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Linear Combination,Span andLinearly 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 ,......., vif V can be expressed as v k k k ..... kwhere k are scalar such that 1 i r
rv v v v
Linear Combination
. v,...., v, v,v ofn Combinatio
Linear a is V then consistent is 1in equation of system theIf 21 k.....kkkv
v,....., v, v,v ofn CombinatioLinear a as V Express 1:follow as is v.....,, v, v,v
orsgiven vect ofn CombinatioLinear a called is V vector a If
r321
r332211
r321
r321
rvvvv
Linear Combination
23
22
21
2
267p
4510p
592p ofn CombinatioLinear
a as 1588p polynomial theExpress 1:
xx
xx
xx
xxEx
1 1 2 2 3 3
2 2 21 2
23
:1 Let p k k k
8 8 15 k (2 9 5 ) k (10 5 4 )
k (7 6 2 )
nSol p p p
x x x x x x
x x
Linear Combination2
1 2 3 1 2 3
21 2 3
1 2 3
1 2 3
1 2 3
8 8 15 (2 k 10k 7 k ) (9k 5k 6k )
(5k 4k 2k ) by comparison we get,
2 k 10 k 7 k 8 9 k 5k 6k 8 5k 4 k 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 Combination1
2 1 3 1
performing R / 2
71 5 42 9 5 6 8
5 4 2 15
performing R 9R , R 5R
71 52 451 0 50 282
5310 212
Linear Combination
2
7251 14
100 25312
3 2
7251 14
100 25479 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 Combination100
3 479
7251 14
100 25676479
71 2 32
51 142 3100 25
6763 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
6132 479
73471 479
by solving above equations
k and
k
which is proven
Linear Combination1
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 7 k 11 5k 7 k 7 k 7
Linear Combination
1
72 43 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 Combination72 4
3 3 310 13 223 3 3
11 1 143 3 3
2
72 43 3 3
13 1110 5
11 1 143 3 3
113 23
1 0
0
Now, R ( 3 /10)
1 0 1
0
Now doing R R
Linear Combination1323
33 2
13
3063 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 Combination13 11
2 310 5
1978 231712 35 115
306 1978 2317123 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 2k
1 1
k k 2k k k 2k5 1
k 3k 2k k1 9
Linear Combination1 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 51 1 2 1
0 0 1 13 2 1 9
Linear Combination2 1 4 1
2
Now, performing R R and R 3R
1 1 2 50 2 4 6
0 0 1 10 1 5 6
Now, doing R / 2
1 1 2 50 1 2 3
0 0 1 10 1 5 6
Linear Combination1 4
1 2 3
Now, R R
1 1 2 50 1 2 3
0 0 1 10 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 Combination2
21
22
23
: 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 21 2
23
: 4 Let p k k k
9 7 15 k (2 4 ) k (1 3 )
k (3 2 5 )
nSol p p p
x x x x x x
x x
Linear Combination2
1 2 3 1 2 3
21 2 3
1 2 3
1 2 3
1 2 3
9 7 15 (2k k 3k ) (k k 2 k )
(4 k 3k 5k ) by comparison we get,
2 k k 3k 9 k k 2 k 7 4 k 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 Combination2
513 3
3 2
513 32 43 3
performing R / 3
1 1 2 7 0 1
0 7 3 13
now performing R 7 R
1 1 2 7 0 1
0 0
Hence, the system is consistent.
Linear Combination
3
3
1 2 3
k 52 3 3
2k 43 3
3
5 22 3 3
2
1
1
k k 2k 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
1613 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 7 R
Linear Combination
1613 32 23 3
33 2
1613 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 1 k 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
Span2
1
22 3 2
21 2 3 2
1 1 2 2 3 3
21 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 22 3
21 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
31 2 3
Now, matrix will be2 0 0
2 0 12 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 .
Span2
1
2 22 3 2
21 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 23 1 2
23
21 2 3 1 2 3 1 2 3
21 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 2 k ) 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 4 k 2 k 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
Span3
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 .
Span2
1
2 22 3 2
21 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 23 1 2
23
21 2 3 1 2 3 1 2 3
21 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 4 k 2k b
Span
1 2
Now, matrix will be1 5 2
1 1 22 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
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
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 4 R and R 2R
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 22 R
Linear Dependence and Linear Independence
311
3
311
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,
Linear Dependence and Linear Independence
1 2 3
32 311
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.
Linear Dependence and Linear Independence
2 2 2
2
2 2 21 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 21 2 3
21 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
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
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 2 R
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