Math Problem Statement
Solution
To prove that is a subspace of , let’s proceed step by step:
Problem Restatement:
Let , where are scalars. Prove that is a subspace of .
Subspace Conditions:
To verify that is a subspace of , we need to check the following three properties:
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The Zero Vector is in :
- The zero vector in is . Substituting into the equation : which is true. Thus, the zero vector is in .
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Closure under Addition:
- Let and . This implies: For their sum , we have: Hence, is closed under addition.
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Closure under Scalar Multiplication:
- Let and be any scalar. Then: For , substituting into the defining equation of : Thus, is closed under scalar multiplication.
Conclusion:
Since satisfies all three conditions, is a subspace of .
Would you like a further breakdown of any specific step or additional examples?
Here are some related questions to explore:
- What is the dimension of in for specific values of ?
- How does the condition define a geometric object in ?
- How would the proof change if were a subset of ?
- Can ever represent a line in ?
- What happens to if ?
Tip: Always confirm closure properties by substituting directly into the defining condition of the subspace.
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Subspaces
Vector Spaces
Formulas
ax + by + cz = 0
Theorems
Subspace Test
Closure under addition and scalar multiplication
Suitable Grade Level
Undergraduate - Linear Algebra