1
$\begingroup$

I am trying to see if I understand the idea of a bases as presented in the text by Friedberg, Insel, and Spence. Please correct any and all misconceptions I have in the following and feel free to add insight if necessary:

1.) The definition of the bases of a vector space is a linearly independent set that generates the vector space. The number of vectors in the base are always the same, where when I say "same" I mean that there can be a one to one correspondence between the two bases. So even in an infinite dimensional vector space, the bases will have the "same" number of vectors"

2.) I find the following exercise too trivial for it to be correct: Prove that a vector space is infinite dimensional if and only if it contains an infinite linearly independent set.

If a vector space $V$ is infinite dimensional, then by definition it has a bases of infinite vectors. On the other hand, if a vector space has an infinite linearly independent set S, then $Span(S)\subset V$ but $Span(S)$ has infinite dimension so $V$ must also.

3.) If two bases are given of the "same" size, then the space generated by them will be of equal "size." Specifically, I will be able to find some 1-1 correspondence between the two spaces that are generated.

4.) Heuristically, all we have to do to find the dimension of a finite dimensional vector space is consider the number of different variables needed to express an element of the space.

5.) If $H \subset W$, where $H$ and $W$ are subspaces of $V$ with bases $\alpha$ and $\beta$, then, even in the infinite case, $\alpha \subset \beta$. In particular $(\alpha \cap \beta) \subset \alpha$ and $\alpha \subset (\alpha \cup \beta)$. Does everything here remain true even when we extend to infinite intersections (and unions). Also what are conditions for $\alpha \cap \beta$ to be a basis for $H \cap W$ (same with union)?

$\endgroup$
  • $\begingroup$ It would be better if you ask one clear question that can be answered. There are too many issues here: first, do you know that every vector space has a base? As far as I know, this only holds if you assume the axiom of choice.. $\endgroup$ – Peter Franek Jun 16 '14 at 19:40
  • $\begingroup$ My book does state that every vector space has a base, although it cannot always be constructed $\endgroup$ – illysial Jun 16 '14 at 19:41
  • $\begingroup$ @illysial Finite dimensional vector spaces (which is the primary focus of linear algebra) always have bases; it is never a problem. If you get into infinite dimensional vector spaces, then they still always have bases, under the assumption of the Axiom of Choice however. Stuff resulting from Choice is generally non-constructive by nature. $\endgroup$ – EuYu Jun 16 '14 at 19:43
  • 1
    $\begingroup$ What do you mean when, in (2), you say "it is too trivial for it to be correct?" (2) is correct! $\endgroup$ – user39280 Jun 16 '14 at 19:44
  • 1
    $\begingroup$ Ok, the inclusion $\alpha\subseteq\beta$ doesn't hold even for one-dimensional spaces; $H=W=V=\mathbb{R}$, $\alpha=\{1\}$, $\beta=\{2\}$ is a counter-example. $\endgroup$ – Peter Franek Jun 16 '14 at 19:48
2
$\begingroup$

Your points 1 and 2 are fine, except that very often a vector space is termed 'infinite dimensional' iff it has no finite basis.

In point 3, simply having a 1-1 correspondence between two spaces is much weaker than being the 'same size' as you say. For example, there is a $1-1$ correspondence between $\mathbb{R}^2$ and $\mathbb{R}$, since they have the same cardinality. But the 'dimension' of $\mathbb{R}^2$ is bigger than the dimension of $\mathbb{R}$, taken as vector spaces over $\mathbb{R}$.

Your point 4 is very loose, and I don't want to comment on the heuristic.

In 5, it's unclear what $\alpha \subset \beta$ means. For example, the space $H$ generated by $(1,0,0)$ is a subspace of the space $W$ generated by $\{(1,1,0), (0,-1,0)\}$, but in terms of set inclusion the former basis is not a subset of the latter basis.

$\endgroup$
  • $\begingroup$ Would point 3 be better if it was a bijective correspondence? $\endgroup$ – illysial Jun 16 '14 at 19:51
  • $\begingroup$ @illysial: I use bijective correspondence is 1-1 correspondence interchangeably. What difference do you mean? $\endgroup$ – davidlowryduda Jun 16 '14 at 19:52
  • $\begingroup$ I mean to say it would be onto as well $\endgroup$ – illysial Jun 16 '14 at 19:55
  • $\begingroup$ @illysial: Yes, I use bijective correspondence, 1-1 correspondence, being both injective and surjective all to mean the same thing. I still don't quite understand what difference you mean? To be clear, there are bijections of $\mathbb{R}^2$ and $\mathbb{R}$. $\endgroup$ – davidlowryduda Jun 16 '14 at 19:58
  • 1
    $\begingroup$ no problem! Good luck! $\endgroup$ – davidlowryduda Jun 16 '14 at 20:05

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.