Math 110 Fall 05 Lectures notes 6 Sep 12 Monday Homework due Thursday Sep 15 1 Sec 1 5 1 justify postponed from last time 2bd 8 9 12 postponed from last time 13 17 2 Recall that that the set of symmetric nxn matrices form a subspace W of M n x n F Find a basis of W What is the dimension of W 3 Sec 1 6 1 justify 5 justify 11 12 13 29 31 Goal for the day Understand bases and dimension Express space V in simplist possible way where every vector in V is a unique linear combination of a set of linear independent vectors called a basis Show that if W has a finite basis then all bases have the same number of vectors and this number is called the dimension of V Def If V span S and S is linearly independent we call S a basis of V Ex V F n then S 1 0 0 0 1 0 0 0 0 1 is called the standard basis ASK WAIT Why is this a basis Ex M m x n F S E 11 E 12 E ij E mn where E ij is a matrix where entry ij is 1 and rest 0 S is also called standard basis for same reason as last example Ex V F 2 S 1 0 1 1 is a basis but not standard ASK WAIT why is this a basis Ex P n F polynomials of degree n over F S 1 x x 2 x n is standard basis Ex P F all polynomials over F S 1 x x 2 is a basis not finite Recall Thm 1 from last time Let V be a vector space over F S a subset Then any v in span S can be written as a unique linear combination 1 of vectors in S if and only if S is linearly independent Corollary a subset S of V is a basis for V if and only if each v in V can be written as a unique linear combination of vectors in S Proof If S is a basis for V then by definition V span S and S is linearly independent By Thm 1 this implies that each v in V can be written as a unique linear combination of S If each v in V is a unique linear combination of S then V span S and by the Thm 1 S is linearly independent so that S is a basis Now we move on to constructing bases and showing if finite they all have to have the same number of vectors the dimension Thm 2 If V span S and S is finite then S contains a finite basis S1 of V Proof If S already independent nothing to show so assume S dependent The idea of the proof is simply to start picking vectors out of S to put in S1 continuing as long as S1 is independent As soon as putting any other vector from S into S1 would make S1 dependent we will show that S1 is a basis We can pick vectors out of S in any order we like and this will produce a basis not always the same one Formally to do an induction pick any nonzero s in S set S1 s S1 is independent why remove s from S S2 S s so we can t pick it again repeat if there exists some t in S2 S1 U t is independent add t to S1 S1 remove t from S2 S2 until we can t find any such t such that then S1 U t S2 t Claim 1 This algorithm for building S1 eventually stops Proof Since S is finite there are only finitely 2 many t to pick and since S is dependent we know we will eventually run out of t s to add Claim 2 When we stop S1 is independent Proof by construction S1 is independent at every step Claim 3 When we stop V span S1 Proof at every step of the algorithm S S1 U S2 When we stop S2 is in Span S1 so span S1 span S1 U S2 span S V The next Theorem will be the main tool for show that all bases have the same dimension Thm 3 Replacement Thm Let V be a vector space over F V generated by G G contains n vectors Let L be any other linearly independent subset of V and suppose it contains m vectors Then m n and there is a subset H of G containing m n vectors such that the n vectors in L U H also span V We defer the proof briefly to present Corollary 1 Let V be a vector space over F with a finite generating set Then every basis of V has the same number of vectors Proof of Corollary 1 If V has a finite generating set then it has a finite basis call it G by Thm 2 Let n be the number of vectors in G Let L be any other finite basis of V containing m vectors By Thm 3 m n Reversing the roles of G and L we get n m So m n Def A vector space V is called finite dimensional if it has a finite basis The number of vectors in the basis is called the dimension of V written dim V By the corollary this number does not depend on the choice of basis so the definition makes sense If V does not have a finite basis it is called infinite dimensional Ex dim F n n dim F 1 3 ASK WAIT if V C F R what is dim V Ex dim M m x n F mn Ex P F is infinite dimensional Ex we say dim 0 0 Proof of Replacement Theorem We use induction on m When m 0 so L is the null set then m 0 n and we can simply choose H G to get the spanning set G G U null set with n vectors Now assume the Thm is true for m we need to prove it for m 1 This means that we assume there is a linearly independent subset L of V where L contains m 1 vectors and have to prove 2 things 1 that m 1 n 2 we can find a set H of n m 1 vectors in G such that span L U H V Write L v 1 v 2 v m 1 Then L v 1 v m has just m vectors is linearly independent too why so by the induction hypothesis we can apply the Thm to L conclude that m n and pick n m vectors out of G to get H u 1 u n m where L U H span V Thus v m 1 is in span L U H V so we can write v m 1 as a linear combination v m 1 a 1 v 1 a …