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How to Prove a Set is a Subspace of a Vector Space

The Math Sorcerer · YouTube · 5:57

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Prove that a plane defined by x₃=x₁+x₂ is a subspace by checking the three standard criteria.

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Chapters

1:10Start with the zero vector3:45Close under addition5:00Close under scalar multiplication

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The vector (0,0,0) satisfies x₃=x₁+x₂, so the set is nonempty. This is the first required condition.

If x₃=x₁+x₂ and y₃=y₁+y₂, then x₃+y₃=(x₁+y₁)+(x₂+y₂). Thus adding any two members preserves the defining relation.

Multiplying x₃=x₁+x₂ by any real scalar gives cx₃=cx₁+cx₂. Together with nonemptiness and addition closure, this proves the set is a subspace.

Knowledge cards

01

Start with the zero vector

The vector (0,0,0) satisfies x₃=x₁+x₂, so the set is nonempty. This is the first required condition.

02

Close under addition

If x₃=x₁+x₂ and y₃=y₁+y₂, then x₃+y₃=(x₁+y₁)+(x₂+y₂). Thus adding any two members preserves the defining relation.

03

Close under scalar multiplication

Multiplying x₃=x₁+x₂ by any real scalar gives cx₃=cx₁+cx₂. Together with nonemptiness and addition closure, this proves the set is a subspace.

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  • Vector spaces ProofAt 1:10
    Why this connection?

    The vector (0,0,0) satisfies x₃=x₁+x₂, so the set is nonempty. This is the first required condition. If x₃=x₁+x₂ and y₃=y₁+y₂, then x₃+y₃=(x₁+y₁)+(x₂+y₂). Thus adding any two members preserves the defining relation. Multiplying x₃=x₁+x₂ by any real scalar gives cx₃=cx₁+cx₂. Together with nonemptiness and addition closure, this proves the set is a subspace.

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