and eigenvectors. First, find A - λI by subtracting λ from the main diagonal. The determinant of A - λI is (1 - λ)((-2 - λ)(4 - λ) - (0 * 3)) = (1 - λ)(-2 - λ)(4 - λ). Setting this equal to zero gives the eigenvalues λ = 1, λ = -2, and λ = 4.
Finding Eigenvectors for λ = 1 [12:54]
For λ = 1, A - λI = [[0, 0, 0], [3, -3, 0], [2, 3, 3]]. From the second row, 3x1 - 3x2 = 0, so x1 = x2. Let x2 = 1, then x1 = 1. From the third row, 2x1 + 3x2 + 3x3 = 0, so 2 + 3 + 3x3 = 0, giving x3 = -5/3. The eigenvector is any multiple of [1, 1, -5/3].
Finding Eigenvectors for λ = -2 [14:08]
For λ = -2, A - λI = [[3, 0, 0], [3, 0, 0], [2, 3, 6]]. From the first row, 3x1 = 0, so x1 = 0. The third row gives 2x1 + 3x2 + 6x3 = 0, which simplifies to 3x2 + 6x3 = 0, so x2 = -2x3. Let x3 = 1, then x2 = -2. The eigenvector is any multiple of [0, -2, 1].
Finding Eigenvectors for λ = 4 [15:04]
For λ = 4, A - λI = [[-3, 0, 0], [3, -6, 0], [2, 3, 0]]. From the first row, -3x1 = 0, so x1 = 0. The second row gives 3x1 - 6x2 = 0, which simplifies to -6x2 = 0, so x2 = 0. The third row gives no new information, so x3 is a free variable. Let x3 = 1. The eigenvector is any multiple of [0, 0, 1].