Question about vector spaces and subsets

[green]

V is a vector space and W is a subset of V. could W be a vector space but not a subspace of V?

 

[fresh_42]

No, at least not if the addition is the same.

 

[fresh_42]

A subspace $W$ of a vector space $V$ over the field $\mathbb{F}$ is a set with the properties $W\subseteq V$ and
$$
c\cdot a\in W \text{ and }a-b\in W \text{ for all }a,b\in W \text{ and }c\in \mathbb{F}.
$$
This makes any subset $W$ that is also a vector space a subspace of $V$ per definition. The only way out of this automatism is to define two different vector space structures on $W$ and $V$.

An example: Consider $V=\mathbb{R}$ and $W=\{0,1\}.$ Then $W\subseteq V$ is a subset of the real vector space $V.$ Now, $W$ is no real subspace, but it is an $\mathbb{F}_2$ vector space. So we have a subset $W$ that is a vector space, but not a subspace of $V$ because both vector space structures are incompatible: $1+_W1=0$ and $1+_V1=2.$

 

[green]

A subspace $W$ of a vector space $V$ over the field $\mathbb{F}$ is a set with the properties $W\subseteq V$ and
$$
c\cdot a\in W \text{ and }a-b\in W \text{ for all }a,b\in W \text{ and }c\in \mathbb{F}.
$$
This makes any subset $W$ that is also a vector space a subspace of $V$ per definition. The only way out of this automatism is to define two different vector space structures on $W$ and $V$.

An example: Consider $V=\mathbb{R}$ and $W=\{0,1\}.$ Then $W\subseteq V$ is a subset of the real vector space $V.$ Now, $W$ is no real subspace, but it is an $\mathbb{F}_2$ vector space. So we have a subset $W$ that is a vector space, but not a subspace of $V$ because both vector space structures are incompatible: $1+_W1=0$ and $1+_V1=2.$

That was very helpful! Thanks.

 

[Office_Shredder]

No, at least not if the addition is the same.

Counterpoint: $\mathbb{Q}\subset \mathbb{R}$, is a 1 dimensional vector space over itself but it's not a subspace of the one dimensional real vector space $\mathbb{R}$. I would argue the addition is the same :)

 

[fresh_42]

Counterpoint: $\mathbb{Q}\subset \mathbb{R}$, is a 1 dimensional vector space over itself but it's not a subspace of the one dimensional real vector space $\mathbb{R}$. I would argue the addition is the same :)

Yes, I should have written 'vector space structure' instead of addition. Laziness is always punished. Your 'counterexample' is the same as mine, only with another field.

 

[PeroK]

V is a vector space and W is a subset of V. could W be a vector space but not a subspace of V?

It's also important to specify the field of scalars, which is part of the definition of a vector space. The Real numbers are a subset of the Complex numbers and a real vector space, but not a complex subspace.

 

[Hornbein]

Counterpoint: $\mathbb{Q}\subset \mathbb{R}$, is a 1 dimensional vector space over itself but it's not a subspace of the one dimensional real vector space $\mathbb{R}$. I would argue the addition is the same :)

Duh, I don't get why not. Care to enlighten me?

 

[Office_Shredder]

Duh, I don't get why not. Care to enlighten me?

I'm going to do things over two dimensions, to make the difference between scalars and vectors a bit more obvious. $\mathbb{Q}^2$ is obviously a 2 dimensional vector space over $\mathbb{Q}$. It is also a subset of $\mathbb{R}^2$. This is both a two dimensional vector space over $\mathbb{R}$, and an infinite dimensional vector space over $\mathbb{Q}$. If you think of it as a vector space over the real numbers, $\mathbb{Q}^2$ is not a subspace because it is not closed under scalar multiplication: $(1,1)\in \mathbb{Q}^2$ but $\pi(1,1)=(\pi,\pi)\notin \mathbb{Q}^2$.