make that "less than or equal to"
suppose we have a linear transformation T:U→V, with dim(U) = n. this means we have a basis B = {u_{1},...,u_{n}}.
i claim T(B) spans T(U).
suppose v is in T(U). this means v = T(u), for some u in U. since B is a basis for U, we have:
u = a_{1}u_{1} +...+ a_{n}u_{n}, for some field elements a_{1},...,a_{n}.
hence v = T(u) = T(a_{1}u_{1} +...+ a_{n}u_{n})
= a_{1}T(u_{1}) +....+ a_{n}T(u_{n}), by the linearity of T.
unfortunately, T(B) is not a basis, usually. in fact, if T(u) = T(u') but u ≠ u', then we have two distinct linear combinations
of the elements of T(B) that are equal, which means that T(u) - T(u') = 0, but u - u' ≠ 0.
now u - u' is a non-zero vector of U, so by the linear independence of B, not all its coefficients in the basis B are 0.
but T(u) - T(u') = T(u - u') is a linear combination of the vectors in T(B), with non-zero coefficients (the same ones as for u - u'),
so in this case T(B) must be a linearly dependent set. of course we can find SOME subset of T(B) that IS a basis for T(U),
but for this set (let's call it C), |C| ≤ |T(B)| = |B|.
but C = dim(T(U)) = dim(im(T)), and B = dim(U).
some easy consequences of this:
if dim(V) < dim(U), T cannot be injective. the null space of T (or kernel of T) has to have some non-zero vectors in it.
if dim(U) < dim(V), then T(U) is a PROPER subspace of V. we can't map U onto a "bigger space" (in a linear way).
intutitively, a linear map does this:
it maps vector spaces to vector spaces. for example, if we start with a 3-dimensional linear space, then T either:
maps to another 3-dimensional linear space (preserving the origin).
shrinks a line down to a point, and so maps the 3-space to a plane (preserving the origin).
shrinks a plane down to a point, and maps the 3-space to a line (preserving the origin).
maps everything to the origin.
********
to see why the dimension of the image of T might be equal to the dimension of the domain, consider the following function:
T:U→U, T(u) = u. this is a linear transformation.