Lesson 1: Eigenvalues and Eigenvectors -- Introduction
Estimated time: 40-50 minutes
Learning Objectives
- Define eigenvalues and eigenvectors of a square matrix
- Understand the geometric meaning of Av = lambda v
- Find eigenvalues and eigenvectors of 2x2 matrices
- Recognize the connection between eigenvalues and the characteristic equation
The Big Idea: Vectors That Only Get Scaled
When you multiply a matrix A by most vectors, the result points in a completely new direction. But some special vectors only get stretched or compressed -- their direction stays the same (or reverses). These are the eigenvectors.
Eigenvalue and Eigenvector: Let A be an n x n matrix. A nonzero vector v is an eigenvector of A if Av = lambda v for some scalar lambda. The scalar lambda is the corresponding eigenvalue.
Important: v must be nonzero. The zero vector trivially satisfies A*0 = lambda*0 for any lambda, so we exclude it.
Geometric Interpretation
If v is an eigenvector with eigenvalue lambda:
- lambda > 1: v is stretched (scaled up)
- 0 < lambda < 1: v is compressed (scaled down)
- lambda = 1: v is unchanged (fixed direction and magnitude)
- lambda = 0: v is sent to the zero vector (collapsed)
- lambda < 0: v is reversed and scaled by |lambda|
Geometric Example
Consider the matrix A = [2 0; 0 3]. This scales the x-component by 2 and the y-component by 3.
A * (1, 0)^T = (2, 0)^T = 2 * (1, 0)^T. So (1, 0) is an eigenvector with eigenvalue 2.
A * (0, 1)^T = (0, 3)^T = 3 * (0, 1)^T. So (0, 1) is an eigenvector with eigenvalue 3.
A * (1, 1)^T = (2, 3)^T which is NOT a scalar multiple of (1, 1). So (1, 1) is NOT an eigenvector.
Finding Eigenvalues: The Key Equation
Starting from Av = lambda v, we rearrange:
For a nonzero v to exist, the matrix (A - lambda I) must be singular:
Characteristic Equation: det(A - lambda I) = 0. The solutions lambda are the eigenvalues of A.
Worked Example: Finding Eigenvalues of a 2x2 Matrix
Find the eigenvalues of A = [4 1; 2 3].
A - lambda I = [4-lambda, 1; 2, 3-lambda].
det(A - lambda I) = (4-lambda)(3-lambda) - (1)(2) = lambda^2 - 7*lambda + 10 = 0.
Factor: (lambda - 5)(lambda - 2) = 0. Eigenvalues: lambda = 5 and lambda = 2.
Finding Eigenvectors
Once you have an eigenvalue lambda, find the eigenvectors by solving (A - lambda I)v = 0.
Worked Example: Complete Eigen-Analysis of a 2x2
For A = [4 1; 2 3] with eigenvalues lambda = 5 and lambda = 2.
lambda = 5: (A - 5I)v = 0. A - 5I = [-1 1; 2 -2]. Row reduce: [1 -1; 0 0]. x1 = x2, so v = t(1, 1). Eigenvector: (1, 1).
lambda = 2: (A - 2I)v = 0. A - 2I = [2 1; 2 1]. Row reduce: [1 1/2; 0 0]. x1 = -x2/2, so v = t(-1, 2). Eigenvector: (-1, 2).
Check: A*(1,1)^T = (5,5)^T = 5*(1,1)^T. A*(-1,2)^T = (-2,4)^T = 2*(-1,2)^T.
Special Cases and Observations
Diagonal Matrices: The eigenvalues of a diagonal matrix are its diagonal entries. The eigenvectors are the standard basis vectors e1, e2, ..., en.
Triangular Matrices: The eigenvalues of a triangular matrix (upper or lower) are also the diagonal entries.
Example: Triangular Matrix
A = [3 1 4; 0 -2 5; 0 0 7]. Eigenvalues are the diagonal entries: lambda = 3, -2, 7.
Key Properties
For an n x n matrix A with eigenvalues lambda_1, ..., lambda_n (counted with multiplicity):
- det(A) = lambda_1 * lambda_2 * ... * lambda_n (product of eigenvalues)
- trace(A) = lambda_1 + lambda_2 + ... + lambda_n (sum of eigenvalues = sum of diagonal entries)
Eigenvalue Zero and Invertibility
An eigenvalue of zero has special significance: if Av = 0v = 0, then v is in the null space of A.
Key Connection: lambda = 0 is an eigenvalue of A if and only if A is not invertible (singular). This is equivalent to det(A) = 0.
Check Your Understanding
1. If Av = 3v for a nonzero vector v, what are the eigenvalue and eigenvector?
2. Find the eigenvalues of A = [5 0; 0 -2].
3. Find the eigenvalues of A = [1 4; 2 3].
4. If A is 3x3 with eigenvalues 1, 2, 3, what is det(A)?
Key Takeaways
- Av = lambda v -- eigenvectors are directions preserved by the transformation
- Find eigenvalues by solving det(A - lambda I) = 0
- Find eigenvectors by solving (A - lambda I)v = 0
- Diagonal/triangular matrices: eigenvalues are the diagonal entries
- det(A) = product of eigenvalues; trace(A) = sum of eigenvalues
- lambda = 0 is an eigenvalue if and only if A is singular