Table of Contents
#flo #ref #hw
1 def of a vector space
- Props of addition and scalar multiplication in FN
- +: comutative, associative, identiy
- every element has an additive inverse
- *: associative, identity
- addition and scalar multiplication, connected by distributive props
- +: comutative, associative, identiy
- let V be a set with an addition and scalar multiplication that satisfy the props,
**addition, scalar multiplication** - addition: assigns an element u+v in V to each pair of elements u, v in V - scalar multiplication: lv with l in f and v in V
**vector space** is V with addition and SCMUL with: - commutativitity - associativity - additive idenitity - additive inverse - multiplicative identity - distibutive properties
- no multiplicative inverse?
- is this how you solve the 0 issue?
- vec, point
- elements of vec space are called vecs or points
- simplest vec space: \(\{0\}\)
- finfin is the set of all seqencues of elements of F
- additive identity: seqnece of all zeros
- vector space can include a set of functions? not quite..
- let S be a set, and FS be the set of functions from S to F
- what?? #review
- let S be the interval [0,1] and F=R
- R^[\0, \1] is the set of real valued function on the interval [0,1]
- ??
- FN -> F1,2,…,n
- Finfin -> F1,2,…
- vector spaces need unique additive inverse
- 0'=0'+0=0+0'=0
- nicer than my proof
- 0'=0'+0=0+0'=0
- unique additive inverse
- w=w+0=w+(v+w')=(w+v)=(w+v)+w'=0+w'=w'
V denotes a vector space over F
1. no multiplicative inverse required? 2. what does the set of functions from S to F mean?
1.1 exercises
- prove that -(-v) = v
- -(-v) = -1(-1v) = (-1*(-1))v = 1v = v
- ab = 0, prove that a or b = 0
- a=0/v = 0, v=0/a = 0
- empty set is not a vector space, it fails to satisfy only of the
reqs. which one?
- no additive idenity
- "there exists an element 0 in v" no there doesn't.
- no additive idenity
homework: KBxSolvingSystems