Chapter 1
Linear Algebra
1.1 Basic Stipulations
a linear space: two sets, three operations.
{X,+} is a Abelian group;
{A,+,∗} is a field;
×:A×X→X.
(×:Z ×X→X has implicated such an object.)
3 operations: +, *, ×. if{x, y} ⊂X,{α, β} ⊂A, then we have 3 combina- tions of these operations:
(α+β)×x= (α×x) + (β×x) (α∗β)×x=α∗(β×x) α×(x+y) = (α×x) + (β×y)
then, we want to know what constructs and properties could such an object (linear space) possess.
1.2 Basic Structure
group and field
dimension and subspace
for 0 of group {X,+}, what does x+y+, ...,+z = 0 means?
∀x∈X,0∗x= 0,0∈A,0∈X;
∀α∈A, α∗0 = 0,0∈X; if α6= 0, x6= 0,then α∗x6= 0;
but, α∗x+β∗y = 0 is possible! when∃α, β, x, y 6= 0.
then, what does that means?
means that
1
2 CHAPTER 1. LINEAR ALGEBRA Lemma 1.2.1 ∀γ ∈ A,{z|z = γ∗x} is a linear space; any such y ∈ {z}; x and y is symmetry in the lemma; {Z} is a subgroup of X.
if ∃ y is not ∈ {z}, then ∀α, β 6= 0, α∗x+β ∗y 6= 0, at this situation, α∗x+β∗y+γ∗z = 0 is possible!
what does this means?
means that:
Lemma 1.2.2 ∀α, β ∈ A,{z|z = α∗x+β ∗y} is a linear space; y and z is symmetry in this lemma; and {z} is a subgroup of X.
let’s go on! if {z}=X, and z can be expressed as z =α∗x+...+β∗y
then we can say{x, ..., y}is bases ofX. the number of the elements of{x, ..., y}
is the dimension number ofX.
because of the formx+y+, ...,+z = 0 can be used to generate all the space X, we name such {x, y, ...z} is linear dependent. and if z = x+...+y, we namex+...+y is a linear combination of z.
Theorem 1.2.3 any vector z of Xcan be expressed as its bases’s unique linear combination.
coordinate transform
ifXis a n-dimension linear space, then any non-linear dependent n elements of Xcan be used as its base.
isomorphism
How to retain the structure of a linear space?
