Finite Dimensional Vector Spaces Flashcards
Linear combination
A linear combination of vectors v₁,…,vₙ is a vector λ₁v₁ + … + λₙvₙ
0 is regarded and a L.C. of …
any set
Subspace spanned by A (span A)
is the set of linear combinations of vectors from A
proposition 5.4 (span A is a subspace)
if A⊆V the the set of vectors span A is a subspace of V
How to find the span of a set of vectors {a,b,c…}
- put all the vectors in a matrix (as rows)
- apply row operations to get into echelon form
- the span of {a,b,c…} is equal to the span of all the non-zero rows
How to determine if a vector is part of a spanning set
if a vector can be written as a L.C. of the vectors in the spanning set then it is inside the span
How to find a vector that is not in a spanning set
- take a vector that is in the span
2. change is last element to 0 (or some other value that isn’t a multiple
Proposition 5.6. if W⊆V then for A⊆W we have
(1) A ⊆ span(A) ⊆ W
(2) A ⊆ span B so span A ⊆ span B
(3) if x is in span A then span A = span A U {x}
linearly dependent
if there is a linear combination λ₁v₁ + … + λₙvₙ = 0 where not all λs are equal to 0
linearly independent
not linearly dependent.
How to prove a set is linearly independent
- take an arbitrary L.C. λ₁v₁ + … + λₙvₙ = 0
- prove all the scalars are equal to 0
OR - put all the vectors in a matrix (as rows)
- use row operations to get into echelon form
- if a row can be made as a linear combination of the others then not linearly independent
if X is linearly independent OR dependent then any subset of X
remains linearly independent or dependent
The empty set is linearly…
independent
Proposition 5.11. linear dependence of subsets
if X is linearly dependent then so is every subset of X
Theorem 5.12. subset of A that is L.I with the same span.
If A⊆V then there is a linearly independent B⊆A s.t. span A = span B
How to find a linearly independent subset B of A with the same spanning set
for each vector in B: if a is linearly independent to current B (start from {}) add a to B. If not linearly independent then discard
Theorem 5.14. subset of A that is L.I with the same span (2nd thrm)
If C⊆A⊆V where C is independent, then there is an independent subset B⊆A containing C for which span A = span B
basis
a subset B⊆V which is linearly independent s.t. spanB=V
The Exchange Lemma
if b∈span(AU{a}) and b∉span(A) then a∈span(AU{b}).
Also if AU{a} is independent then so is AU{b}
Theorem 5.20. Sizes of spans.
if A,B⊆V are independent and A⊆spanB then |A|≤|B|
span S
is the set of all linear combinations of the vectors in S
If C is independent and A is a basis of V then B…
is a basis of V containing C
dimension
If B is a basis of V then |B| is the dimension of V
Corollary 5.21. size of bases.
if A,B⊆V are bases of V then |A|=|B|