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Saturday, June 13, 2026

MA25C02| Linear algebra|Vector sapce | Linear Dependence Explained: Why v₁ = (−1,2,4) and v₂ = (5,−10,−20) Are Dependent in ℝ³"

Linear Dependence of Vectors in R³ | Engineering Mathematica
📐 Linear Algebra A/M-24 · R-21 ✔ Solved

Why are \(v_1\) & \(v_2\)
Linearly Dependent
in \(\mathbb{R}^3\)?

A complete walkthrough — definition, step-by-step proof, geometric interpretation, and key takeaways for the exam.

📖 Topic: Vector Spaces 4 min read 📌 Unit I — Vectors in ℝn

Tutorial Video: Linear Independence and Linear Dependence
01 · Background

What is Linear Dependence?

Before solving, recall the core definition. This is what examiners expect you to state before your working.

📋 Definition — Linear Dependence

A set of vectors \(\{v_1, v_2, \ldots, v_k\}\) in \(\mathbb{R}^n\) is linearly dependent if there exist scalars \(c_1, c_2, \ldots, c_k\), not all zero, such that $$c_1\,v_1 + c_2\,v_2 + \cdots + c_k\,v_k = \mathbf{0}.$$ For two vectors, this simplifies to: one is a scalar multiple of the other.

The key phrase is "not all zero." If every constant must equal zero, the vectors are linearly independent.


02 · Worked Example

Proving Linear Dependence

⚙ Example — Linear Dependence in \(\mathbb{R}^3\)

Question: Explain why the following vectors are linearly dependent.

Vector v₁
( −1,  2,  4 )
Vector v₂
( 5,  −10,  −20 )

Solution:

1
Check for a constant ratio between components

Divide each component of \(v_2\) by the corresponding component of \(v_1\):

\(\dfrac{5}{-1} = -5,\qquad \dfrac{-10}{2} = -5,\qquad \dfrac{-20}{4} = -5\)

The ratio is constant = −5, confirming \(v_2\) is a scalar multiple of \(v_1\).

2
Express \(v_2\) explicitly in terms of \(v_1\)
\(v_2 = -5\,(-1,\;2,\;4) = -5\,v_1\)
3
Write the linear dependence relation

Rearranging \(v_2 = -5\,v_1\) gives:

\(5\,v_1 + v_2 = \mathbf{0}\)

Here \(c_1 = 5\) and \(c_2 = 1\) are not both zero — the definition is satisfied. ✓

v₁ = ( −1, 2, 4 )
× (−5)
v₂ = ( 5, −10, −20 )
Conclusion
Since \(v_2 = -5\,v_1\), the vectors \(v_1\) and \(v_2\) are linearly dependent in \(\mathbb{R}^3\).
💡
Exam tip: For two vectors, the fastest check is the ratio test — if \(\tfrac{v_2[i]}{v_1[i]}\) equals the same constant for every index \(i\), the vectors are linearly dependent. Here every ratio equals \(-5\).

03 · Geometric Meaning

What Does This Look Like?

🔭 Geometric Interpretation

Linearly dependent vectors in \(\mathbb{R}^3\) are collinear — they lie along the same line through the origin. No matter how many such vectors you collect, they can never span beyond that line. Their combined span contributes only 1 dimension, not 2.

Two linearly independent vectors span a full plane (a 2-D subspace). You need three linearly independent vectors to span all of \(\mathbb{R}^3\).


04 · Summary

Key Takeaways

Three equivalent confirmations that \(\{v_1, v_2\}\) is linearly dependent:

📊
Matrix Rank
rank\([v_1 \mid v_2] = 1\), not 2
📐
Span
Span = a line (1-D subspace of ℝ³)
🔢
Determinant
Any 2×2 submatrix has det = 0
⚖️
Scalar Multiple
\(v_2 = -5\,v_1\) directly

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