0
$\begingroup$

What is a proper subspace of an $n$-dimensional $\Bbb R$-vectorspace $V$? It is a subset of $V$ that is a vectorspace in its own right correct? How would we denote these vectorspaces, surely as a subset they aren't $\Bbb R$-vectorspaces of their own.

Could we take the $\Bbb Q$-vectorspace as a subspace of $V$?

$\endgroup$
1
  • 1
    $\begingroup$ When you talk about subspaces the scalar field doesn't change. If $V$ is a vectorspace over $\mathbb{R}$ then all subsets of $V$ that are also vector spaces over $\mathbb{R}$ are subspaces of $V$. $\endgroup$
    – R_D
    Oct 3 '15 at 12:09
2
$\begingroup$

Consider the case when $n=1$. $\mathbb{R}$ is an $\mathbb{R}$ - vector space. The only vector subspaces of it are $\{0\}$ and $\mathbb{R}$ itself. $\mathbb{Q}$ is a subset of $\mathbb{R}$ and is a $\mathbb{Q}$ - vector space but is not an $\mathbb{R}$ - vector space.

EDIT : I realized that I did not answer your entire question but only answered the last part so here is the rest of the answer -

You are right when you say subspaces of an $n$ - dimensional $\mathbb{R}$ - vector space $V$ are subsets of $V$ that are vector spaces in their own right. However when you say it is a vector space in its own right the scalar field is always fixed. So a better way to say it is that subspaces of $V$ are subsets of $V$ that are $\mathbb{R}$ - vector spaces in their own right.

Secondly, proper subspaces of $V$ are proper subsets that are also $\mathbb{R}$ - vector spaces. So a proper subspace of $V$ is a $k$ - dimensional subspace of $V$ where $k < n$. (If $k = n$ then the subspace is no longer proper). How would you visualize such a space? Well if you think of $V$ ar $\mathbb{R}^n$ then the $0$ - dimensional subspace is just $\{0\}$, the $1$ - dimensional subspaces are lines through the origin, $2$ - dimensional are planes through the origin and so on (This is also mentioned in another answer)

$\endgroup$
1
  • $\begingroup$ Thank you, that does indeed make sense. $\endgroup$ Oct 3 '15 at 13:34
2
$\begingroup$

Indeed a proper subset of V that is a $\mathbb{R}$ vector space is a proper subspace and the scalar field is fixed. Note in $\mathbb{R}^2$ {0}, and the lines passing through the origin are the only subspaces of $\mathbb{R}^2$. Similarly in $\mathbb{R}^n$ the only subspaces are {0}, the lines passing through the origin, planes containing the origin, the n-k (k=1,...,n-3) hyperplanes containing the origin.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.