(V, +, *) vector space.

Let v be {v1, ..., vn} where vi \in V.
v is a basis of V if
(1) v is linearly independent
(2) span{v} = V.

Lemma: Suppose v = {v1, ..., vj} is a linearly independent set in V and
w = {w1, w2, ..., wk} s.t. span{w} = V. Then j \leq k.

Since w spans V,

[ v1 | v2 | ... | vj ] = [ w1 | w2 | ... | wk ] A,
        B                       C
where A is a k-by-j matrix.

Suppose k < j.
    Then Ax = 0 has nontrivial solutions because more unknowns (j) than
    outputs.
    Therefore, there exists x0 s.t. Ax0 = 0 with x0 \neq 0.

B = CA \to Bx0 = CAx0 = 0, but Bx0 = 0 with nontrivial x0 is not
possible because B is defined to be linearly independent.
Therefore, k \geq j.

Thm: If v and w are each bases of V, then k = j by application of the
previous lemma (k\geq j and j\geq k)
Let dim V = k = j.

Suppose that dim V = n.
    Any linearly independent set in V will have at most n vectors, and
    Any spanning set in V must have at least n vectors.
Converses
    Every set with more than n vectors in V is linearly dependent.
    Every set with less than n vectors in V does not span V.
(2')For linearly independent set w which doesn't span V, there exists a
vector which may be added to V outside the span, and w \cup {v} is also
linearly independent.
    [I think this is equivalent to saying "if w is linearly independent,
    then w1 \not\in span{w \setminus {w1}}