From 32f4af5f369fa9f0b2988ecad7797f4bec3661c3 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Tue, 21 Sep 2021 17:12:46 -0400 Subject: notes and homework --- li/06_basis | 40 ++++++++++++++++++++++++++++++++++++++++ 1 file changed, 40 insertions(+) create mode 100644 li/06_basis (limited to 'li/06_basis') diff --git a/li/06_basis b/li/06_basis new file mode 100644 index 0000000..2a103b7 --- /dev/null +++ b/li/06_basis @@ -0,0 +1,40 @@ +(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}} -- cgit