In this presentation, the topic of Linear Combination, Span and Linearly Independent and Linearly Dependent Sets have been discussed. The sums for each topic have been given to understand the concept clearly for viewers.

1. 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

2. 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    
 

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

4. 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
  
          
 

5. 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
 
 
 
  

6. 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

 
 
 
  
 
 
 
  
 
 
 
 
 
 
 
  

7. 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

 
 
 
 
 
  
 
 
 
 
 
 

8. 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
 
 
 
 
 
 

    
   
 

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

10. 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
        
   
   
   

11. 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

 
 
 
  

 
 
 
 
 
  

12. 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


 

 
 
 
 
 
 
 
 
 
 
 
 

13. 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
 
 
  
  
 
  
 
  
  

 

14. 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.
   
  

  


15. 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
  
     
      
          

16. 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
   
   
  
   

 
 
 
  
 
  

17. 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
  
 
 
 
  
 
    

 
 
 
  
 
    

18. 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 .
 
 
 
 
  
 
   


19. 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
  
          
 

20. 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
 
 
 
  

21. 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
 
   
 
 
  
  
   
 
 
  

22. 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.

   
 
 
  
 
  
 
 
  


23. 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
    
  
 
  
 
 
    
  

24. 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

        
  
     


25. 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
      
 
   
   
   

26. 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
 
 
 
  
 
  
 
 
  
  

27. 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
  
 
  
   
 
  
 
  
   
 

28. 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
 
 
  
   
 
  
 
  
  

 

29. 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.
   

   


30. 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


31. 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
    
     
 
  
   

32. 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 .

 
 
 
  
 
 

33. 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
- Choose an arbitary vector b b +b +b P
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

    
   

34. 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

 
 
 
  
 
 
 
 
  
  

35. 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 .
 
 
 
   
 
 
 
 
   


36. 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
- Choose an arbitary vector b b +b +b P
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

    
   

37. 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 .

38. 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 


39. Linear Dependence and Linear
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
  
    
         
   
  
   

40. Linear Dependence and Linear
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

  
 
 
  
 
  
 
 
  
  

41. Linear Dependence and Linear
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
 
 
  
  
 
 
 
 
  
 

42. Linear Dependence and Linear
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,
  
 
 
  

  
 
 
  


43. Linear Dependence and Linear
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.
   
  
 
 
 


44. Linear Dependence and Linear
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
       
       
  
  
   

45. Linear Dependence and Linear
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

 
 
 
  
 
 
 
 
  
  

46. Linear Dependence and Linear
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
 
 
  
  
 
 
 
 
  
 

47. Linear Dependence and Linear
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
  
 
 
  
  
  
   
        
      
