Is it possible to have a vector space with restricted scalars?

Click For Summary
SUMMARY

The discussion centers on the question of whether a set of vectors defined as all vectors \( a = (a_1, ..., a_n) \) in \( \mathbb{R}^n \) with the condition \( a_1 \geq 0 \) forms a subspace of \( \mathbb{R}^n \) for \( n \geq 3 \). The conclusion is that this set does not satisfy the criteria for being a subspace because it fails the closure under scalar multiplication. Specifically, while the vector \( (1, 1, ..., 1) \) is included in the set, its additive inverse \( (-1, -1, ..., -1) \) is not, violating the necessary conditions for a vector space.

PREREQUISITES
  • Understanding of vector spaces and subspaces in linear algebra.
  • Familiarity with the properties of closure under addition and scalar multiplication.
  • Knowledge of the field properties required for scalar multipliers in vector spaces.
  • Basic comprehension of the notation and concepts in \( \mathbb{R}^n \).
NEXT STEPS
  • Review the properties of vector spaces and subspaces in linear algebra.
  • Study the implications of closure under scalar multiplication in vector spaces.
  • Examine the definition of fields and their role in vector space theory.
  • Explore examples of sets that do and do not form vector spaces based on defined conditions.
USEFUL FOR

Students of linear algebra, educators teaching vector space concepts, and anyone involved in mathematical research or applications requiring a solid understanding of vector spaces and their properties.

jolly_math
Messages
51
Reaction score
5
Homework Statement
Is the following set of vectors a = (a1, ... a_n) in Rn subspaces of Rn (n≥3): all a such that a1 ≥ 0 ?
Relevant Equations
i) For vectors x and y in the subspace, x + y is in the subspace.
ii) For any vector x in the subspace, cx is in the subspace.
I don't understand the solution: that for (1, ..., 1) the additive inverse is (-1, ..., -1), so the condition is not satisfied (and it is not a subspace).

Which condition is not met?

Thank you.
 
Physics news on Phys.org
jolly_math said:
Homework Statement:: Is the following set of vectors a = (a1, ... a_n) in Rn subspaces of Rn (n≥3): all a such that a1 ≥ 0 ?
Relevant Equations:: i) For vectors x and y in the subspace, x + y is in the subspace.
ii) For any vector x in the subspace, cx is in the subspace.

I don't understand the solution: that for (1, ..., 1) the additive inverse is (-1, ..., -1), so the condition is not satisfied (and it is not a subspace).

Which condition is not met?

Thank you.
The one you listed as ii) in your relevant equations. If u is in the set, cu also has to be in the set as one criterion of being a subspace. The vector u = <1, 1, ..., 1> is in your set, but -1u is not in the set, so the set isn't a subspace of ##\mathbb R^n##.
 
  • Like
  • Informative
Likes   Reactions: jolly_math, berkeman and topsquark
The subset defined is not a vector space so it can not be a subspace. Check the properties of a vector space and you will see what they are talking about.
 
  • Like
Likes   Reactions: jolly_math
FactChecker said:
The subset defined is not a vector space so it can not be a subspace. Check the properties of a vector space and you will see what they are talking about.
Since the given set is a subset of something that is a vector space, it's not necessary to confirm that all vector space properties are satisfied. All that needs to be checked are closure under vector addition and closure under scalar multiplication. For this problem, it's the latter property that fails.
 
  • Like
Likes   Reactions: PeroK
Mark44 said:
Since the given set is a subset of something that is a vector space, it's not necessary to confirm that all vector space properties are satisfied. All that needs to be checked are closure under vector addition and closure under scalar multiplication. For this problem, it's the latter property that fails.
Good point. It is not clear to me that the problem statement does not imply a restriction on the scalar field of reals. The definition of the subset might imply a restriction on the field so that the scalar multipliers are no longer a field. That might be where the original confusion occurred.
 
FactChecker said:
Good point. It is not clear to me that the problem statement does not imply a restriction on the scalar field of reals. The definition of the subset might imply a restriction on the field so that the scalar multipliers are no longer a field. That might be where the original confusion occurred.

The multipliers have to be a field to have a vector space. Trying to use something that's not a field gives surprisingly annoying behavior.
 
Office_Shredder said:
The multipliers have to be a field to have a vector space. Trying to use something that's not a field gives surprisingly annoying behavior.
Right. I should have mentioned that in my post (In editing, I deleted more than I meant to). It is not valid to interpret the scalars as being restricted so that they are not a field. But I think that might be what the OP had in mind.
 

Similar threads

Replies
8
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
29
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K