Questions: Vectors in R^n: Addition and Scalar Multiplication

Question 1

Vector v = (3, -2) is multiplied by the scalar -2. What is the resulting vector?

• A. (-6, 4)
• B. (6, -4)
• C. (-3, 2)
• D. (-6, -4)

Answer: (A)

Scalar multiplication applies component-wise: -2 × (3, -2) = (-2 × 3, -2 × -2) = (-6, 4). Option B applies +2 instead of -2. Option C negates without scaling (multiplies by -1 only). Option D incorrectly doubles while keeping the negative sign on the second component.

Question 2

Two copies of vector u = (2, 5) are drawn at different starting points in the plane. Are they the same vector?

• A. No — vectors are identified by their starting point, so different placements are different vectors
• B. Only if one copy is placed at the origin
• C. Yes — a vector is defined by its direction and magnitude, not its location
• D. It depends on whether the vectors are parallel

Answer: (C)

Vectors represent displacements — a direction and magnitude — not fixed positions. Two vectors are equal if and only if they have the same components, regardless of where they are drawn. This is the key distinction between a vector and a point. Option A confuses position vectors (rooted at the origin) with general free vectors.

Question 3

A vector's components uniquely determine a fixed location in the coordinate plane.

• T. True
• F. False

Answer: False

Vectors represent displacements, not fixed positions. A vector (3, 2) means 'move 3 right and 2 up' — this displacement can be applied starting anywhere. The same vector can be placed at any point in the plane. Only a position vector (defined as the displacement from the origin to a specific point) corresponds to a unique location. Confusing vectors with points is one of the most persistent early errors in linear algebra.

Question 4

The sum of any vector v and its negation -v equals the zero vector (0, 0, ..., 0).

• T. True
• F. False

Answer: True

For any v = (v₁, v₂, ..., vₙ), adding v + (-v) gives (v₁ - v₁, v₂ - v₂, ..., vₙ - vₙ) = (0, 0, ..., 0) by component-wise subtraction. The zero vector is the additive identity in ℝⁿ: adding it changes nothing, and it is the unique result of adding any vector to its own negation. This is one of the eight vector space axioms.

Question 5

Explain why vector addition in ℝⁿ automatically satisfies commutativity and associativity without requiring a separate proof for each dimension n.

Model answer: Because vector addition in ℝⁿ is performed component-wise, and each component is a real number. Real number addition is already commutative (a + b = b + a) and associative ((a + b) + c = a + (b + c)). These properties are inherited independently in each coordinate slot. Since all algebraic axioms reduce to statements about individual real-number components, and real numbers satisfy them, ℝⁿ satisfies all vector space axioms automatically for any n — no matter how many dimensions.

This is the deeper reason why working in high-dimensional spaces like ℝ¹⁰⁰ feels algebraically manageable: it is the same arithmetic of real numbers applied independently in each coordinate. The vector space structure is lifted from the reals, not invented fresh.