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.
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?
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.
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).