Term 2.5 Content Flashcards
What is a linear combination
Given a list of vectors v1, . . . , vm ∈ Rn, a linear combination of v1, . . . , vm is an expression of the form x1v1 +···+xmvm
with x1,…,xm ∈R.
how do you check if a vector can be written as a linear combinations of other vectors?
use Gaussian elimination. We first write down the corresponding augmented matrix and then use elementary row operations to transform it to reduced echelon form.
What are the properties of linear combinations
- the zeroth vector can always be written as a linear combination
- Any of the vectors vi (for 1 ≤ i ≤ m) can be written as a linear
combination of v1, . . . , vm. - If a and b can be written as linear combinations of v1, . . . , vm,
then so can a + b, and also αa for any α ∈ R.
What does it mean if a vectors v1…vm are linearly independent
c1v1 + c2v2…cmvm = 0, where all coefficients are 0
What is the algorithm to test for linear independence
Given a list of vectors v1,…,vm in Rn, form the n×m matrix
whose columns are v1, . . . , vm. Apply Elementary Row Opera- tions to put this matrix in REF. If the REF matrix has a leading entry in every column then v1, . . . , vm are linearly independent; otherwise they are linearly dependent (and we can find a linear dependence between them by putting the matrix in RREF and reading off the general solution of the corresponding system of linear equations).
what’s an easy way to prove a vector isn’t linearly independent
Let v1,…,vm be a list of vectors in Rn. If m > n, then v1, . . . , vm cannot be linearly independent.
What does a mean for vectors to span a space
Given a list of vectors v1, . . . , vm in Rn, we say that v1,…,vm span Rn if every b ∈ Rn can be written (in at least one way) as a linear combination of v1, . . . , vm.
How can you test if vectors span
To test whether v1,…,vm ∈ Rn span Rn, form the n×m ma-
trix whose columns are v1,…,vm, and apply Elementary Row Operations to put it in REF. If there are no zero rows in this REF matrix (so every row contains a leading entry) then v1,…,vm span Rn. If one or more rows of the REF matrix consists entirely of zeros, then v1,…,vm do not span Rn. (We can find a specific b which cannot be written as a linear combination of v1, . . . , vm by the method of Example 5.22).
what’s an easy way to prove vectors won’t span
Let v1,…,vm be a list of vectors in Rn. If m < n, then v1,…,vm cannot span Rn.
what does it mean. for vectors to form a basis
it means they are both linearly independent and span
Main Theorem for Rn.
- Every basis of Rn contains exactly n vectors.
- For n vectors v1, . . . , vn in Rn, the following are equivalent
-v1-vn: span, are linearly independent and are a basis
Let K=R or C. Av ector space over K is a set V together with an addition function V × V → V (so we have v + w ∈ V for all v, w ∈ V) and a scalar multiplication function K × V → V (so we have αv for all α ∈ K and v ∈ V) such that the following conditions hold:
u+(v+w) = (u+v)+w
u+v=v+u
there is a zero element
for each v∈V ,there is an element −v ∈ V so that v +(−v) = 0;
α(βv)=(αβ)v
1v = v
α(u+v)=(αu)+(αv)
(α+β)v=(αv)+(βv)
what is the definition of a subspace
A subspace of V is a subset W ⊆ V such that
(a) 0∈W;
(b) v+w∈Wforallv,w∈W;
(c) αw∈Wforallα∈Kandw∈W.
(Thus W is a subspace if it contains 0 and is closed under the two basic operations on V: addition and scalar multiplication.)
Let V, W be vector spaces over K (= R or C). A function θ : V → W is called a linear map (or linear transforma- tion) if
θ(u+v) = θ(u)+θ(v) for all u, v ∈ V;
θ(αv)=αθ(v) for all α∈K and all v∈V.
Let V be a vector space over K with a basis v1,…,vm
consisting of m elements. Then there is a linear isomorphism θ : V → Km.
Let V be a vector space over K with a basis v1,…,vm
consisting of m elements. Then there is a linear isomorphism θ : V → Km.
