Vector Decomposition via Algebraic Elimination: Solve Linear Combinations

Solving Vector Decomposition with Algebraic Elimination

Vector decomposition determines the coefficients needed to express a target vector as a linear combination of basis vectors. This answers two critical questions:

1. Existence: Is the target vector within the span of the basis?
2. Composition: What specific coefficients create this linear combination?

After analyzing this tutorial, I've observed that learners often grasp the geometric intuition but struggle with systematic computation. We'll bridge that gap using algebraic elimination—a foundational technique applicable from 2D to n-dimensional spaces.

Core Methodology: System of Linear Equations

Given basis vectors u and v, and target vector w = xu + yv:

1. Expand into component equations:
   • w₁ = x·u₁ + y·v₁
   • w₂ = x·u₂ + y·v₂
2. Solve the system using elimination:
   • Align coefficients: Multiply equations to match terms
   • Eliminate variables: Subtract equations to isolate one unknown
   • Back-substitute: Solve for remaining variables

Key Insight: This process directly tests if w ∈ span(u, v). If no solution exists, w lies outside the span.

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Step-by-Step Case Studies

Case 1: 2D Vectors (Successful Decomposition)

Given:
u = [1, 2], v = [3, 1], w = [9, 8]
Equations:

1. 9 = 1x + 3y
2. 8 = 2x + 1y

Solution:

1. Multiply equation (2) by 3:
   24 = 6x + 3y
2. Subtract equation (1) from this:
   (24 - 9) = (6x - x) + (3y - 3y) → 15 = 5x → x=3
3. Substitute x=3 into equation (1):
   9 = 1(3) + 3y → y=2

Verification: 3×[1,2] + 2×[3,1] = [3+6, 6+2] = [9,8] ✓

Case 2: Collinear Vectors (No Solution)

Given:
c = [1, 2], d = [2, 4], q = [3, 5]
Equations:

1. 3 = 1x + 2y
2. 5 = 2x + 4y

Attempt:
Multiply equation (1) by 2: 6 = 2x + 4y
Subtract equation (2): (6-5) = (2x-2x) + (4y-4y) → 1 = 0 (Contradiction)

Conclusion: q ∉ span(c, d) since collinear bases only span their common line.

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3D Vector Decomposition

Given:
a=[2,0,1], b=[1,2,3], c=[0,1,2], w=[5,7,11]
System:

1. 5 = 2x + 1y + 0z
2. 7 = 0x + 2y + 1z
3. 11 = 1x + 3y + 2z

Solution Strategy:

1. Eliminate z from equations (2) and (3):
   • Multiply (2) by 2: 14 = 4y + 2z
   • Subtract (3): (14-11) = 4y - 3y + 2z - 2z → 3 = y
2. Substitute y=3 into (1) and (2):
   • (1): 5 = 2x + 3 → x=1
   • (2): 7 = 2(3) + z → z=1
3. Verify with (3): 1(1) + 3(3) + 2(1) = 1+9+2=12 ≠11 → Mistake!

Correction:

• Substitute y=3 into (3): 11 = x + 3(3) + 2z → x + 2z = 2
• From (1): 2x + 3 = 5 → x=1
• Then 1 + 2z = 2 → z=0.5
  Final: w = 1a + 3b + 0.5c

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Real-World Application: Resource Allocation

Beatrix's Dilemma (Budget & Nutrition Constraints):

• Money: £30 = 2*(cake slices) + 3*(lemonade liters)
• Calories: 3000 = 100*(cake) + 500*(lemonade)

Solution:

1. Multiply money equation by 50: 1500 = 100x + 150y
2. Subtract calorie equation:
   (1500 - 3000) = (100x - 100x) + (150y - 500y)
   -1500 = -350y → y ≈ 4.29
3. Substitute into money equation:
   30 = 2x + 3(4.29) → x ≈ 8.56

Engineering Insight: As an aerospace engineer, Beatrix would later use this same method to optimize material combinations, balancing weight, cost, and strength parameters.

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Critical Pitfalls and Pro Tips

1. Collinear Vectors: Always check if basis vectors are parallel before solving.
2. Inconsistent Systems: If elimination yields contradictions (e.g., 1=0), the target isn't in the span.
3. Scaling Sensitivity: In 3D+, solve for one variable at a time and verify all equations.

Actionable Checklist:

1. Write explicit component equations
2. Multiply equations to align coefficients
3. Eliminate variables via subtraction
4. Back-substitute systematically
5. Validate with vector arithmetic

Recommended Tools:

• Beginners: Desmos (visualizes 2D/3D spans)
• Advanced Users: Python NumPy (linalg.solve for systems)

"Which step in elimination do you find most error-prone? Share your challenges in comments!"

Key Takeaway: Vector decomposition via algebraic elimination isn't just academic—it's a foundational tool for optimizing real-world constraints.