vectorspacesunit-2ppt-201129154737 (2).pdf

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

Definition : Basis
A basis of a vector space V is an ordered set of linearly
independent (non-zero) vectors that spans V.
Notation: 1 , , n
β 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

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
 
 

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 ik
k


e
kth component of ei =
1
0
i k
for
i k


 



𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉2 𝑅 𝑜𝑟 𝑅2
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
𝑛 𝑅 𝑜𝑟 𝑅𝑛

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 , 6
x x x
2 2 3
1,1 ,1 ,1
x 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.

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
 
  β
v β then  
i i i
i
c d
 
 β 0
∴ L.I.  uniqueness

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

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 𝑆𝑘 𝑎𝑛𝑑 𝐿 𝑆𝑘 = 𝑉. 𝐻𝑒𝑛𝑐𝑒 𝑆𝑘 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉

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 𝑆1𝑐𝑎𝑛 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

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 𝑆𝑘 ⊂ 𝑆 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑆𝑘 𝑖𝑠 𝐿. 𝐼.
∴ 𝐿 𝑆𝑘 = 𝐿 𝑆 = 𝑉
𝑆𝑘 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑠𝑒𝑡 𝑜𝑓 𝑆 𝑓𝑜𝑟𝑚𝑖𝑛𝑔 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉.

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 𝐚𝟏, 𝐚𝟐, 𝐚𝟑, … , 𝐚𝐧 ∈ 𝐅
such that
𝛂 = 𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝐧𝛂𝐧
𝛂 = 𝐛𝟏𝛂𝟏 + 𝐛𝟐𝛂𝟐 + 𝐛𝟑𝛂𝟑 + ⋯ + 𝐛𝐧𝛂𝐧
If there exists other set of scalars 𝐛𝟏, 𝐛𝟐, 𝐛𝟑, … , 𝐛𝐧 ∈ 𝐅 such that
then 𝐚𝟏= 𝐛𝟏, 𝐚𝟐 = 𝐛𝟐, 𝐚𝟑= 𝐛𝟑,…, 𝐚𝐧= 𝐛𝐧

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 = 𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝐢𝛂𝐢 + ⋯ + 𝐚𝐧𝛂𝐧
for 𝐚𝟏, 𝐚𝟐, 𝐚𝟑, … , 𝐚𝐧 ∈ 𝐅
then the set 𝐚𝟏, 𝐚𝟐, 𝐚𝟑, … , 𝐚𝐧 are called the coordinates

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

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

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

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

⇒ 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.

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 𝑆𝑛 𝑎𝑛𝑑 𝑆𝑚 𝑏𝑒 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉 𝐹 𝑤ℎ𝑒𝑟𝑒
𝑆𝑚 = α1, α2, α3, … , αm
𝑆𝑛 = β1, β2, β3, … , βn
𝑂𝑏𝑣𝑖𝑜𝑢𝑠𝑙𝑦 𝑏𝑜𝑡ℎ 𝑆𝑛 𝑎𝑛𝑑 𝑆𝑚 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 𝑆𝑛 𝑎𝑠 𝑏𝑎𝑠𝑖𝑠 𝑎𝑛𝑑 𝑆𝑚 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡 𝑜𝑓 𝑉
⇒ 𝐿 𝑆𝑛 = 𝑉 𝑎𝑛𝑑 𝑛 𝑆𝑛 = 𝑛
⇒ 𝑛 𝑆𝑛 = 𝑛
⇒ 𝑛 𝑆𝑚 = 𝑚
∴ 𝐵𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑆𝑚 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓 𝑉
⇒ 𝑚 ≤ 𝑛 −− − 1
𝑖𝑖 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑆𝑚 𝑎𝑠 𝑏𝑎𝑠𝑖𝑠 𝑎𝑛𝑑 𝑆𝑛 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡 𝑜𝑓 𝑉
⇒ 𝐿 𝑆𝑚 = 𝑉 𝑎𝑛𝑑 𝑛 𝑆𝑚 = 𝑚
∴ 𝐵𝑦 𝑏𝑎𝑠𝑖𝑠 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑆𝑛 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

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

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

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 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

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 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

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

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 :-
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, … , α𝑚 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, … , α𝑚, 𝛽
𝑆𝑖𝑛𝑐𝑒 𝑆 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑠𝑒𝑡 𝑜𝑓 𝐿𝐼. 𝑣𝑒𝑐𝑡𝑜𝑟𝑠, 𝑆1 𝑖𝑠 𝐿. 𝐷.
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝐚𝟏, 𝐚𝟐, 𝐚𝟑, … , 𝐚𝒎, 𝒃 ∈ 𝐅, not all Zeros such that
𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝒎𝛂𝒎 + 𝒃𝜷 = 𝟎

𝐿𝑒𝑡 𝑏 = 0, 𝑡ℎ𝑒𝑛 𝑤𝑒 ℎ𝑎𝑣𝑒
𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝒎𝛂𝒎 + 𝟎. 𝜷 = 𝟎
𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝒎𝛂𝒎 = 𝟎
⇒ 𝐚𝟏= 𝟎, 𝐚𝟐 = 𝟎, 𝐚𝟑 = 𝟎, … , 𝐚𝒎 = 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
𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝒎𝛂𝒎 + 𝒃. 𝜷 = 𝟎
⇒ 𝒃. 𝜷 = − 𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝒎𝛂𝒎
⇒ 𝜷 = −𝒃−𝟏 𝐚𝟏𝛂𝟏 + 𝐚𝟐𝛂𝟐 + 𝐚𝟑𝛂𝟑 + ⋯ + 𝐚𝒎𝛂𝒎
⇒ 𝜷 = (−𝒃−𝟏𝐚𝟏)𝛂𝟏 + (−𝒃−𝟏𝐚𝟐)𝛂𝟐 + (−𝒃−𝟏𝐚𝟑)𝛂𝟑 + ⋯ + (−𝒃−𝟏𝐚𝒎)𝛂𝒎
⇒ 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

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, … , 𝛾𝑘 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, … , 𝛾𝑘, 𝛼1, 𝛼2, … , 𝛼𝑚
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, … , 𝛾𝑘, 𝛽1, 𝛽2, … , 𝛽𝑛
⇒ dim 𝑊1 = 𝑘 + 𝑚
⇒ dim 𝑊2 = 𝑘 + 𝑛

∴ 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, … , 𝛾𝑘, 𝛼1, 𝛼2, … , 𝛼𝑚, 𝛽1, 𝛽2, … , 𝛽𝑛 = 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
𝑐1𝛾2 + 𝑐2𝛾2 + ⋯ + 𝑐𝑘𝛾𝑘 + 𝑎1𝛼1 + 𝑎2𝛼2 + ⋯ + 𝑎𝑚𝛼𝑚 + 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 = 0
… . . . (𝐼)
⇒ 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 = − 𝑐1𝛾2 + 𝑐2𝛾2 + ⋯ + 𝑐𝑘𝛾𝑘 + 𝑎1𝛼1 + 𝑎2𝛼2 + ⋯ + 𝑎𝑚𝛼𝑚
= 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵1
∈ 𝑊1
⇒ 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 ∈ 𝑊1 … … … … … 1
⇒ 0𝛾2 + 0𝛾2 + ⋯ + 0𝛾𝑘 + 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛
= 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓𝐵2
𝐴𝑔𝑎𝑖𝑛
⇒ 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 ∈ 𝑊2 … … … … … 2

𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 ∈ 𝑊1 ∩ 𝑊2
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑏𝑦 1 𝑎𝑛𝑑 (2)
𝑎𝑠 𝑊1 ∩ 𝑊2 = 𝐿 𝑆 ,
𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠
𝑙. 𝑐. 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑆 𝑜𝑓𝑊1 ∩ 𝑊2
𝐿𝑒𝑡 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 = 𝑑1𝛾2 + 𝑑2𝛾2 + ⋯ + 𝑑𝑘𝛾𝑘
𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑛𝛽𝑛 − 𝑑1𝛾2 − 𝑑2𝛾2 − ⋯ − 𝑑𝑘𝛾𝑘 = 0
𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵2 = 0
⇒ 𝑏1 = 0, 𝑏2 = 0, … , 𝑏𝑛 = 0, 𝑑1 = 0, 𝑑2 = 0, … , 𝑑𝑘 = 0
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒𝑠𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑖𝑛 (𝐼)
𝑐1𝛾2 + 𝑐2𝛾2 + ⋯ + 𝑐𝑘𝛾𝑘 + 𝑎1𝛼1 + 𝑎2𝛼2 + ⋯ + 𝑎𝑚𝛼𝑚 = 0
𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵1 = 0
𝑐1 = 0, 𝑐2 = 0, … , 𝑐𝑘 = 0, 𝑎1 = 0, 𝑎2 = 0, … , 𝑎𝑚 = 0
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝐼 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡
𝑐1 = 0, 𝑐2 = 0, … , 𝑐𝑘 = 0, 𝑎1 = 0, 𝑎2 = 0, … , 𝑎𝑚 = 0, 𝑏1 = 0, 𝑏2 = 0, … , 𝑏𝑛 = 0
∴ 𝑆′ 𝑖𝑠 𝐿. 𝐼 𝑠𝑒𝑡.

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 )

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

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

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

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, … , 𝛼𝑛 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, … , 𝛼𝑛, 𝛽1, 𝛽2, … , 𝛽𝑚 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 + 𝛽𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 𝑜𝑓
𝑉
𝑊

𝐢 𝐓𝐨 𝐏𝐫𝐨𝐯𝐞 𝐒′ 𝐢𝐬 𝐋. 𝐈.
𝑇ℎ𝑒 𝑧𝑒𝑟𝑜 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓
𝑉
𝑊
𝑖𝑠 𝑊.
𝑁𝑜𝑤 𝑏1 𝑊 + 𝛽1 + 𝑏2 𝑊 + 𝛽2 + ⋯ + 𝑏𝑚 𝑊 + 𝛽𝑚 = 𝑊
⟹ 𝑊 + 𝑏1𝛽1 + 𝑊 + 𝑏2𝛽2 + ⋯ + 𝑊 + 𝑏𝑚𝛽𝑚 = 𝑊
⟹ 𝑊 + 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑚𝛽𝑚 = 𝑊
⟹ 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑚𝛽𝑚 ∈ 𝑊
𝐵𝑢𝑡 𝑎𝑛𝑦 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑊 𝑖𝑠 𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵.
𝐿𝑒𝑡 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑚𝛽𝑚 = 𝑎1𝛼1 + 𝑎2𝛼2 + ⋯ + 𝑎𝑛𝛼𝑛
⟹ 𝑏1𝛽1 + 𝑏2𝛽2 + ⋯ + 𝑏𝑚𝛽𝑚 − 𝑎1𝛼1 − 𝑎2𝛼2 − ⋯ − 𝑎𝑛𝛼𝑛 = 0
⟹ 𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐿. 𝐼. 𝑠𝑒𝑡 𝑆 = 0
𝑎1 = 0, 𝑎2 = 0, … , 𝑎𝑛 = 0, 𝑏1 = 0, 𝑏2 = 0, … , 𝑏𝑚 = 0
⟹ 𝑇ℎ𝑒 𝑠𝑒𝑡 𝑆′ 𝑖𝑠 𝐿. 𝐼.

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 = 𝑐1𝛼1 + 𝑐2𝛼2 + ⋯ + 𝑐𝑛𝛼𝑛 + 𝑑1𝛽1 + 𝑑2𝛽2 + ⋯ + 𝑑𝑚𝛽𝑚
= 𝛾 + 𝑑1𝛽1 + 𝑑2𝛽2 + ⋯ + 𝑑𝑚𝛽𝑚
𝑤ℎ𝑒𝑟𝑒 𝛾 = 𝑐1𝛼1 + 𝑐2𝛼2 + ⋯ + 𝑐𝑛𝛼𝑛
𝛾 = (𝑙. 𝑐. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐵)
⇒ 𝛾 ∈ 𝑊
𝑓𝑜𝑟 𝛼 ∈ 𝑉, 𝑊 + 𝛼 ∈
𝑉
𝑊
𝑊 + 𝛼 = 𝑊 + 𝛾 + 𝑑1𝛽1 + 𝑑2𝛽2 + ⋯ + 𝑑𝑚𝛽𝑚
= 𝑊 + 𝑑1𝛽1 + 𝑑2𝛽2 + ⋯ + 𝑑𝑚𝛽𝑚
⇔ 𝑊 = 𝑊 + 𝛾
= 𝑑1(𝑊 + 𝛽1) + 𝑑2(𝑊 + 𝛽2) + ⋯ + 𝑑𝑚 𝑊 + 𝛽𝑚
= (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 .

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vectorspacesunit-2ppt-201129154737 (2).pdf