Lesson 1: Change of Basis and Similar Matrices
Estimated time: 40-50 minutes
Learning Objectives
- Understand coordinates relative to different bases
- Compute the change-of-basis (transition) matrix
- Define similar matrices and explain their shared properties
- Connect change of basis to diagonalization
Coordinates Relative to a Basis
A basis B = {b1, b2, ..., bn} provides a coordinate system for R^n. Every vector x can be uniquely written as x = c1*b1 + c2*b2 + ... + cn*bn, and the coordinate vector is [x]_B = (c1, c2, ..., cn).
Coordinate Vector: If B = {b1, ..., bn} is a basis for R^n and x = c1*b1 + ... + cn*bn, then [x]_B = (c1, ..., cn) is the B-coordinate vector of x.
Worked Example 1
B = {(1, 1), (1, -1)}. Express x = (5, 3) in B-coordinates.
Solve c1(1,1) + c2(1,-1) = (5, 3). System: c1 + c2 = 5 and c1 - c2 = 3. Adding: 2c1 = 8, c1 = 4. c2 = 1.
[x]_B = (4, 1). Check: 4(1,1) + 1(1,-1) = (4,4) + (1,-1) = (5,3).
The Change-of-Basis Matrix
Change-of-Basis Matrix: The matrix P_B = [b1 | b2 | ... | bn] converts B-coordinates to standard coordinates: x = P_B * [x]_B.
To go the other direction: [x]_B = P_B^{-1} * x.
Worked Example 2
B = {(1, 1), (1, -1)}. P_B = [1 1; 1 -1]. P_B^{-1} = (1/2)[1 1; 1 -1] = [1/2 1/2; 1/2 -1/2].
For x = (5, 3): [x]_B = P_B^{-1} * (5,3)^T = (4, 1)^T. Matches Example 1.
Transition between two bases: To convert from B-coordinates to C-coordinates:
[x]_C = P_C^{-1} * P_B * [x]_B. The matrix P_C^{-1} * P_B is the transition matrix from B to C.
Worked Example 3
B = {(1,0), (0,1)}, C = {(1,1), (1,-1)}. Find the transition matrix from B to C.
P_B = I (standard basis). P_C = [1 1; 1 -1]. P_C^{-1} = [1/2 1/2; 1/2 -1/2].
Transition: P_C^{-1} * P_B = P_C^{-1} = [1/2 1/2; 1/2 -1/2].
For x = (3, 1): [x]_C = [1/2 1/2; 1/2 -1/2](3,1)^T = (2, 1)^T. Check: 2(1,1) + 1(1,-1) = (3,1).
Similar Matrices
Similar Matrices: A and B are similar if there exists an invertible P such that B = P^{-1}AP. They represent the same linear transformation in different bases.
Similar Matrices Share:
- The same eigenvalues (with the same multiplicities)
- The same determinant
- The same trace
- The same rank
- The same characteristic polynomial
Worked Example 4
A = [4 -2; 1 1]. Find a diagonal matrix similar to A.
Eigenvalues: lambda^2 - 5*lambda + 6 = (lambda-2)(lambda-3). lambda = 2, 3.
lambda = 2: v1 = (1, 1). lambda = 3: v2 = (2, 1). P = [1 2; 1 1].
D = P^{-1}AP = [3 0; 0 2] (eigenvalues on diagonal). A and D are similar.
Check shared properties: det(A) = 4+2 = 6 = 3*2 = det(D). trace(A) = 5 = 3+2 = trace(D).
Change of Basis for Linear Transformations
If T has standard matrix A, then the matrix of T relative to basis B is:
Key Insight: Diagonalization IS change of basis. When A = PDP^{-1}, the matrix D = P^{-1}AP is the matrix of the same transformation in the eigenvector basis. In the eigenvector basis, the transformation is just scaling along each axis.
Worked Example 5
T(x,y) = (4x-2y, x+y). Standard matrix A = [4 -2; 1 1]. In the eigenvector basis B = {(1,1), (2,1)}:
[T]_B = P^{-1}AP = [2 0; 0 3]. In this basis, T just scales the first coordinate by 2 and the second by 3.
Why Change Basis?
The right basis can make a complicated transformation look simple:
- Diagonalization: eigenvector basis turns A into a diagonal matrix
- Principal axes: basis of eigenvectors of a symmetric matrix orthogonalizes a quadratic form
- Fourier basis: decomposes signals into frequency components
- SVD: finds the optimal bases for both domain and range
Check Your Understanding
1. If B = {(2, 1), (1, 1)} and [x]_B = (3, -1), find x in standard coordinates.
2. If A and B are similar, must they have the same eigenvalues?
3. How is diagonalization related to change of basis?
Key Takeaways
- Coordinate vector [x]_B expresses x relative to basis B
- Change-of-basis matrix P_B converts B-coords to standard: x = P_B [x]_B
- Similar matrices B = P^{-1}AP share eigenvalues, det, trace, rank
- Diagonalization is change of basis to the eigenvector basis
- Choosing the right basis can dramatically simplify a problem