Introduction to Linear Algebra [LA1]
In this series of posts, I`ll be writing about some basics of Linear Algebra [LA] so we can learn together. The book I`ll be using as the material is:
Cabral, M. & Goldfeld, P. (2012). Curso de Álgebra Linear: Fundamentos e Aplicações. Third Edition.
1. Vectors
A vector in ℝⁿ is an ordered list with n real numbers, which means that the vectors (1,2) and (2,1) are different. The common notation used to represent vectors is
A vector (3, 2) ∈ ℝ², as seen above, can be represented as an arrow that goes from (0, 0) to (3, 2). The same vector has infinite representations, we can see below that we have the same vector (3, 2) in different places of the plane.
1.1 Operation with vectors
a) Adding Vectors:
Example: u = (1, 1), v = (2, 2). u + v = (3, 3).
The same can be applied to matrices, given matrix M ∈ M₂x₂, we have
And given M, N ∈ M₂x₂, we define the addition as
b) The Null Vector:
We define the null vector as 0 = (0, 0, 0, …, 0).
As we can see, v + 0 = 0 + v = v .
c) Scalar multiplication:
Given vector
and scalar k ∈ ℝ, the multiplication of k and u goes as follows:
Example: u = (2, 4), k = 2. ku = (4, 8).
Based on those definitions, we know that two vectors u and v are parallels if
u = kv.
The same rules of multiplication applies to matrices, given λ ∈ ℝ, we have
1.2 Vector Space
Formally, with a set V, the operations of +(addition) and *(multiplication), we need to have the following conditions to obtain the vector space:
a) For all u, v ∈ V and λ ∈ ℝ, we have u + v ∈ ℝ and λ*u ∈ V.
b) + properties:
- For all u, v ∈ V, we see that u + v = v + u
- For all u, v, w ∈ V, we see that u + (v + w) = (v + u) + w
- There is a 0 ∈ V, the null vector, that for all v ∈ V we have v + 0 = 0 + v = v
- For all v ∈ V, there is a element -v ∈ V that v + (-v) = (-v) + v = 0
c) * properties:
- For all u ∈ V and λ, μ ∈ ℝ we verify that (λ + μ) * u = λ*u + μ*u
and (λμ)*u = λ*(μ*u).
- For all u ∈ V we have 1*u = u.
d) Mixed Properties (+ and *):
- For all u, v ∈ V and λ ∈ ℝ we have that λ*(u + v) = λ*u + λ*v
2.1 Vector Subspace
On some occasions, we have a vector space that is a subset of some larger vector space, with the same operations of addition and scalar product. If we have a vector space V with + and * operations, if W is a subset of V, we verify that:
- for all u, v ∈ W, we have u + v ∈ W,
- for all u ∈ W and λ ∈ ℝ, we have λ*u ∈ W.
2. Linear Combination and Span
2.1 Linear Combination
Therefore, the same vector can be a linear combination of infinite different vectors. E.g. (3, 3) = 3(1, 1) + 0(-2, -2) = 1(1, 1) -2(-2, -2).
2.2 Span
Example: span{(1,0), (0,1)} are all elements from ℝ², because given (a,b) ∈ ℝ², (a, b) = a(1, 0) + b(0, 1).
Span{(1, 1, 1), (-1, -1, -1)} is equal to span{(1, 1, 1)}, because (-1, -1, -1) is -1(1, 1, 1). Is this case, we say that the vector (-1, -1, -1) is redundant, or that it does not add anything new to span{(1, 1, 1), (-1, -1, -1)}.
2.3 Linearly (in)dependent
Example:
The set of vectors {(1, -2, 1), (1, 0, 1), (1, 1, 1)} is linearly dependent because (1, -2, 1) = 3(1, 0, 1) -2(1, 1, 1). Therefore, (1, -2, 1) is redundant.
2.3 Basis
- β vectors generate V, that is, all vector v ∈ V can be written in the form of
- The vectors of β are linearly independent
Examples:
Considering β = {(1, 1, 1), (1, 2, 2), (1, 3, 3), (1, 2, 1), (2, 1, 1)}, we can see that it is not a basis, because the vectors are not linearly independent.
However, it is possible to obtain a base of ℝ³ with some vectors of β, we just have to eliminate some vectors. We can see that β’ = {(1, 1, 1), (1, 2, 2), (1, 2, 1)} is a basis.
2.4 Dimension
Examples:
I) span{(1, 0, 0), (0, 1, 0), (0, 0, 1)} has 3 dimensions.
II) span{(1, 1, 1), (1, 0, 1), (0, 1, 0)} has 2 dimensions.
We end this lesson here, I hope it was easy to follow. If you have anything to add or comment, feel free to contact me!
Contact
Stay in touch via LinkedIn ou via email rafavsbastos@gmail.com
I put all my codes on https://github.com/rafavsbastos
