Linear Algebra-Linear Transformations

Question

Find the associated matrix (i.e. matrix with respect to the standard basis ) for the linear transformation on R2 representing reflection about the line 3y=4x followed by rotation by 3π ⁄2 radians counterclockwise.

You may use that ref_L(x)=2proj_L(x)-X.

Answer 1

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Answer 2

Hi there,

Here is a step-by-step breakdown of how to solve this:

Solution: Composite Linear Transformation in ℝ²

Given:

1. Reflection about the line 3y = 4x
2. Rotation by 3π/2 radians counterclockwise

Step 1: Find Reflection Matrix (using 𝑟𝑒𝑓ₗ(𝑥) = 2𝑝𝑟𝑜𝑗ₗ(𝑥) − 𝑥)

1. Normalize the line direction vector:
2. Line direction: 𝒗 = (4,3)
3. Unit vector: 𝒖 = (4/5, 3/5)
4. Projection matrix onto 𝒖:
5. P=𝒖𝒖T= (16/2512/2512/259/25) P=uuT= (16/2512/25​12/259/25​)
6. Reflection matrix:
7. R=2P−I= (7/2524/2524/25−7/25) R=2P−I= (7/2524/25​24/25−7/25​)

Step 2: Find Rotation Matrix (3π/2 CCW)

Q= (01−10) Q= (0−1​10​)

Step 3: Compute Composite Transformation

Matrix multiplication (Q∘R):

A=QR= (24/25−7/25−7/25−24/25) A=QR= (24/25−7/25​−7/25−24/25​)

Verification:

Testing 𝒆₁ = (1,0):

1. Reflection: (7/25, 24/25)
2. Rotation: (24/25, -7/25) ✓ matches first column of A

Why Students Struggle with This Problem:

1. Projection Confusion: Misunderstanding 𝑝𝑟𝑜𝑗ₗ(𝑥) = (𝒖⋅𝑥) 𝒖
2. Composition Order: Applying rotation before reflection
3. Sign Errors: Especially in rotation matrix entries

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1. Visual Demonstrations: Animated geometric transformations
2. Error Prevention Checklists: My signature "3-Point Verification"
3. Application Focus: Computer graphics/pathology examples

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Answer 3

The line L is given by 3y=4x, so a direction vector v = [3 4] (as a vertical vector), with its length ||v|| = 5. So the unit vector along L is u = 1/5 * [3 4] = [3/5 4/5] (again as a vertical vector).

For any x ∈ R², the orthogonal projection of x onto L is projL(x)=(u^Tx) u, which in matrix form is projL​(x)=(uu^T)x. So M_proj​=uu^T = [[9/25 12/25], [12/25 16/25]] (each inner bracket is a row).

Now we use ref_L​(x)=2proj_L​(x)−x. Hence the reflection matrix is M_ref​=2M_proj​−I = 2 * [[9/25 12/25], [12/25 16/25]] - [[1 0], [0 1]] = [[-7/25 24/25], [24/25 7/25]].

A counterclockwise rotation by 3π/2 is R_3π/2 = [[cos(3π/2) −sin(3π/2)], [sin(3π/2) cos(3π/2)]] = [[0 1], [-1 0]].

Applying the reflection first, then rotation we have M=R_3π/2​ * M_ref.

M = [[0 1], [-1 0]] * [[-7/25 24/25], [24/25 7/25]] = [[24/25 7/25], [7/25 -24/25]].