2
$\begingroup$

I'm having some issues thinking about this one. I'm going to think aloud here. If $V$ is infinite dimensional, this means that there infinitely many basis vectors. I know the choice of the representative of the space doesn't matter. Likewise for $S$, only any vector we choose from $S$ will also already be in $V$.

Now, $\frac{V}{S}$ is a space of vectors in V modulo the vectors in $S$. This only means that the difference of say $v-s \in \frac{V}{S}$, right? I could really use some help here. Thanks.

$\endgroup$
  • 2
    $\begingroup$ Consider the space of infinite real-valued sequences and the subspace thereof that consists of those sequences that are zero in every even coordinate. $\endgroup$ – David Mitra Feb 4 '14 at 14:40
  • $\begingroup$ Okay, I think I see what you're saying here. So this means that V/S isn't necessarily finite. "modding out" the subspace of sequences that consist of zero in every coordinate do not change anything. Right? $\endgroup$ – Calculus08 Feb 4 '14 at 14:48
3
$\begingroup$

Let $a_1,\cdots, a_n, \cdots $ is a basis for $V$. Let $S$ be the vector subspace spanned by $a_2, \cdots, a_{2k}, \cdots$, then $V/S$ is spanned by $a_1+S, \cdots, a_{2k+1}+S. \cdots, $, which is infinite dimension.

$\endgroup$
1
$\begingroup$

Consider the space $V$ of all functions $\Bbb R\to\Bbb R$ as a vector space over $\Bbb R.$ The set $P$ of all polynomial functions is an infinite-dimensional subspace, but (for example) consider the set $F$ of functions $f:\Bbb R\to\Bbb R$ such that $\{x\in\Bbb R:f(x)\ne 0\}$ is finite. Then $F$ is again infinite-dimensional, and can readily be shown to be isomorphic to its image modulo $P,$ so $V/P$ is again infinite-dimensional.

$\endgroup$

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.