Linear Combination, Span And Linearly Independent, Dependent Set

Linear Combination,
Span and
Linearly Independent and
Linearly Dependent
-by Dhaval Shukla(141080119050)
Abhishek Singh(141080119051)
Abhishek Singh(141080119052)
Aman Singh(141080119053)
Azhar Tai(141080119054)
-Group No. 9
-Prof. Ketan Chavda
-Mechanical Branch
-2nd Semester

Linear Combination
1 2 3 r
1 1 2 2 3 3 r
i
A vector V is called a Linear Combination of
vectors v , v , v ,......., v
if V can be expressed as
v k k k ..... k
where k are scalar such that 1 i r
rv v v v    
 

Linear Combination
 

  
.v,....,v,v,vofnCombinatio
LinearaisVthenconsistentis1inequationofsystemtheIf2
1k.....kkkv
v,.....,v,v,vofnCombinatioLinearaasVExpress1
:followasisv.....,,v,v,v
orsgiven vectofnCombinatioLinearacalledisVvectoraIf
r321
r332211
r321
r321
 rvvvv

Linear Combination
2
3
2
2
2
1
2
267p
4510p
592pofnCombinatioLinear
aas1588ppolynomialtheExpress1:
xx
xx
xx
xxEx




1 1 2 2 3 3
2 2 2
1 2
2
3
:1 Let p k k k
8 8 15 k (2 9 5 ) k (10 5 4 )
k (7 6 2 )
n
Sol p p p
x x x x x x
x x
  
          
 

Linear Combination
2
1 2 3 1 2 3
2
1 2 3
1 2 3
1 2 3
1 2 3
8 8 15 (2k 10k 7k ) (9k 5k 6k )
(5k 4k 2k )
by comparison we get,
2k 10k 7k 8
9k 5k 6k 8
5k 4k 2k 15
now, turning the above equatio
x x x
x
         
 

   
   
   
 ns into an Augmented Matrix:
82 10 7
89 5 6
155 4 2
 
 
 
  

Linear Combination
1
2 1 3 1
performing R / 2
7
1 5
42
9 5 6 8
5 4 2 15
performing R 9R , R 5R
7
1 5
2 4
51
0 50 28
2
5
31
0 21
2

 
 
 
  
 
 
 
  
 
 
 
 
 
 
 
  

Linear Combination
2
7
2
51 14
100 25
31
2
3 2
7
2
51 14
100 25
479 169
100 25
1
performing R ( )
50
1 5 4
0 1
0 21 5
now performing R 21R
1 5 4
0 1
0 0

 
 
 
 
 
  
 
 
 
 
 
 

Linear Combination
100
3 479
7
2
51 14
100 25
676
479
7
1 2 32
51 14
2 3100 25
676
3 479
performing R ( )
1 5 4
0 1
0 0 1
Hence, here sysem is consistent
k 5k k 4
k k
k
 
 
 
 
 
 

    
   
 

Linear Combination
613
2 479
7347
1 479
by solving above equations
k and
k
which is proven

  
  


Linear Combination
1
2 3
1 1 2 2 3 3
1 2 3
: 2 Express v (6,11,6) as Linear Combination of v (2,1,4),
v (1, 1,3), v (3,2,5).
: 2
- Let v k v k v k v
(6,11,6) k (2,1,4) k (1, 1,3) k (3,2,5)
(6,11,6
n
Ex
Sol
 
  
  
   
1 2 3 1 2 3 1 2 3
1 2 3
1 2 3
1 2 3
) (2k k 3k ) (k k 2k ) (4k 3k 5k )
2k k 3k 6
k 2k 7k 11
5k 7k 7k 7
        
   
   
   

Linear Combination
1
72 4
3 3 3
2 1 3 1
Therefore,
3 2 4 7
2 2 7 12
5 7 7 7
Performing R / 3
1
2 2 7 12
5 7 7 7
Now, performing R 2R and R 5R

 
 
 
  

 
 
 
 
 
  

Linear Combination
72 4
3 3 3
10 13 22
3 3 3
11 1 14
3 3 3
2
72 4
3 3 3
13 11
10 5
11 1 14
3 3 3
11
3 23
1
0
0
Now, R ( 3/10)
1
0 1
0
Now doing R R


 

 
 
 
 
 
 
 
 
 
 
 
 

Linear Combination
1
3
2
3
3
3 2
1
3
306
3 23
1 1 2 11
0 1 6
0 0 4
Now, performing R ( )
1 1 2 11
0 1 6
0 0 1 1
So, we get
k
 
 
  
  
 
  
 
  
  

 

Linear Combination
13 11
2 310 5
1978 23171
2 35 115
306 1978 23171
23 5 115
k k
k and k
Now,
(7,12,7)= (3,2,5) (2, 2,7) (4,6,7)
(7,12,7)=(7,12,7)
Which is proven.
   
  

  


Linear Combination
1 2 3
1 1 2 2 3 3
1 2
5 1
:3 Express the matrix A= as a Linear Combination
1 9
1 1 1 1 2 2
of A , A and A .
0 3 0 2 1 1
:3 Let A=k A k A k A
5 1 1 1 1 1
k k
1 9 0 3 0 2
n
Ex
Sol
 
  
     
            
 
    
         
3
1 2 3 1 2 3
3 1 2 3
2 2
k
1 1
k k 2k k k 2k5 1
k 3k 2k k1 9
  
     
      
          

Linear Combination
1 2 3
1 2 3
3
1 2 3
k k 2k 5
-k k 2k 1
-k 1
3k 2k k 9
The Augmented Matrix will be
1 1 2 5
1 1 2 1
0 0 1 1
3 2 1 9
   
   
  
   

 
 
 
  
 
  

Linear Combination
2 1 4 1
2
Now, performing R R and R 3R
1 1 2 5
0 2 4 6
0 0 1 1
0 1 5 6
Now, doing R / 2
1 1 2 5
0 1 2 3
0 0 1 1
0 1 5 6
  
 
 
 
  
 
    

 
 
 
  
 
    

Linear Combination
1 4
1 2 3
Now, R R
1 1 2 5
0 1 2 3
0 0 1 1
0 0 3 3
The system is Inconsistent. Therefore the given
matrix A is not the linear combination of all three
matrices A , A , A .
 
 
 
 
  
 
   


Linear Combination
2
2
1
2
2
2
3
: 4 Express the polynomial p 9 7 15 as a
Linear Combination of p 2 4
p 1 3
p 3 2 5
Ex x x
x x
x x
x x
   
  
  
  
1 1 2 2 3 3
2 2 2
1 2
2
3
: 4 Let p k k k
9 7 15 k (2 4 ) k (1 3 )
k (3 2 5 )
n
Sol p p p
x x x x x x
x x
  
          
 

Linear Combination
2
1 2 3 1 2 3
2
1 2 3
1 2 3
1 2 3
1 2 3
9 7 15 (2k k 3k ) (k k 2k )
(4k 3k 5k )
by comparison we get,
2k k 3k 9
k k 2k 7
4k 3k 5k 15
now, turning the above equations into
x x x
x
         
 

   
   
   
 an Augmented Matrix:
92 1 3
71 1 2
154 3 5
 
 
 
  

Linear Combination
1 2
2 1 3 1
performing R R
1 1 2 7
2 1 3 9
4 3 5 15
performing R 2R , R 4R
1 1 2 7
0 3 1 5
0 7 3 13
 
   
 
 
  
  
   
 
 
  

Linear Combination
2
51
3 3
3 2
51
3 3
2 4
3 3
performing R / 3
1 1 2 7
0 1
0 7 3 13
now performing R 7R
1 1 2 7
0 1
0 0
Hence, the system is consistent.

   
 
 
  
 
  
 
 
  


Linear Combination
3
3
1 2 3
k 5
2 3 3
2k 4
3 3
3
5 2
2 3 3
2
1
1
k k 2 k 7
k
k 2
k
k 1
k 1 2( 2) 7
k 2
    
  
 
  
 
 
    
  

Linear Combination
2 2 2
2
2 2
Now,
9 7 15 =( 2)(2 4 ) (1)(1 3 )
( 2)(3 2 5 )
9 7 15 = 9 7 15
Which is proven.
x x x x x x
x x
x x x x

        
  
     


Linear Combination
1 2 3
1 1 2 2 3 3
1 2 3
:5 Check whether the following v (6,11,6) as Linear
Combination of v (2,1,4), v (1, 1,3), v (3,2,5).
:5
- Let v k v k v k v
(6,11,6) k (2,1,4) k (1, 1,3) k (3,
n
Ex
Sol

   
  
   
1 2 3 1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
2,5)
(6,11,6) (2k k 3k ) (k k 2k )
(4k 3k 5k )
2k k 3k 6
k k 2k 11
4k 3k 5k 6
      
 
   
   
   

Linear Combination
1 2
2 1 3 1
2 1 3 6
1 1 2 11
4 3 5 6
Now, R R
1 1 2 11
2 1 3 6
4 3 5 6
Now doing R 2R and R 4R
 
 
 
  
 
  
 
 
  
  

Linear Combination
2
161
3 3
3 2
1 1 2 11
0 3 1 16
0 7 3 38
Now, R / ( 3)
1 1 2 11
0 1
0 7 3 38
Now doing R 7R
  
 
  
   
 
  
 
  
   
 

Linear Combination
161
3 3
2 2
3 3
3
3 2
161
3 3
3
1 1 2 11
0 1
0 0
Now, performing R ( )
1 1 2 11
0 1
0 0 1 1
So, we get
k 1
 
 
  
   
 
  
 
  
  

 

Linear Combination
2 1k 5 and k 4
Now,
(6,11,6)=4(3,2,5) ( 5)(2, 2,7) 1(4,6,7)
(6,11,6)=(6,11,6)
Which is proven.
   

   


Span
 
 
1 2 3
1 2 3
The set of all the vectors that are linear combination
of the vectors in the set S= v , v , v ,....., v is
called span of S and denoted by Span S or span
v , v , v ,....., v .
r
r


Span
2
1
2
2 3 2
2
1 2 3 2
1 1 2 2 3 3
2
1 2 3 1
:6 Determine whether the polynomial p 2 ,
p 1 , p 2 span P .
:6
- Choose an arbitary vector b b +b +b P
b=k p k p k p
b +b +b ) k (2
n
Ex x
x x x
Sol
x x
x x
 
    
 
 
  2 2
2 3
2
1 2 3 1 1 3 1 2 3
1 1
1 3 2
1 2 3 3
) k (1 ) k (2 )
b +b b ) (2k ) (2k k ) (2k 3k k )
2k b
2k k b
2k 3k k b
x x x x
x x
    
     
 
  
   

Span
3
1 2 3
Now, matrix will be
2 0 0
2 0 1
2 3 1
det(A)=6 0
Here det(A) 0 therefore matrix is non-Singular
therefore the system is consistent. And so, the
vectors v , v , v span R .

 
 
 
  
 
 

Span 2
1
2 2
2 3 2
2
1 2 3 2
1 1 2 2 3 3
1 2
:7 Determine whether the polynomial p 1 2 ,
p 5 4 , p 2 2 2 span P .
:7
b=k p k p k p
b +b +
n
Ex x x
x x x x
Sol
x x
x
  
       
 
 
2 2 2
3 1 2
2
3
2
1 2 3 1 2 3 1 2 3
2
1 2 3
1 2
b k (1 2 ) k (5 4 )
k ( 2 2 2 )
b +b +b (k 5k 2k ) ( k k 2k )
(2k 4 k 2k )
k 5k 2k
x x x x x
x x
x x x
x
      
  
      
  
   3 1
1 2 3 2
1 2 3 3
b
k k 2k b
2k 4k 2k b

    
   

Span
1
2
3
1 3
3
2
1
2 1 3 1
Therefore,
2 1 2 4 b
1 0 1 1 b
1 1 0 1 b
Performing R R
1 1 0 1 b
1 0 1 1 b
2 1 2 4 b
Now, performing R R and R 2R

 
 
 
  
 
 
 
 
  
  

Span
3
2 3
1 3
3 2
3
2 3
1 2 3
1 2 3 4 2
1 1 0 1 b
0 1 1 2 b b
0 1 0 2 b 2b
Now, R R
1 1 0 1 b
0 1 1 2 b b
0 0 1 4 b b b
The system is consistent for all choices of b.
Therefore vectors p ,p ,p ,p span P .
 
 
 
   
 
 
 
 
   


Span
2
1
2 2
2 3 2
2
1 2 3 2
1 1 2 2 3 3
1 2
:8 Determine whether the polynomial p 1 2 ,
p 5 4 , p 2 2 2 span P .
:8
b=k p k p k p
b +b +
n
Ex x x
x x x x
Sol
x x
x
  
       
 
 
2 2 2
3 1 2
2
3
2
1 2 3 1 2 3 1 2 3
2
1 2 3
1 2
b k (1 2 ) k (5 4 )
k ( 2 2 2 )
b +b +b (k 5k 2k ) ( k k 2k )
(2k 4 k 2k )
k 5k 2k
x x x x x
x x
x x x
x
      
  
      
  
   3 1
1 2 3 2
1 2 3 3
b
k k 2k b
2k 4k 2k b

    
   

Span
1 2
Now, matrix will be
1 5 2
1 1 2
2 4 2
det(A)=0
Here det(A)=0. Therefore matrix is Singular
therefore the system is consistent for some choices
of b. And so, the polynomials p , p

 
   
  


3 2, p span P .

Linear Dependence and Linear
Independence
 1 2 3
1 1 2 2 3 3
1
Let S= v , v , v ,...., v be the non-empty set
such that k v k v k v ...... k v 0 (1)
S is called Linearly Independent set if the system
of equation (1) has trivial solutions (means k 0
r
r r

     


2
,
k 0,....., k 0).
S is called Linearly dependent then the system of
equation (1) has non-trivial solution (means at least
one scalar which is non-zero).
r 


Independence
1 2 3
:9 Check whether the following vectors are
Linearly Independent or Linearly Dependent. (4,1, 2),
( 4,10,2), (4,0,1).
:9
- v (4,1, 2), v ( 4,10,2), v (4,0,1)
n
Ex
Sol


    
1 1 2 2 3 3
1 2 3
1 2 3 1 2 1 2 3
1 2 3
1 2
1 2 3
- Let k v k v k v 0
0 k (4,1, 2) k ( 4,10,2) k (4,0,1)
0 (4k 4k 4k ) (k 10k ) ( 2k 2k k )
4k 4k 4k 0
k 10k 0
-2k 2k k 0
  
    
         
   
  
   

Independence
1 2
2 1 3 1
Therefore,
4 4 4 0
1 10 2 0
2 2 1 0
Performing R R
1 10 2 0
4 4 4 0
2 2 1 0
Now, performing R 4R and R 2R

  
 
 
  
 
  
 
 
  
  

Independence
2
3 2
1 1 2 0
0 2 2 0
0 1 1 0
Performing R / ( 2)
1 1 2 0
0 1 1 0
0 1 1 0
Now, performing R 22R
 
 
  
  
 
 
 
 
  
 

Independence
3
11
3
3
11
1 10 2 0
0 1 0
0 0 3 0
Performing R / 3
1 10 2 0
0 1 0
0 0 1 0
Now,
  
 
 
  

  
 
 
  


Independence
1 2 3
3
2 311
3
2
1
1 2 3
k 10k 2k 0
k k 0
k 0
k 0
k 0
Here k , k , k all are of zero values. Therefore
the system of equation has trivial solution.
Therefore it is Linearly Independent.
   
  
 
 
 


Independence
 2 2 2
2
2 2 2
1 2 3
1 1 2 2 3 3
:10 S= 2 , 2 ,2 2 3 Check whether S is
Linearly Independent or Linearly Dependent in P .
:10
- p 2 , p 2 , p 2 3
- Let k p k p k p 0
n
Ex x x x x x x
Sol
x x x x x x
    
      
  
2 2 2
1 2 3
2
1 3 1 2 3 1 2 3
1 3
1 2
1 2 3
0 k (2 ) k ( 2 ) k (2 2 3 )
0 (2k 2k ) (k k 2k ) (k 2k 3k )
2k 2k 0
k 10k 0
k 2k 3k 0
x x x x x x
x x
       
       
  
  
   

Independence
1 2
2 1 3 1
Therefore,
2 0 2 0
1 1 2 0
1 2 3 0
Performing R R
1 1 2 0
2 0 2 0
1 2 3 0
Now, performing R 2R and R R

 
 
 
  
 
 
 
 
  
  

Independence
2
3 2
1 1 2 0
0 2 2 0
0 1 1 0
Performing R / ( 2)
1 1 2 0
0 1 1 0
0 1 1 0
Now, performingR 2R
 
 
  
  
 
 
 
 
  
 

Independence
2 3
1 2 3
3
2
1
1
1
2
3
k k 0
k +k +2k 0
- taking k t 0
k t
k ( t)+2t=0
k t
k 1
k t 1
k 1
Here the system has trivial solution.
Therefore it is Linearly Dependent
  
 
 
  
  
  
   
        
      


Linear Combination, Span And Linearly Independent, Dependent Set

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