Let V be a vector space over an infinite field, that cannot be spanned by a single vector, then V has an infinite number of subspaces.

I've been trying to prove this for an hour now, and I've got to submit my homework tomorrow. Any suggestions?

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- May 12th 2012, 12:17 PMloui1410Infinite subspaces?
Let V be a vector space over an infinite field, that cannot be spanned by a single vector, then V has an infinite number of subspaces.

I've been trying to prove this for an hour now, and I've got to submit my homework tomorrow. Any suggestions? - May 12th 2012, 01:00 PMDevenoRe: Infinite subspaces?
since V cannot be spanned by a single vector, dim(V) ≥ 2. let {u,v} be a basis for a 2-dimensional subspace of V. consider the subspaces spanned by au + v, for a in F.

- May 12th 2012, 01:06 PMloui1410Re: Infinite subspaces?
That's what I thought, but thing is - how can I write this formally?

And why au + v is actually a subspace of {u,v} and not equal to it? - May 12th 2012, 01:26 PMHallsofIvyRe: Infinite subspaces?
First, what you are writing is not correct and is not what Deveno said. "au+ v" is NOT a subspace, it is a single vector. It

**spans**a one-dimensional subspace and the condition is that V is not one-dimensional. Taking different values for the number "a" gives different subspaces. - May 12th 2012, 04:18 PMloui1410Re: Infinite subspaces?
How can "au+ v" span a one-dimensional subspace? Which vector space does it form on its own? Can you perhaps give an actual example?

Thanks again! - May 13th 2012, 11:57 AMDevenoRe: Infinite subspaces?
for any non-zero vector x in a vector space V over a field F, the set S = {ax : a in F} is a subspace of V:

1. S is closed under addition: for ax,bx in S we have: ax + bx = (a+b)x, and a+b is surely some element of F.

2. S is non-empty: it contains the non-zero vector x. equivalently: since 0 is an element of F, 0x = 0 (the 0-vector of V), so the 0-vector of V is contained in S.

3. S is closed under scalar multiplication: let ax be in S, and let c be any scalar in F. then c(ax) = (ca)x, and ca is surely an element of F.

such a subspace (which is one-dimensional) looks like a line passing through the 0-vector in the direction of the vector x.

(you CAN use a 0 vector, but then S = {a0: a in F} = {0}, which is only 0-dimensional (but IS a subspace). using a non-zero vector avoids this complication. note that if {u,v} are basis vectors of a 2-dimensional subspace of V, they are, by definition, non-zero vectors).

Ok, an example. let V = R^{2}. consider the basis {(1,0),(0,1)} (so we are using u = (1,0), and v = (0,1)). so a vector of the form au + v is: a(1,0) + (0,1) = (a,1). so for different values of a, we get different subspaces:

if a = 1, we get span({(1,1)}), or the line y = x.

if a = 0, we get span({(0,1)}), the y-axis (the line x = 0).

if a = 2, we get span({(2,1)}), or all vectors of the form (2y,y), which is to say, the line y = (1/2)x.

if a = 1/2, we get span({(1/2,1)}), or all vectors of the form (y/2,y), which is to say the line y = 2x.

in general, we get the line passing through (0,0) and (a,1), which has slope (1-0)/(a - 0) = 1/a, and of course, has the y-intercept 0 (it passes through the origin), so the line y = x/a = (1/a)x. surely you can see that we have an infinite number of such lines (pick any value of a from the real numbers).