# An example of a non-subspace

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.

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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?

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.

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?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

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.

Orodruin
Staff Emeritus
Homework Helper
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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.

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.

Ray Vickson
Homework Helper
Dearly Missed
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.

Orodruin
Staff Emeritus
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.