An example of a non-subspace

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  • #1
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Hi,
I hope you guys help me with this exercise in the book "Linear Algebra Done Right"

Homework Statement


" Give an example of a nonempty subset ##U## of ##R^2## such that ##U## is closed under addition and under taking additive inverses (meaning ##−u## ##∈## ##U## whenever ##u## ##∈## ##U##), but ##U## is not a subspace of ##R^2##."

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3. The Attempt at a Solution [/B]
From what I understood from the question, the subset must not be closed under multiplication because it is closed under addition as well as the additive inverses which imply that ##0## vector must exist.
So the only thing that would make it non-subspace is the multiplication.
Hope you guys correct my understanding and help me with the solution.
 

Answers and Replies

  • #2
Orodruin
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Your understanding of the problem seems fine. Now you just have to construct such a subset. I suggest taking a non-zero vector and then just adding the elements that you have to add based on the problem description. What does that give you?
 
  • #3
Ray Vickson
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Hi,
I hope you guys help me with this exercise in the book "Linear Algebra Done Right"

Homework Statement


" Give an example of a nonempty subset ##U## of ##R^2## such that ##U## is closed under addition and under taking additive inverses (meaning ##−u## ##∈## ##U## whenever ##u## ##∈## ##U##), but ##U## is not a subspace of ##R^2##."

Homework Equations


3. The Attempt at a Solution [/B]
From what I understood from the question, the subset must not be closed under multiplication because it is closed under addition as well as the additive inverses which imply that ##0## vector must exist.
So the only thing that would make it non-subspace is the multiplication.
Hope you guys correct my understanding and help me with the solution.

Well, it would certainly be closed under integer multiplication, because ##7 v## is just ##v+v+v+v+v+v+v## and your set is closed under addition. Thus, you need to avoid multiplication by non-integer rationals or by irrational reals.
 
  • #4
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Well, it would certainly be closed under integer multiplication, because ##7 v## is just ##v+v+v+v+v+v+v## and your set is closed under addition. Thus, you need to avoid multiplication by non-integer rationals or by irrational reals.
I understand now the problem. So I can say that the set is the vectors ##(x_1,x_2)## where ##x_1,x_2## are integers only (negative and positive). When we check the multiplication closure by multiplying by rational number for example, we'll get a vector that doesn't belong to ##U##, correct?
I wrote it as: ##U=\left\{\left(x_1,x_2\right)\in R^2 : x_1,x_2\in Z\right\}##
What do you think?
 
  • #5
Orodruin
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I understand now the problem. So I can say that the set is the vectors ##(x_1,x_2)## where ##x_1,x_2## are integers only (negative and positive). When we check the multiplication closure by multiplying by rational number for example, we'll get a vector that doesn't belong to ##U##, correct?
I wrote it as: ##U=\left\{\left(x_1,x_2\right)\in R^2 : x_1,x_2\in Z\right\}##
What do you think?
This is just one example of such a subset. There are many many more, but the task was to give an example so ...

Edit: The more minimalistic construction is to note that the set needs to contain at least one non-zero element ##x \in \mathbb R^2## apart from also containing the zero vector. In order for the set to be closed under addition, it also needs to contain all integer multiples of ##x##, but once you have included those the set satisfies everything you required of it. It does not matter what ##x## is as long as it is non-zero.
 
  • #6
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This is just one example of such a subset. There are many many more, but the task was to give an example so ...
It would be great If you could give more examples because I want to fully understand the subspaces subject.
 
  • #7
Orodruin
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It would be great If you could give more examples because I want to fully understand the subspaces subject.
See my edit.
 
  • #8
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See my edit.
Thank you very much for this, I have now a good understanding of the problem. The things you said in your edit made a lot of sense to me. I really thank you very much. Thanks to @Ray Vickson for the help as well.
 
  • #9
Ray Vickson
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I understand now the problem. So I can say that the set is the vectors ##(x_1,x_2)## where ##x_1,x_2## are integers only (negative and positive). When we check the multiplication closure by multiplying by rational number for example, we'll get a vector that doesn't belong to ##U##, correct?
I wrote it as: ##U=\left\{\left(x_1,x_2\right)\in R^2 : x_1,x_2\in Z\right\}##
What do you think?

This works, but it is not the only possible example. Another example would be to choose "vectors" ##u_1 = (a_1,b_1)## and ##u_2 = (a_2, b_2)## for some real ##a_i, b_i##. Then a "lattice"built using ##u_1## and ##u_2## would work as well. This would be the set of point of the form ##n_1 u_1 +n_2 u_2 =(n_1 a_1 + n_2 a_2, n_1 b_1 + n_2 b_2), ## where ##n_1, n_2 \in \mathbb{Z}.## That could give a lattice consisting of stacked quadrilaterals instead of squares.
 
  • #10
Orodruin
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This works, but it is not the only possible example. Another example would be to choose "vectors" ##u_1 = (a_1,b_1)## and ##u_2 = (a_2, b_2)## for some real ##a_i, b_i##. Then a "lattice"built using ##u_1## and ##u_2## would work as well. This would be the set of point of the form ##n_1 u_1 +n_2 u_2 =(n_1 a_1 + n_2 a_2, n_1 b_1 + n_2 b_2), ## where ##n_1, n_2 \in \mathbb{Z}.## That could give a lattice consisting of stacked quadrilaterals instead of squares.
This is also one example, but far from the only possibility. The simplest construction being the one described in #5.
 

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