Chapter 4 (Vector Spaces) Flashcards
When is a set closed under op *
If a * b is in A (for all a and b in A)
Axiom 1
u + v is in v (closed under +)
Axiom 2
u + v = v + u (commutative)
Axiom 3
u + (v + w) = (u + v) + w (associative)
Axiom 4
0 + u = u + 0 = u
–> 0 is an object in V (zero vector)
Axiom 5
u + (-u) = (-u) + u = 0
–> each u in V has a has a -u in V
Axiom 6
ku is in V (closed under scalar multiplication
Axiom 7
k(u+v) = ku + kv
Axiom 8
(k+m)u = ku + mu
Axiom 9
k(mu) = (km)u
Axiom 10
1u = u
What are the conditions for W to be a subspace of V
If under + and scalar multiplication on V, W is also a vector space
REMEMBER!
- If axioms 2, 3, 7, 8, 9 are valid in V –> also valid in W
- Just have to check 1 and 6
1 –> u + v is in W
6 –> ku is in W
How can vector W be a linear combination of vectors v1,v2,…,vr
If w = k1v1 + k2v2, … +…krvr
–> linear combination of v1,v2,…vr
What is a space spanned by v1,v2…,vr (Set S)
Subspace W of all linear combinations of v1,v2,…vr
–> W = span(S) = {k1v1 + k2v2 …+krvr)
span {v1} and eq
line through origin
–> span{v1} = tv1 (k1v1)
–> line // to v1
span {v1, v2} and eq
plane through origin // to v1 and v2
–> span {v1, v2} = {t1v1 + t2v2} (k1v1+ k2v2)
What is a linearly independent set (2)
- If k1v1 + k2v2 +…+ knvn = 0 has only the trivial soln
–> if k1=0, k2 =0, …, kn =0
–> spanned space (subspace) has only the trivial soln - (v1 NOT = kv2)
If no other vector is expressed as a linear combination of other vectors
What is a linearly dependent set (2)
- If k1v1 + k2v2 +…+ knvn = 0 has other soln than trivial
–> space spanned (subspace) has other solutions - v1 = kv2
If at least one vector can be expressed as a linear combination of other vectors
Give an example of a linearly dependent and independent set
- Linearly dependent: Finite set containing a zero vector
- Linearly independent: Set with 2 vectors that are not scalar multiples of each other
