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MathidiaMathematics Academy
Vector spaces
Lesson 21 of 75
Introduction

Vector-space structure

Objective. Verify vector-space and subspace axioms.

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  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

A vector space is a set V with addition and scalar multiplication satisfying axioms such as closure, associativity, commutativity of addition, existence of a zero vector, additive inverses, and distributivity of scalar multiplication over vector and scalar addition. A subspace is a subset of V that is itself a vector space under the same operations, which can be verified with the subspace test: it contains the zero vector, and is closed under addition and scalar multiplication. For W={(x,y,z)∈ℝ³ : x+y+z=0}, the zero vector (0,0,0) satisfies the equation, and adding two elements or scaling one preserves the equation, so W is a subspace. Checking all required axioms, not just closure, matters when verifying an unfamiliar structure is a vector space.

Worked example

Verify that W={(x,y,z)∈ℝ³ : x+y+z=0} is a subspace of ℝ³ using the vector-space subspace test.

Verify that W={(x
Illustrated card showing Verify that W={(x
  1. 1Check the zero vector: (0,0,0) satisfies 0+0+0=0, so zero is in W.
  2. 2Check closure under addition: if (a,b,c),(d,e,f)∈W then a+b+c=0 and d+e+f=0, so their sum (a+d)+(b+e)+(c+f)=0, meaning the sum lies in W.
  3. 3Check closure under scalar multiplication: for scalar k, k(a+b+c)=k·0=0, so k(a,b,c)∈W.
  4. 4All three subspace conditions hold, so W is a subspace of ℝ³.

Answer: W is a subspace of ℝ³

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Common mistake. Students verify only closure under addition and scalar multiplication when checking a subspace, forgetting to confirm the zero vector actually belongs to the subset, which is essential since a set like {(x,y):x+y=1} fails despite being closed-looking.

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