Solving: Vector Subspaces Question in R3

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SUMMARY

The set T = {(x, y, z) : -1 ≤ x + y + z ≤ 1} is definitively not a vector subspace of R3. This conclusion is established by demonstrating that T fails the closure property under scalar multiplication. Specifically, the vector (1, 0, 0) is an element of T, but its scalar multiple 2(1, 0, 0) = (2, 0, 0) does not satisfy the condition for T, as 2 > 1. Therefore, T cannot be classified as a vector subspace.

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  • Understanding of vector spaces and their properties
  • Knowledge of scalar multiplication in R3
  • Familiarity with inequalities and their implications in vector spaces
  • Basic linear algebra concepts
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  • Study the properties of vector subspaces in linear algebra
  • Learn about closure properties of vector sets
  • Explore examples of vector subspaces in R3
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Students of linear algebra, educators teaching vector space concepts, and anyone seeking to deepen their understanding of vector subspaces in R3.

markovchain
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Could someone please help me with the following question with a guided step by step answer:

Show that T = (x, y, z) : -1 ≤ x + y + z ≤ 1
is not a vector subspace of R3

Thanks!
 
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markovchain said:
Could someone please help me with the following question with a guided step by step answer:

Show that T = (x, y, z) : -1 ≤ x + y + z ≤ 1
is not a vector subspace of R3

Thanks!


$$(1,0,0)\in T\,\,\,but\,\,\,2(1,0,0)=(2,0,0)\notin T$$

Thus T cannot be v. subspace as it isn't closed under scalar multiplication

DonAntonio
 

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