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