# Sets of Vectors as Subspaces

• jolly_math
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.

#### jolly_math

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.

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##.

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.

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.

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.