Linear Combination, Span And Linearly Independent, Dependent Set

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


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


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


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


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


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


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


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


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,


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.


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


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


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


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

Engineering

Transcript of Linear Combination, Span And Linearly Independent, Dependent Set