The document discusses the concepts of basis and dimension in linear algebra, defining a basis as an ordered set of linearly independent vectors that spans a vector space. It includes examples of bases in different vector spaces and explains the criteria for a subset to be a basis, as well as the concept of finite-dimensional vector spaces. Theorems and proofs regarding the uniqueness of representations and the relationship between different bases are also provided.

1. LINEAR ALGEBRA
BASIS AND DIMENSION
MANIKANTA SATYALA
Department of Mathematics
VSM COLLEGE(A), Ramachandrapuram

2. Definition : Basis
A basis of a vector space V is an ordered set of linearly
independent (non-zero) vectors that spans V.
Notation: 1 , , nβ β
Definition :- Basis
A subset S of a vector space V(F) is said to be the basis of V, if
i) S is linearly independent
ii) The linear span of S is V i.e., L(S)=V
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

3. Example :
2 1
,
4 1
B
   
    
   
is a basis for R2
B is L.I. :
2 1 0
4 1 0
a b
     
      
     
→
2 0
4 0
a b
a b
 
 
→
0
0
a
b


B spans R2:
2 1
4 1
x
a b
y
     
      
     
→
2
4
a b x
a b y
 
 
→
 
1
2
2
a y x
b x y
 
 
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

4. Example :
1 2
,
1 4
B
   
     
   
is a basis for R2 that differs from B only in order.
Definition : Standard / Natural Basis for Rn
1 0 0
0 1 0
, , ,
0 0 1
n
     
     
     
     
     
     
1 2, , , n e e e
 i ikk
ekth component of ei =
1
0
i k
for
i k

 

MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

5. 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉2 𝑅 𝑜𝑟 𝑅2
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉3 𝑅 𝑜𝑟 𝑅3
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉4 𝑅 𝑜𝑟 𝑅4
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉𝑛 𝑅 𝑜𝑟 𝑅 𝑛
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

6. Example:
For the function space
 cos sin ,a b a b  
a natural basis is cos , sin 
Another basis is cos sin , 2cos 3sin    
Proof is straightforward.
Example :
For the function space of cubic polynomials P3 ,
a natural basis is 2 3
1, , ,x x x
Other choices can be
3 2
, 3 , 6 , 6x x x
2 2 3
1,1 ,1 ,1x x x x x x     
Proof is again straightforward.
Rule: Set of L.C.’s of a L.I. set is L.I. if each L.C. contains a different vector.
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

7. Example : Matrices
Find a basis for this subspace of M22 : 2 0
0
a b
a b c
c
   
     
   
2
,
0
b c b
b c
c
    
   
   
Solution:
1 1 2 0
,
0 0 1 0
b c b c
     
      
     
∴ Basis is
1 1 2 0
,
0 0 1 0
   
   
   
( Proof of L.I. is
left as exercise )
Theorem :
In any vector space, a subset is a basis if and only if each vector in the space can be
expressed as a linear combination of elements of the subset in a unique way.
Let i i i i
i i
c d   βv β then  i i i
i
c d  β 0
∴ L.I.  uniqueness
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

8. 𝟐. 𝐅𝐢𝐧𝐢𝐭𝐞 𝐃𝐢𝐦𝐞𝐧𝐭𝐢𝐨𝐧𝐚𝐥 𝐕𝐞𝐜𝐭𝐨𝐫 𝐬𝐩𝐚𝐜𝐞
Definition :
A vector space V(F) is said to be finite dimensional if it
has a finite basis
or
A vector space V(F) is said to be finite dimensional if
there is a finite subset S in V such that L(S)=V
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

9. Theorem :-
𝐼𝑓 𝑉 𝐹 𝑖𝑠 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑠𝑒𝑡 𝑜𝑓 𝑉
Proof :- Since V F is finite dimentional vector space
By the definition of finite dimentional vector space
there exists a finite set S such that L S = V
Let S = α1, α2, α3, … , αn
Assume that S does not contains 0 vector
If S is L.I., then S is a Basis set of V.
If S is L.D., then there exists a vector 𝛼𝑖 ∈ S which can be expressed as
linear combination of its preceding vectors
Omitting vector 𝛼𝑖 from S
Let S1 = α1, α2, α3, … , αi−1, αi+1, … , αn ⇒ S1 ⊂ S
𝐵𝑦 𝑘𝑛𝑜𝑤𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 L S1 = L(S)
𝑁𝑜𝑤 𝐿 𝑆 = 𝑉 ⇒ 𝐿 𝑆1 = 𝑉
𝐼𝑓 𝑆1 𝑖𝑠 𝐿. 𝐼. 𝑠𝑒𝑡 𝑡ℎ𝑒𝑛 𝑆1 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉.
𝐼𝑓 𝑆1 𝑖𝑓 𝑖𝑠 𝐿. 𝐷. 𝑡ℎ𝑒𝑛 𝑝𝑟𝑜𝑐𝑒𝑒𝑑𝑖𝑛𝑔 𝑎𝑠 𝑎𝑏𝑜𝑣𝑒 𝑓𝑜𝑟 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑒𝑝𝑠,
𝑤𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑙𝑒𝑓𝑡 𝑤𝑖𝑡ℎ 𝑎 𝐿. 𝐼. 𝑠𝑒𝑡 𝑆 𝑘 𝑎𝑛𝑑 𝐿 𝑆 𝑘 = 𝑉. 𝐻𝑒𝑛𝑐𝑒 𝑆 𝑘 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS

10. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem :-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑎𝑛𝑑 S = α1, α2, α3, … , αm
𝑎 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑉. 𝑇ℎ𝑒𝑛 𝑒𝑖𝑡ℎ𝑒𝑟 𝑆 𝑖𝑡𝑠𝑒𝑙𝑓 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑜𝑟 𝑆 𝑐𝑎𝑛 𝑏𝑒
𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑓𝑜𝑟𝑚 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
Proof :-
Since V F is finite dimentional vector space, it has a finite basis let it be B
Given S = α1, α2, α3, … , αm a linearly independent subset of V.
Let B = β1, β2, β3, … , βn
Now consider the set S1 = S ∪ B = α1, α2, α3, … , αm, β1, β2, β3, … , βn
𝑐𝑙𝑒𝑎𝑟𝑙𝑦 𝐿 𝐵 = 𝑉
𝐸𝑎𝑐ℎ 𝛼 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝛽′ 𝑠 𝑎𝑠 𝐵 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
⇒ 𝑆1 𝑖𝑠 𝐿. 𝐷.
𝐻𝑎𝑛𝑐𝑒 𝑠𝑜𝑚𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑆1 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓
𝑖𝑡𝑠 𝑝𝑟𝑒𝑐𝑒𝑒𝑑𝑖𝑛𝑔 𝑣𝑒𝑐𝑡𝑜𝑟.
𝑇ℎ𝑖𝑠 𝑣𝑒𝑐𝑡𝑜𝑟 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑎𝑛𝑦 𝑜𝑓 𝛼′ 𝑠, 𝑠𝑖𝑛𝑐𝑒 𝑆 𝑖𝑠 𝐿. 𝐼. 𝑠𝑜 𝑡ℎ𝑖𝑠 𝑣𝑒𝑐𝑡𝑜𝑟 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑠𝑜𝑚𝑒 𝛽𝑖
L S1 = L S ∪ B = V 𝑎𝑠 L S ∪ B = 𝐿 𝑆 ∪ 𝐿 𝐵 = 𝐿 𝑆 ∪ 𝑉 = 𝑉

11. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
S2 = S ∪ B = α1, α2, α3, … , αm, β1, β2, β3, … , βi−1, β𝑖+1 … , βn = S1 − βi
now delete the βi from S1
𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑛𝑒𝑤 𝑠𝑒𝑡
𝑜𝑏𝑣𝑖𝑜𝑢𝑠𝑙𝑦 L S2 = L S1 = V.
𝐼𝑓 𝑆2 𝑖𝑠 𝐿. 𝐼, 𝑡ℎ𝑒𝑛 𝑆2 𝑓𝑜𝑟𝑚𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑎𝑛𝑑 𝑖𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑠𝑒𝑡 𝑜𝑓 𝑆
𝐼𝑓 𝑆2 𝑖𝑠 𝐿. 𝐷. 𝑡ℎ𝑒𝑛 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑒 𝑡ℎ𝑖𝑠 𝑝𝑟𝑜𝑐𝑒𝑑𝑢𝑟𝑒 𝑡𝑖𝑙𝑙 𝑤𝑒 𝑔𝑒𝑡 𝑎 𝑠𝑒𝑡 𝑆 𝑘 ⊂ 𝑆 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑆 𝑘 𝑖𝑠 𝐿. 𝐼.
∴ 𝐿 𝑆 𝑘 = 𝐿 𝑆 = 𝑉
𝑆 𝑘 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑠𝑒𝑡 𝑜𝑓 𝑆 𝑓𝑜𝑟𝑚𝑖𝑛𝑔 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉.

12. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝐍𝐨𝐭𝐞: −𝟏
Every basis is a spanning set but converse need not true
If S is Basis of V then L S = V
but If L S = V for S ⊂ V ⇏ S
𝐍𝐨𝐭𝐞: −2
Let S = α1, α2, α3, … , αn be a basis set of a finite dimesional vector space V F
Then for every α ∈ V there exists a unique set of scalars 𝐚 𝟏, 𝐚 𝟐, 𝐚 𝟑, … , 𝐚 𝐧 ∈ 𝐅
such that
𝛂 = 𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚 𝐧 𝛂 𝐧
𝛂 = 𝐛 𝟏 𝛂 𝟏 + 𝐛 𝟐 𝛂 𝟐 + 𝐛 𝟑 𝛂 𝟑 + ⋯ + 𝐛 𝐧 𝛂 𝐧
If there exists other set of scalars 𝐛 𝟏, 𝐛 𝟐, 𝐛 𝟑, … , 𝐛 𝐧 ∈ 𝐅 such that
then 𝐚 𝟏= 𝐛 𝟏, 𝐚 𝟐 = 𝐛 𝟐, 𝐚 𝟑= 𝐛 𝟑,…, 𝐚 𝐧= 𝐛 𝐧

13. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
3. COORDINATES
Definition : Coordinates
Let S = α1, α2, α3, … , αn be a basis set of a finite dimesional vector space V F .
Let β ∈ V be given by
𝛃 = 𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚𝐢 𝛂𝐢 + ⋯ + 𝐚 𝐧 𝛂 𝐧
for 𝐚 𝟏, 𝐚 𝟐, 𝐚 𝟑, … , 𝐚 𝐧 ∈ 𝐅
then the set 𝐚 𝟏, 𝐚 𝟐, 𝐚 𝟑, … , 𝐚 𝐧 are called the coordinates

14. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 1,1,2 , 1.2.5 , 5,3,4 𝑜𝑓 𝑅3 𝑅 𝑑𝑜 𝑛𝑜𝑡 𝑓𝑜𝑟𝑚 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑅3 𝑅
Example:
Solution:
𝐺𝑖𝑣𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 1,1,2 , 1.2.5 , 5,3,4 𝑜𝑓 𝑅3 𝑅
𝐿𝑒𝑡 𝑎, 𝑏, 𝑐 ∈ 𝑅 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝑎. 1,1,2 + 𝑏. 1.2.5 + 𝑐. 5,3,4 = 0 = (0,0,0)
𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑚𝑎𝑡𝑟𝑖𝑥 𝑓𝑟𝑜𝑚 𝑎𝑏𝑜𝑣𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑖𝑠
𝑇ℎ𝑒𝑛 𝑤𝑒 𝑔𝑒𝑡
𝑎 + 𝑏 + 5𝑐 = 0
𝑎 + 2𝑏 + 3𝑐 = 0
2𝑎 + 5𝑏 + 4𝑐 = 0
1 1 5
1 2 3
2 5 4
By reducing the matrix to echelon form
𝑒𝑐ℎ𝑜𝑙𝑒𝑛 𝑓𝑜𝑟𝑚
𝑎 𝑏 𝑐
0 𝑑 𝑒
0 0 𝑓
𝑥
𝑦
𝑧
𝑅1
𝑅2
𝑅3
𝑅3 → 𝑅3 − 2𝑅2
1 1 5
1 2 3
0 1 −2

15. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝑅2 → 𝑅2 − 𝑅1
1 1 5
0 1 −2
0 1 −2
𝑅3 → 𝑅3 − 𝑅2
1 1 5
0 1 −2
0 0 0
Since there are only 2 non zero rows and 3 unknowns
Hence the given vectors are L.D.
Therefore given vectors don’t form basis

16. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝐍𝐨𝐭𝐞: −𝟑
Given set of vectors are L. I. if
1. In the coefficient matrix
𝑁𝑜 𝑜𝑓 𝑈𝑛𝑘𝑛𝑜𝑤𝑛𝑠 = 𝑅𝑎𝑛𝑘 𝑜𝑓 𝑀𝑎𝑡𝑟𝑖𝑥
Rank of Matrix = No of non − zero rows
2. In the systerm of equations all coefficients are zeros
𝑎 + 𝑏 + 5𝑐 = 0
𝑎 + 2𝑏 + 3𝑐 = 0
2𝑎 + 5𝑏 + 4𝑐 = 0
⇒ 𝑎 = 𝑏 = 𝑐 = 0
Given set of vectors are L. D. if
1. In the coefficient matrix
𝑁𝑜 𝑜𝑓 𝑈𝑛𝑘𝑛𝑜𝑤𝑛𝑠 ≠ 𝑅𝑎𝑛𝑘 𝑜𝑓 𝑀𝑎𝑡𝑟𝑖𝑥
2. In the systerm of equations all coefficients are zeros
𝑎 + 𝑏 + 5𝑐 = 0
𝑎 + 2𝑏 + 3𝑐 = 0
2𝑎 + 5𝑏 + 4𝑐 = 0
⇒ 𝑎, 𝑏, 𝑐 𝑛𝑜𝑡 𝑎𝑙𝑙 𝑧𝑒𝑟𝑜𝑠

17. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑠𝑒𝑡 1,0,0 , 1,1,0 , 1,1,1 𝑖𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝐶3 𝐶 .
𝐻𝑒𝑛𝑐𝑒 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 3 + 4𝑖, 6𝑖, 3 + 7𝑖 𝑖𝑛𝐶3 𝐶
Solution:
Example:
𝐿𝑒𝑡 𝑆 = 1,0,0 , 1,1,0 , 1,1,1
𝐿𝑒𝑡 𝑎, 𝑏, 𝑐 ∈ C 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝑎. 1,0,0 + 𝑏. 1.1.0 + 𝑐. 1,1,1 = 0 = (0,0,0)
𝐵𝑦 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑤𝑒 𝑔𝑒𝑡
𝑇ℎ𝑒𝑛 𝑤𝑒 𝑔𝑒𝑡
𝑎 + 𝑏 + 𝑐 = 0
0 + 𝑏 + 𝑐 = 0
0 + 0 + 𝑐 = 0
𝑐 = 0, 𝑏 = 0, 𝑎 = 0
∴ 𝑔𝑖𝑣𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝐿. 𝐼.
𝐿𝑒𝑡 𝛾 ∈ 𝐶3 𝛾 = 𝑥, 𝑦, 𝑧 𝑤ℎ𝑒𝑟𝑒 𝑥, 𝑦, 𝑧 ∈ 𝐶
𝑁𝑜𝑤 𝑥, 𝑦, 𝑧 = 𝑝. 1,0,0 + 𝑞. 1.1.0 + 𝑟. 1,1,1 𝑓𝑜𝑟 𝑝, 𝑞, 𝑟 ∈ 𝐶
= 𝑝, 0,0 + 𝑞, 𝑞, 0 + (𝑟, 𝑟, 𝑟)
= 𝑝 + 𝑞 + 𝑟, 𝑞 + 𝑟, 𝑟

18. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
⇒ 𝑥 = 𝑝 + 𝑞 + 𝑟
𝑦 = 𝑞 + 𝑟
𝑧 = 𝑟
⇒ 𝑟 = 𝑧
𝑞 = 𝑦 − 𝑧
𝑝 = 𝑥 − 𝑦
∴ 𝛾 = 𝑥, 𝑦, 𝑧 = 𝑝. 1,0,0 + 𝑞. 1.1.0 + 𝑟. 1,1,1
𝛾 = (𝑥 − 𝑦). 1,0,0 + (𝑦 − 𝑧). 1.1.0 + 𝑧. 1,1,1
= 𝐿𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑆
∈ 𝐿(𝑆)
∴ 𝐶3 𝐶 ⊆ 𝐿 𝑆 −− − (1)
𝐴𝑠 𝑆 ⊂ 𝐶3 ⇒ 𝐿 𝑆 ⊂ 𝐶3 −− − 2
𝑓𝑟𝑜𝑚 1 & 2
𝐶3 𝐶 = 𝐿 𝑆 & 𝑆 𝑖𝑠 𝐿. 𝐼.
∴ 𝑆 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝐶3 𝐶
𝑁𝑜𝑤 𝑖𝑓 𝑥, 𝑦, 𝑧 = 3 + 4𝑖, 6𝑖, 3 + 7𝑖
𝑇ℎ𝑒𝑛 𝑝 = 3 − 2𝑖, 𝑞 = −3 − 𝑖 & 𝑟 = 3 + 7𝑖
𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟.

19. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem 3:-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒.
𝑇ℎ𝑒𝑛 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑏𝑎𝑠𝑒𝑠 𝑜𝑓 𝑉 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠
Proof :- 𝐿𝑒𝑡 𝑆 𝑛 𝑎𝑛𝑑 𝑆 𝑚 𝑏𝑒 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝐹 𝑤ℎ𝑒𝑟𝑒
𝑆 𝑚 = α1, α2, α3, … , αm
𝑆 𝑛 = β1, β2, β3, … , βn
𝑂𝑏𝑣𝑖𝑜𝑢𝑠𝑙𝑦 𝑏𝑜𝑡ℎ 𝑆 𝑛 𝑎𝑛𝑑 𝑆 𝑚 𝑎𝑟𝑒 𝐿. 𝐼. 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝑉
𝐵𝑦 𝐵𝑎𝑠𝑖𝑠 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑤𝑒𝑤 𝑐𝑎𝑛 𝑠𝑎𝑦 𝑡ℎ𝑎𝑡
𝑛𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝐿. 𝐼 𝑠𝑒𝑡 ≤ 𝑛𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝐵𝑎𝑠𝑖𝑠
𝑖 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑆 𝑛 𝑎𝑠 𝑏𝑎𝑠𝑖𝑠 𝑎𝑛𝑑 𝑆 𝑚 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡 𝑜𝑓 𝑉
⇒ 𝐿 𝑆 𝑛 = 𝑉 𝑎𝑛𝑑 𝑛 𝑆 𝑛 = 𝑛
⇒ 𝑛 𝑆 𝑛 = 𝑛
⇒ 𝑛 𝑆 𝑚 = 𝑚
∴ 𝐵𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑆 𝑚 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
⇒ 𝑚 ≤ 𝑛 −− − 1
𝑖𝑖 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑆 𝑚 𝑎𝑠 𝑏𝑎𝑠𝑖𝑠 𝑎𝑛𝑑 𝑆 𝑛 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡 𝑜𝑓 𝑉
⇒ 𝐿 𝑆 𝑚 = 𝑉 𝑎𝑛𝑑 𝑛 𝑆 𝑚 = 𝑚
∴ 𝐵𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑆 𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
⇒ 𝑛 ≤ 𝑚 −− − 2
𝑓𝑟𝑜𝑚 1 & 2 𝑛 = 𝑚
𝑇ℎ𝑢𝑠 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑏𝑎𝑠𝑒𝑠 𝑜𝑓 𝑉 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

20. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
4. DIMENSION OF A VECTOR SPACE
Definition : DIMENSION OF A VECTOR SPACE
Let V F be the finite dimensional vector space. The Number of elements
in any basis of V is called the dimension of V and denoted by dim V
𝐿𝑒𝑡 𝑆 = α1, α2, α3, … , α10 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉(𝐹)
For Example :-
𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑉 𝐹 = dim 𝑉 = 𝑛 𝑆 = 𝑁𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑏𝑎𝑠𝑖𝑠 𝑆 = 10
𝐍𝐨𝐭𝐞: −𝟒
𝑇ℎ𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑛𝑢𝑙𝑙 𝑠𝑝𝑎𝑐𝑒 = dim 0 = 𝑧𝑒𝑟𝑜
𝐼𝑓 𝑆 = 1,0,0 , 0,1,0 , (0,0,1) 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉3(𝐹)
𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓𝑉3(𝐹) = dim 𝑉3(𝐹) = 𝑛 𝑆 = 𝑁𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑏𝑎𝑠𝑖𝑠 𝑆 = 3

21. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem 4:-
𝐸𝑣𝑒𝑟𝑦 𝑠𝑒𝑡 𝑜𝑓 𝑛 + 1 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑖𝑛 𝑎𝑛 𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑖𝑠
𝐿𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡
Proof :-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑛 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑖. 𝑒., dim 𝑉(𝐹) = 𝑛
𝐿𝑒𝑡 𝑆 = α1, α2, α3, … , αn+1 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑉 𝐹 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 (𝑛 + 1) 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
𝐼𝑓 𝑆 𝑖𝑠 𝐿. 𝐼 𝑡ℎ𝑒𝑛 𝐵𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
𝑆 𝑖𝑡𝑠𝑒𝑙𝑓 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑜𝑟 𝑆 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑓𝑜𝑟𝑚 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛 𝑜𝑓 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡
𝑇ℎ𝑒 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑎𝑛𝑦 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑉
𝐼𝑛 𝑎𝑛𝑦 𝑜𝑓 𝑡ℎ𝑒𝑠𝑒 𝑡𝑤𝑜 𝑐𝑎𝑠𝑒𝑠 𝑆 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑛 + 1 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
𝐵𝑢𝑡 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑚𝑢𝑠𝑡 ℎ𝑎𝑣𝑒 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠.
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑆 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑎 𝐿. 𝐼.
𝐻𝑒𝑛𝑐𝑒 𝑆 𝑖𝑠 𝐿. 𝐷.
𝑖. 𝑒., 𝐸𝑣𝑒𝑟𝑦 𝑠𝑒𝑡 𝑜𝑓 𝑛 + 1 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑖𝑛 𝑎𝑛 𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑖𝑠
𝐿𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡

22. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem 5:-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛 𝑛.
𝑇ℎ𝑒𝑛 𝑎𝑛𝑦 𝑠𝑒𝑡 𝑜𝑓 𝑛 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑖𝑛 𝑉 𝑓𝑜𝑟𝑚𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
Proof :-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑛 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑖. 𝑒., dim 𝑉(𝐹) = 𝑛
𝐿𝑒𝑡 𝑆 = α1, α2, α3, … , αn 𝑏𝑒 𝐿. 𝐼 𝑠𝑒𝑡 𝑜𝑓 𝑉 𝐹 𝑎𝑛𝑑 𝑛 𝑆 = 𝑛
𝐼𝑓 𝑆 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝐵𝑎𝑠𝑖𝑠
𝑡ℎ𝑒𝑛 𝐵𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
𝑆 𝑖𝑡𝑠𝑒𝑙𝑓 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑜𝑟 𝑆 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑓𝑜𝑟𝑚 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
𝐼𝑛 𝑎𝑛𝑦 𝑠𝑢𝑐ℎ 𝑐𝑎𝑠𝑒 𝑆 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
𝐵𝑢𝑡 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑚𝑢𝑠𝑡 ℎ𝑎𝑣𝑒 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠.
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑜𝑢𝑟 Supposition is wrong
𝑎𝑛𝑑 𝑆 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑎 𝑏𝑎𝑠𝑖𝑠

23. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem 6:-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛 𝑛.
𝐿𝑒𝑡 𝑆 𝑏𝑒 𝑎 𝑠𝑒𝑡 𝑜𝑓 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑉 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝐿 𝑆 = 𝑉. 𝑇ℎ𝑒𝑛 𝑆 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉(𝐹)
Proof :-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑛 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑖. 𝑒., dim 𝑉(𝐹) = 𝑛
𝐿𝑒𝑡 𝑆 = α1, α2, α3, … , αn 𝑏𝑒 𝑎 𝑠𝑒𝑡 𝑜𝑓 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑉 𝐹 𝑎𝑛𝑑 𝐿 𝑆 = 𝑉
𝐼𝑓 𝑆 𝑖𝑠 𝐿. 𝐼. 𝑠𝑒𝑡 𝑎𝑛𝑑 𝑎𝑠 𝐿 𝑆 = 𝑉 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑆 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝐵𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉(𝐹)
𝐼𝑠 𝑆 𝑖𝑠 𝐿. 𝐷. 𝑠𝑒𝑡 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎 𝑝𝑟𝑜𝑝𝑒𝑟 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑆 𝑓𝑜𝑟𝑚𝑖𝑛𝑔 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉(𝐹)
𝐼𝑛 𝑠𝑢𝑐ℎ 𝑐𝑎𝑠𝑒 𝑤𝑒 𝑔𝑒𝑡 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
𝐴𝑠 𝑒𝑣𝑒𝑟𝑦 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝑚𝑢𝑠𝑡 𝑐𝑜𝑛𝑡𝑎𝑖𝑛 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
𝑆𝑜, 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑆 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝐿. 𝐷.
𝐻𝑎𝑛𝑐𝑒 𝑆 𝑖𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉

24. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
5. DIMENSION OF A SUBSPACE
Definition : DIMENSION OF A SUBSPACE
Let V F be the finite dimensional vector space and W F be the subspace of V F
The Number of elements in any basis of W F is called the dimension of W
and denoted by dim W
𝐿𝑒𝑡 𝑆 = α1, α2, α3, … , α7 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊(𝐹)
For Example :-
𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑊 𝐹 = dim 𝑊 = 𝑛 𝑆 = 𝑁𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑏𝑎𝑠𝑖𝑠 𝑆 = 7
𝐼𝑓 𝑆 = 1,0,0 , 0,1,0 , (0,0,1) 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊3(𝐹)
= dim 𝑊3(𝐹)= 𝑛 𝑆 = 𝑁𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑏𝑎𝑠𝑖𝑠 𝑆 = 3

25. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem 7:-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛 𝑛 𝑎𝑛𝑑 𝑊 𝑏𝑒 𝑡ℎ𝑒
𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑉. 𝑇ℎ𝑒𝑛 𝑊 𝑖𝑠 𝑎 𝑓𝑖𝑛𝑖𝑡 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑤𝑖𝑡ℎ dim 𝑊 ≤ 𝑛.
Proof :-
𝐿𝑒𝑡 𝑉 𝐹 𝑏𝑒 𝑎 𝑛 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑖. 𝑒., dim 𝑉(𝐹) = 𝑛
𝑒𝑎𝑐ℎ 𝑛 + 1 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑉 𝑓𝑜𝑟𝑚 𝑎𝑛 𝐿𝐷
𝐺𝑖𝑣𝑒𝑛 𝑊 𝑖𝑠 𝑎 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑉 𝐹 𝑒𝑎𝑐ℎ 𝑠𝑒𝑡 𝑜𝑓 𝑛 + 1 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑖𝑛 𝑊 𝑖𝑠 𝑎 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑉
𝑎𝑛𝑑 ℎ𝑒𝑛𝑐𝑒 𝐿. 𝐷.
𝑇ℎ𝑢𝑠 𝑎𝑛𝑦 𝐿. 𝐼. 𝑠𝑒𝑡 𝑜𝑓 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑖𝑛 𝑊 𝑐𝑎𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑚𝑜𝑠𝑡 𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠.
𝐿𝑒𝑡 𝑆 = α1, α2, α3, … , α 𝑚 𝑏𝑒 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝐿. 𝐼. 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑊, 𝑤ℎ𝑒𝑟𝑒 𝑚 ≤ 𝑛.
𝑁𝑜𝑤 𝑤𝑒 𝑠ℎ𝑎𝑙𝑙 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑆 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊.
𝐹𝑜𝑟 𝑎𝑛𝑦 𝛽 ∈ 𝑊 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟
𝑆1 = α1, α2, α3, … , α 𝑚, 𝛽
𝑆𝑖𝑛𝑐𝑒 𝑆 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑠𝑒𝑡 𝑜𝑓 𝐿𝐼. 𝑣𝑒𝑐𝑡𝑜𝑟𝑠, 𝑆1 𝑖𝑠 𝐿. 𝐷.
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝐚 𝟏, 𝐚 𝟐, 𝐚 𝟑, … , 𝐚 𝒎, 𝒃 ∈ 𝐅, not all Zeros such that
𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚 𝒎 𝛂 𝒎 + 𝒃𝜷 = 𝟎

26. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝐿𝑒𝑡 𝑏 = 0, 𝑡ℎ𝑒𝑛 𝑤𝑒 ℎ𝑎𝑣𝑒
𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚 𝒎 𝛂 𝒎 + 𝟎. 𝜷 = 𝟎
𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚 𝒎 𝛂 𝒎 = 𝟎
⇒ 𝐚 𝟏= 𝟎, 𝐚 𝟐 = 𝟎, 𝐚 𝟑 = 𝟎, … , 𝐚 𝒎 = 𝟎 𝒂𝒔 𝑺 𝒊𝒔 𝑳. 𝑰.
𝑇ℎ𝑖𝑠 𝑝𝑟𝑜𝑣𝑒𝑠 𝑡ℎ𝑎𝑡 𝑆1 𝑖𝑠 𝐿. 𝐼. 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑡𝑟𝑎𝑑𝑖𝑐𝑡𝑖𝑜𝑛.
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑏 ≠ 0, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑏−1 ∈ 𝐹 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑏𝑏−1 = 1
𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚 𝒎 𝛂 𝒎 + 𝒃. 𝜷 = 𝟎
⇒ 𝒃. 𝜷 = − 𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚 𝒎 𝛂 𝒎
⇒ 𝜷 = −𝒃−𝟏 𝐚 𝟏 𝛂 𝟏 + 𝐚 𝟐 𝛂 𝟐 + 𝐚 𝟑 𝛂 𝟑 + ⋯ + 𝐚 𝒎 𝛂 𝒎
⇒ 𝜷 = (−𝒃−𝟏 𝐚 𝟏)𝛂 𝟏 + (−𝒃−𝟏 𝐚 𝟐)𝛂 𝟐 + (−𝒃−𝟏 𝐚 𝟑)𝛂 𝟑 + ⋯ + (−𝒃−𝟏 𝐚 𝒎)𝛂 𝒎
⇒ 𝜷 = 𝑳𝒊𝒏𝒆𝒓 𝒄𝒐𝒎𝒃𝒊𝒏𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒐𝒇 𝑺
⇒ 𝜷 ∈ 𝑳(𝑺) ⇒ 𝑳 𝑺 = 𝑾
𝐴𝑠 𝑆 𝑖𝑠 𝐿. 𝐼. 𝑠𝑒𝑡 . 𝐻𝑎𝑛𝑐𝑒 𝑆 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊
∴ 𝑊 𝑖𝑠 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑤𝑖𝑡ℎ dim 𝑊 ≤ 𝑛

27. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem 8:-
𝐿𝑒𝑡 𝑊1 𝑎𝑛𝑑 𝑊2 𝑏𝑒 𝑡𝑤𝑜 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒𝑠 𝑜𝑓 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑉 𝐹 .
𝑇ℎ𝑒𝑛 dim 𝑊1 + 𝑊2 = dim 𝑊1 + dim 𝑊2 − dim(𝑊1 ∩ 𝑊2 )
Proof :-
𝐿𝑒𝑡 𝑊1 𝑎𝑛𝑑 𝑊2 𝑏𝑒 𝑡𝑤𝑜 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒𝑠 𝑜𝑓 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑉 𝐹 .
𝑆𝑖𝑛𝑐𝑒 𝑊1 𝑎𝑛𝑑 𝑊2 𝑎𝑟𝑒 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒𝑠 𝑜𝑓 𝑉,
𝑊1 + 𝑊2 𝑎𝑛𝑑 𝑊1 ∩ 𝑊2 𝑎𝑟𝑒 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒𝑠 𝑜𝑓 𝑉
𝐿𝑒𝑡 dim(𝑊1 ∩ 𝑊2 ) = 𝑘 𝑎𝑛𝑑
𝑆 = 𝛾1, 𝛾2, 𝛾3, … , 𝛾 𝑘 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊1 ∩ 𝑊2
𝑐𝑙𝑒𝑎𝑟𝑙𝑦 𝑆 ⊆ 𝑊1 𝑎𝑛𝑑 𝑆 ⊆ 𝑊2
𝑆𝑖𝑛𝑐𝑒 𝑆 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡 𝑎𝑛𝑑 𝑆 ⊆ 𝑊1 𝑡ℎ𝑒𝑛 𝑏𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
𝑆 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑓𝑜𝑟𝑚 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊1
𝐵1 = 𝛾1, 𝛾2, 𝛾3, … , 𝛾 𝑘, 𝛼1, 𝛼2, … , 𝛼 𝑚
𝑆𝑖𝑛𝑐𝑒 𝑆 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡 𝑎𝑛𝑑 𝑆 ⊆ 𝑊2 𝑡ℎ𝑒𝑛 𝑏𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
𝑆 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑓𝑜𝑟𝑚 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊2
𝐵2 = 𝛾1, 𝛾2, 𝛾3, … , 𝛾 𝑘, 𝛽1, 𝛽2, … , 𝛽 𝑛
⇒ dim 𝑊1 = 𝑘 + 𝑚
⇒ dim 𝑊2 = 𝑘 + 𝑛

28. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
∴ dim 𝑊1 + dim 𝑊2 − dim 𝑊1 ∩ 𝑊2
𝐴𝑠 dim(𝑊1 ∩ 𝑊2 ) = 𝑘 , dim 𝑊1 = 𝑘 + 𝑚 𝑎𝑛𝑑 dim 𝑊2 = 𝑘 + 𝑛
= 𝑘 + 𝑚 + 𝑘 + 𝑛 − 𝑘 = 𝑘 + 𝑚 + 𝑛
𝑁𝑜𝑤 𝑤𝑒 𝑠ℎ𝑎𝑙𝑙 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑠𝑒𝑡
𝑆′ = 𝛾1, 𝛾2, 𝛾3, … , 𝛾 𝑘, 𝛼1, 𝛼2, … , 𝛼 𝑚, 𝛽1, 𝛽2, … , 𝛽 𝑛 = 𝐵1 ∪ 𝐵2 𝑖𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊1 + 𝑊2
𝑎𝑛𝑑 ℎ𝑎𝑛𝑐𝑒 dim 𝑊1 + 𝑊2 = 𝑘 + m + 𝑛
𝐢 𝐓𝐨 𝐩𝐫𝐨𝐯𝐞 𝐭𝐡𝐚𝐭 𝐒′ 𝐢𝐬 𝐋. 𝐈.
𝑁𝑜𝑤
𝑐1 𝛾2 + 𝑐2 𝛾2 + ⋯ + 𝑐 𝑘 𝛾 𝑘 + 𝑎1 𝛼1 + 𝑎2 𝛼2 + ⋯ + 𝑎 𝑚 𝛼 𝑚 + 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 = 0
… . . . (𝐼)
⇒ 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 = − 𝑐1 𝛾2 + 𝑐2 𝛾2 + ⋯ + 𝑐 𝑘 𝛾 𝑘 + 𝑎1 𝛼1 + 𝑎2 𝛼2 + ⋯ + 𝑎 𝑚 𝛼 𝑚
= 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵1
∈ 𝑊1
⇒ 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 ∈ 𝑊1 … … … … … 1
⇒ 0𝛾2 + 0𝛾2 + ⋯ + 0𝛾 𝑘 + 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛
= 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓𝐵2
𝐴𝑔𝑎𝑖𝑛
⇒ 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 ∈ 𝑊2 … … … … … 2

29. 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 ∈ 𝑊1 ∩ 𝑊2
MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑏𝑦 1 𝑎𝑛𝑑 (2)
𝑎𝑠 𝑊1 ∩ 𝑊2 = 𝐿 𝑆 ,
𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠
𝑙. 𝑐. 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑆 𝑜𝑓𝑊1 ∩ 𝑊2
𝐿𝑒𝑡 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 = 𝑑1 𝛾2 + 𝑑2 𝛾2 + ⋯ + 𝑑 𝑘 𝛾 𝑘
𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑛 𝛽 𝑛 − 𝑑1 𝛾2 − 𝑑2 𝛾2 − ⋯ − 𝑑 𝑘 𝛾 𝑘 = 0
𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵2 = 0
⇒ 𝑏1 = 0, 𝑏2 = 0, … , 𝑏 𝑛 = 0, 𝑑1 = 0, 𝑑2 = 0, … , 𝑑 𝑘 = 0
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒𝑠𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑖𝑛 (𝐼)
𝑐1 𝛾2 + 𝑐2 𝛾2 + ⋯ + 𝑐 𝑘 𝛾 𝑘 + 𝑎1 𝛼1 + 𝑎2 𝛼2 + ⋯ + 𝑎 𝑚 𝛼 𝑚 = 0
𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵1 = 0
𝑐1 = 0, 𝑐2 = 0, … , 𝑐 𝑘 = 0, 𝑎1 = 0, 𝑎2 = 0, … , 𝑎 𝑚 = 0
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝐼 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡
𝑐1 = 0, 𝑐2 = 0, … , 𝑐 𝑘 = 0, 𝑎1 = 0, 𝑎2 = 0, … , 𝑎 𝑚 = 0, 𝑏1 = 0, 𝑏2 = 0, … , 𝑏 𝑛 = 0
∴ 𝑆′ 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡.

30. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝐢𝐢 𝐓𝐨 𝐩𝐫𝐨𝐯𝐞 𝐋 𝐒′ = 𝐖𝟏 + 𝐖𝟐
𝐸𝑣𝑒𝑟𝑦 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆′ 𝑖𝑠 𝑎 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑊1 + 𝑊2 𝑖. 𝑒., 𝑆′ ⊂ 𝑊1 + 𝑊2
∴ 𝐿 𝑆′ ⊆ 𝑊1 + 𝑊2
𝑙𝑒𝑡 𝛿 ∈ 𝑊1 + 𝑊2.
∴ 𝛿 = 𝛼 + 𝛽 𝑤ℎ𝑒𝑟𝑒 𝛼 ∈ 𝑊1 𝑎𝑛𝑑 𝛽 ∈ 𝑊2.
𝛿 = 𝑙. 𝑐 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵1 + 𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵2 .
= 𝑙. 𝑐 𝑜𝑓 𝛾′ 𝑠 𝑎𝑛𝑑 𝛼′ 𝑠 + 𝑙. 𝑐. 𝑜𝑓 𝛾′ 𝑠 𝑎𝑛𝑑 𝛽′ 𝑠 .
= 𝑙. 𝑐 𝑜𝑓 𝛾′ 𝑠, 𝛼′ 𝑠 𝑎𝑛𝑑 𝛽′ 𝑠
= 𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑆′
∈ 𝐿 𝑆′
∴ 𝛿 ∈ 𝐿(𝑆′) ⇒ 𝑊1 + 𝑊2 ⊆ 𝐿 𝑆′
∴ L S′ = W1 + W2
𝐻𝑒𝑛𝑐𝑒 𝑆′ 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 W1 + W2
∴ dim 𝑊1 + 𝑊2 = 𝑘 + 𝑚 + 𝑛
𝐻𝑒𝑛𝑐𝑒 dim 𝑊1 + 𝑊2 = dim 𝑊1 + dim 𝑊2 − dim(𝑊1 ∩ 𝑊2 )

31. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
6. QUOTIENT SPACE
6.1 COSET
𝐿𝑒𝑡 𝑊 𝑏𝑒 𝑎 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑎 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑉 𝐹 , 𝑡ℎ𝑒𝑛 𝑓𝑜𝑟 𝑎𝑛𝑦
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝛼 ∈ 𝑉, 𝑡ℎ𝑒 𝑠𝑒𝑡
𝑊 + 𝛼 = {𝑥 + 𝛼 / 𝑥 ∈ 𝑊}
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑅𝑖𝑔ℎ𝑡 𝐶𝑜𝑠𝑒𝑡 𝑜𝑓 𝑊 𝑖𝑛 𝑉, 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑏𝑦 𝛼
𝐿𝑒𝑡 𝑊 𝑏𝑒 𝑎 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑎 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑉 𝐹 , 𝑡ℎ𝑒𝑛 𝑓𝑜𝑟 𝑎𝑛𝑦
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝛼 ∈ 𝑉, 𝑡ℎ𝑒 𝑠𝑒𝑡
𝛼 + 𝑊 = {𝛼 + 𝑥 / 𝑥 ∈ 𝑊}
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝐿𝑒𝑓𝑡 𝐶𝑜𝑠𝑒𝑡 𝑜𝑓 𝑊 𝑖𝑛 𝑉, 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑏𝑦 𝛼
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦
Note:- 𝑖𝑓 𝑉, + 𝑖𝑠 𝑎 𝑎𝑏𝑒𝑙𝑖𝑎𝑛 𝑔𝑟𝑜𝑢𝑝 𝑜𝑓 𝑠𝑢𝑏𝑔𝑟𝑜𝑢𝑝 𝑊, + , 𝑡ℎ𝑒𝑛 𝑏𝑦
𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑥 + 𝛼 = 𝛼 + 𝑥 ∀ 𝑥 ∈ 𝑊 𝑎𝑛𝑑 𝛼 ∈ 𝑉
⇒ 𝑊 + 𝛼 = 𝛼 + 𝑊
𝐻𝑒𝑛𝑐𝑒 𝑊 + 𝛼 𝑖𝑠 𝑠𝑖𝑚𝑝𝑙𝑦 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑠 𝑐𝑜𝑠𝑒𝑡 𝑜𝑓 𝑊 𝑖𝑛 𝑉, 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑏𝑦 𝛼

32. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
COSET PROPERTIES
1. for 0 ∈ V, 0 + W = W
∴ W is itself a coset in V, generated by 0
2. for 𝑥 ∈ W, 𝑥 + W = W
∴ Coset 𝑥 + W = Coset W
4. if α + W and β + W are two cosets of W in V then
𝛼 + 𝑊 = 𝛽 + 𝑊 ⇔ 𝛼 − 𝛽 ∈ 𝑊
3. any two cosets of W in V are either identical or disjoint
𝑖. 𝑒. , 𝐸𝑖𝑡ℎ𝑒𝑟 𝛼 + 𝑊 = 𝛽 + 𝑊, 𝑜𝑟 𝛼 + 𝑊 ∩ 𝛽 + 𝑊 ≠ ϕ
6. QUOTIENT SPACE

33. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
6.2 QUOTIENT SET
6. QUOTIENT SPACE
𝐿𝑒𝑡 𝑊 𝑏𝑒 𝑎 𝑆𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑉 𝐹 . 𝑇ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑐𝑜𝑠𝑒𝑡𝑠 𝑜𝑓 𝑊 𝑖𝑛 𝑉 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑏𝑦
𝑉
𝑊
= 𝑊 + 𝛼 , ∀𝛼 ∈ 𝑉
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑠 𝑄𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑠𝑒𝑡
𝐿𝑒𝑡 𝑊 𝑏𝑒 𝑎 𝑆𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑉 𝐹 . 𝑇ℎ𝑒𝑛 𝑡ℎ𝑒 𝑄𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑠𝑒𝑡
𝑉
𝑊
𝑖𝑠 𝑠𝑎𝑖𝑑 𝑡𝑜 𝑏𝑒 𝑄𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑠𝑝𝑎𝑐𝑒
𝑖𝑓 𝑡ℎ𝑎𝑡 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑠𝑒𝑡 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑎 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑓𝑖𝑒𝑙𝑑 𝐹
𝑓𝑜𝑟 𝑡ℎ𝑒 𝑉𝑒𝑐𝑡𝑜𝑟 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑊 + 𝛼 + 𝑊 + 𝛽 = 𝑊 + 𝛼 + 𝛽 ∀ 𝛼, 𝛽 ∈ 𝑉 𝑎𝑛𝑑
𝑓𝑜𝑟 𝑡ℎ𝑒 𝑉𝑒𝑐𝑡𝑜𝑟 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑎 𝑊 + 𝛼 = 𝑊 + 𝑎𝛼
𝑁𝑜𝑡𝑒: − 𝐼𝑛 𝑄𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑠𝑝𝑎𝑐𝑒 𝑊 + 0 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦 𝑖𝑛
𝑉
𝑊

34. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
Theorem 9:-
𝐿𝑒𝑡 𝑊 𝑏𝑒 𝑎 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝑉 𝐹 , 𝑡ℎ𝑒𝑛
𝐝𝐢𝐦
𝑽
𝑾
= 𝐝𝐢𝐦 𝑽 − 𝐝𝐢𝐦 𝑾 .
Proof :-
𝑆𝑖𝑛𝑐𝑒 𝑉 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙, 𝑊 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑠𝑒𝑡 𝐵 = 𝛼1, 𝛼2, … , 𝛼 𝑛 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑊
∴ dim 𝑊 = 𝑛
𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑒𝑡 𝐵 𝑖𝑠 𝐿. 𝐼. 𝑠𝑒𝑡 𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑓𝑜𝑟𝑚 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉. 𝐿𝑒𝑡 𝑡ℎ𝑒 𝑠𝑒𝑡
𝑆 = 𝛼1, 𝛼2, … , 𝛼 𝑛, 𝛽1, 𝛽2, … , 𝛽 𝑚 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
∴ dim 𝑉 = 𝑛 + 𝑚
∴ dim 𝑉 − dim 𝑊 = (𝑛 + 𝑚) − 𝑛 = 𝑚
𝑁𝑜𝑤 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑆′ = 𝑊 + 𝛽1, 𝑊 + 𝛽2, … , 𝑊 + 𝛽 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓
𝑉
𝑊

35. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝐢 𝐓𝐨 𝐏𝐫𝐨𝐯𝐞 𝐒′ 𝐢𝐬 𝐋. 𝐈.
𝑇ℎ𝑒 𝑧𝑒𝑟𝑜 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓
𝑉
𝑊
𝑖𝑠 𝑊.
𝑁𝑜𝑤 𝑏1 𝑊 + 𝛽1 + 𝑏2 𝑊 + 𝛽2 + ⋯ + 𝑏 𝑚 𝑊 + 𝛽 𝑚 = 𝑊
⟹ 𝑊 + 𝑏1 𝛽1 + 𝑊 + 𝑏2 𝛽2 + ⋯ + 𝑊 + 𝑏 𝑚 𝛽 𝑚 = 𝑊
⟹ 𝑊 + 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑚 𝛽 𝑚 = 𝑊
⟹ 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑚 𝛽 𝑚 ∈ 𝑊
𝐵𝑢𝑡 𝑎𝑛𝑦 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑊 𝑖𝑠 𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵.
𝐿𝑒𝑡 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑚 𝛽 𝑚 = 𝑎1 𝛼1 + 𝑎2 𝛼2 + ⋯ + 𝑎 𝑛 𝛼 𝑛
⟹ 𝑏1 𝛽1 + 𝑏2 𝛽2 + ⋯ + 𝑏 𝑚 𝛽 𝑚 − 𝑎1 𝛼1 − 𝑎2 𝛼2 − ⋯ − 𝑎 𝑛 𝛼 𝑛 = 0
⟹ 𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐿. 𝐼. 𝑠𝑒𝑡 𝑆 = 0
𝑎1 = 0, 𝑎2 = 0, … , 𝑎 𝑛 = 0, 𝑏1 = 0, 𝑏2 = 0, … , 𝑏 𝑚 = 0
⟹ 𝑇ℎ𝑒 𝑠𝑒𝑡 𝑆′ 𝑖𝑠 𝐿. 𝐼.

36. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS
𝐢𝐢 𝐓𝐨 𝐏𝐫𝐨𝐯𝐞 𝐋 𝐒′ =
𝐕
𝐖
.
𝐴𝑠 𝑆′ ⊂
𝑉
𝑊
⇒ 𝐿(𝑆′) ⊆
𝑉
𝑊
(𝟏)
𝑆𝑖𝑛𝑐𝑒 𝑆 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 , 𝑓𝑜𝑟 𝛼 ∈ 𝑉 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑆
𝑖. 𝑒.,
𝛼 = 𝑐1 𝛼1 + 𝑐2 𝛼2 + ⋯ + 𝑐 𝑛 𝛼 𝑛 + 𝑑1 𝛽1 + 𝑑2 𝛽2 + ⋯ + 𝑑 𝑚 𝛽 𝑚
= 𝛾 + 𝑑1 𝛽1 + 𝑑2 𝛽2 + ⋯ + 𝑑 𝑚 𝛽 𝑚
𝑤ℎ𝑒𝑟𝑒 𝛾 = 𝑐1 𝛼1 + 𝑐2 𝛼2 + ⋯ + 𝑐 𝑛 𝛼 𝑛
𝛾 = (𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵)
⇒ 𝛾 ∈ 𝑊
𝑓𝑜𝑟 𝛼 ∈ 𝑉, 𝑊 + 𝛼 ∈
𝑉
𝑊
𝑊 + 𝛼 = 𝑊 + 𝛾 + 𝑑1 𝛽1 + 𝑑2 𝛽2 + ⋯ + 𝑑 𝑚 𝛽 𝑚
= 𝑊 + 𝑑1 𝛽1 + 𝑑2 𝛽2 + ⋯ + 𝑑 𝑚 𝛽 𝑚
⇔ 𝑊 = 𝑊 + 𝛾
= 𝑑1(𝑊 + 𝛽1) + 𝑑2(𝑊 + 𝛽2) + ⋯ + 𝑑 𝑚 𝑊 + 𝛽 𝑚
= (𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑆′)
⇒
𝑉
𝑊
⊆ 𝐿(𝑆′)
⇒ 𝑊 + 𝛼 ∈ 𝐿 𝑆′
(𝟐)
∴ 𝐿 𝑆′ =
𝑉
𝑊
𝑖. 𝑒., 𝑆′ 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓
𝑉
𝑊
. ∴ dim
𝑉
𝑊
= 𝑚 = dim 𝑉 − dim 𝑊 .

37. MANIKANTA SATYALA || LINEAR ALGBRA || BASIS AND DIMENSIONS