Linear Algebra II
The definition of genius is taking the complex and making it simple
-Albert Einstein
Hi! As I had promised in the previous post I will be discussing on Dependence of vector spaces and few other concepts
LINEAR DEPENDENCY:
A finite set of vectors {a1 , a2, a3…,an} of a vector space V over a field F is said to be linearly dependent in V if there exists scalars c1 , c2 , c3 , … cn not all zero, in F such that
c_1a_1+ c_2a_2+….+c_na_n= ϴ… (1)
And said to be independent in V if the equality (1) is satisfied only when c_1 =c_2 =….=c_n =0
SUBSPACES
Now we are all more or less acquainted with the term subset. To put it simple a set which is a part of(or the total)the universal set.
Similarly a subspace is equivalent to a non empty subset of the given vector space and is closed under vector addition and scalar multiplication on the given vector space.
Linear Span
If S be a finite subset of vector space V over a finite field F. If every vector V can be expressed as a linear combination of vectors from S, then we say S spans V or S generates V and we write L(S)=V
where L(S)is a set of all linear combinations by the vectors of S
Below I have discussed a proof,
Basis
If V be a vector space over a field F. A set S of vectors in V is said to be a basis of V if
•S is linearly independent in V
- S generates V
A FEW PROPOSITIONS ON BASIS OF A VECTOR SPACE:
•There exists a basis for every finitely generated vector space.
•Replacement Theorem
•If {a1 , a2, a3…,an} be a basis of a finite dimensional vector space V over a field F, then any linearly independent set of vectors in V contains at most n vectors
•Any 2 basis of a finite dimensional vector space V have the same number of vectors
- Extension Theorem
Below I’ve tried to illustrate the concept via an example
Prove that the set S={(1,0,1),(0,1,1),(1,1,0)}is a basis of R3
Let a1 = (1,0,1)
a2=(0,1,1) and a3 = (1,1,0)
c1 , c2 , c3 are considered real numbers
c1a1+c2a2+c3a3= ϴ
Then c1(1,0,1)+ c2 (0,1,1)+ c3 (1,1,0)= (0,0,0)
Giving c1 +c2 =0, c2 , c3 =0 , c1 +c3 =0
Hence c1=c2=c3 =0
Hence S is linearly independent
Let ξ= (a,b,c) be an arbitrary vector of R3
Let ξ= r1a1+r2a2+r3a3 for real r1,r2,r3
Then r1+r2= =a, r2 +r3 =b, r1+r3 =c
This is a non homogenous system of 3 equations in r1,r2,r3
The coefficient determinant of the system is
By Cramer’s Rule there exists an unique solution for r1,r2,r3
This proves that ξ ⋲ L(S)
Again S ⊂R3 and L(S) being the smallest subspace of R3 containing S, L(S) ⊂R3
Consequently L(S) =R3
Hence S fulfills both the conditions for being the basis of R3
Dimension
The number of vectors in a basis of a vector space V is said to be the dimension (or rank) of V and is denoted by dimV
The null space {ϴ} is said to be of dimension 0.
A FEW PROPOSITIONS ON DIMENSION OF A VECTOR SPACE:
•Let V be a vector space of dimension n over a field F. Then any linearly independent set of n vectors of V is a basis of V.
- Let V be a vector space of dimension n over a field F. Then any subset of n vectors of V that generates V is a basis of V.
Below I’ve tried to illustrate the concept of Dimension via an example
Find the dimension of the subspace S of defined by
S ={(x,y,z) ⋲ R3 2x+y-z=0}
Let P=(a,b,c) ⋲ S
Then 2a+b+c=0
c=2a+b
Now, P=(a,b,c)
=(a,b,2a+b)
=a(1,0,2)+b(0,1,1)
=Aa+Bb where A=(1,0,2) and B=(0,1,1)
…………….
c1 =c2 =0…. {A,B} linearly independent in S
{A,B} is the basis of R3
Since {A,B} contains two elements
Hence S=2
I hope this short synopsis has helped you to grasp the basic concepts and acted as a cheat sheet for your upcoming exams.
In the next post I will try to dig deeper and talk about rank nullity theorem! For understanding you should check out my post on Linear Transformation. I will be updating that content also .Do post your comments which help me understand how I can improve my work❤
I have used the following books as Reference
- Published by Levant Books — HIGHER ALGERA- Abstract and Linear-
Author — S K Mapa
A geometric approach to Linear Algebra by S Kumaresan
Links used-
•https:en.m.wikipedia.org
•https://www.quora.com
