# A Basis Question

1. Jul 20, 2008

### e(ho0n3

Let B be a basis of a vector space V. If U is a subspace of V, is it true that a subset of B may serve as a basis for U?

2. Jul 20, 2008

### LukeD

Let V be R^2. One basis of R^2 is {(1,0),(0,1)} (we'll call this B). Let U be {(x,y): y = 2x}. This is a subspace of R^2. One basis of U is {(1,2)}, and every basis of this space contains a multiple of the vector (1,2). Since no multiple of (1,2) is in B, there is no subset of B that is a basis of U.

However, it is possible to have a subset of B be a basis for U. For instance, if B were {(1,2),(2,1)} (this is indeed a basis of V), then {(1,2)} is a basis for U and a subset of B.

3. Jul 20, 2008

### n_bourbaki

This question is so trivially false that it tells me that you didn't bother to think about any concrete examples at all. The first thing you do if you're not sure about something is you see what happens in a case you can work out by hand, if at all possible. A trivial consequence of this statement would be that a vector space only had a finite number of subspaces, for heaven's sake. I apologise for sounding harsh.

Last edited: Jul 20, 2008
4. Jul 20, 2008

### tiny-tim

Hi e(ho0n3!

Hint: take a really easy example to visualise: let U be ordinary space (R3), and let B be (1,0,0) (0,1,0) and (0,0,1).

Can you draw a two-dimensional subspace which doesn't include any of B?

5. Jul 20, 2008

### HallsofIvy

Staff Emeritus
Be careful exactly what you are saying. If U is a proper subspace of V, then a basis for V cannot be a basis for U because it will span a space, V, larger than U.

What is true is that there always exists a basis for V that contains a basis for U.

6. Jul 20, 2008

### tiny-tim

Hi HallsofIvy!

A basis, yes, but not necessarily the given basis.

7. Jul 20, 2008

### e(ho0n3

I have a hard time coming up with examples or counterexamples. Thanks guys.

8. Jul 20, 2008

### n_bourbaki

Does that mean you didn't understand LukeD's example? Or you don't understand that in a vector space of dimension 2 with basis {u,v} then not evey subspace is spanned by either u or v, which is surely the most obvious and trivial example you should have first thought of. Or you do, but just didn't think of them yourself and you're explaining why?

Surely it is clear that if I have n (linearly independent) vectors, then subsets of these span exactly 2^n possible vector subspaces? And that almost all vector spaces have a lot more subspaces than that?

Last edited: Jul 20, 2008
9. Jul 20, 2008

### e(ho0n3

This one.

Are you asking? The first one is clear to me. The second one isn't.

10. Jul 20, 2008

### n_bourbaki

In one thread you're asking about how to distribute operators over tensor products in relation to quantum computing, and in another you're do not know that a vector space of dimension at least 2 (over something like the field of complex numbers) has infinitely many distinct subspaces? This bothers me.

Consider R^2. How many lines through the origin are there?

11. Jul 20, 2008

### e(ho0n3

It bothers me a lot more.

Countless. I understand now. Thus, for any n-dimensional space V, since it is isomorphic to Rn, it contains a subspace isomorphic to R2, and since R2 contains infinitely many subspaces, V has infinitely many subspaces. Right?

12. Jul 20, 2008

### n_bourbaki

any n-dimensional real vector space. But a similar analysis for any field will tell you something. But the 2 wasn't important: R^n contains infinitely many proper subspaces of any dimension.