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+(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}}