Basis of Vector Space

Basis of Vector SpaceThe set of vectors in a vector space V is called a basis for V if

(1) S is linearly independent

(2) S spans V

Note: Basis of a vector space is not unique

Theorem: If is a basis for a vector space V then every vector in V can be expressed as a linear combination of the vectors in S in exactly one way.

1 2, , ..., nS v v v

1 2, , ..., nS v v v

Ex: 1. Show that the vectors 1 1, 0, 0e i , 2 0,1, 0e j and 3 0, 0,1e k form a basis

for 3R

Let 1 2 3, ,b b b b be an arbitrary vector in 3R and can be expressed as a linear combination of

the given vectors.

1 1 2 2 3 3b k e k e k e

1 2 3 1 2 3, , 1, 0, 0 0,1, 0 0, 0,1b b b k k k

1 2 3, ,k k k

Thus, 1 2 3,e e and e are linearly independent

Hence,

1 2 3,e e and e form a basis for 3R and is known as standard or natural basis for 3R

Equating corresponding components, 1 1 2 2 3 3, ,b k b k b k

Since for each choice of 1 2 3, ,b b b some scalars 1 2 3, ,k k k exists, the given vectors span 3R

Now consider,

1 1 2 2 3 3 0k e k e k e

1 2 31, 0, 0 0,1, 0 0, 0,1 0k k k

1 2 3, , 0k k k

Equating corresponding components, 1 2 30, 0, 0k k k

Ex: 2. Show that the set 21, , , ..., nS x x x is a basis for the vector space nP .

Each polynomial p in nP can be written as

2

0 1 2 ... n

np a a x a x a x

Which is a linear combination of the vectors 21, , , ..., .nx x x Thus, the set S spans nP .

Now consider,

2

1 2 3 1... 0n

nk k x k x k x

Equating corresponding components, 1 2 3 10, 0, 0, ..., 0nk k k k

Thus, the set S is linearly independent.

Hence, the set S is a basis for nP

and is known as standard or natural basis for nP .

Ex: 3. Show that the set 1 2 3, ,S v v v is a basis for 3R , where 1 1, 0, 0v , 2 2, 2, 0v and

3 3, 3, 3v .

Let 1 2 3, ,b b b b be an arbitrary vector in 3R and can be expressed as a linear combination of

the given vectors.

1 1 2 2 3 3 (1)b k e k e k e

1 2 3 1 2 3, , 1, 0, 0 2, 2, 0 3, 3, 3b b b k k k

1 2 3 2 3 32 3 , 2 3 , 3k k k k k k

Ex: 4. Let V be the space spanned by 2

1 cos ,v x 2

2 sin ,v x 3 cos2 .v x show that (1)

1 2 3, ,S v v v is not a basis for V. (2) Find a basis for V.

(1) From trigonometry, we have

2 2cos sin cos 2x x x

1 2 3. . (1)i e v v v

This shows that 3v

can be expressed as a linear combination of 1v

and 2v . Therefore,

1 2 3, ,v v v are linearly dependent.

Hence, S is not a basis for V.

(2) Since from equation (1), any one vector can be expressed as the linear combination of

the remaining two, so any two of vectors 1 2 3, ,v v v will form a basis.

Equating corresponding components,

1 2 3 1

2 3 2

3 3

2 3

2 3 (2)

3

k k k b

k k b

k b

Coefficient matrix,

1 2 3

0 2 3

0 0 3

A

1 2 3

det ( ) 0 2 3

0 0 3

A

6 0

Since the determinant of the coefficient matrix obtained from equation (2) is non-zero, the set

S spans V.

To prove that S is linearly independent, we need to show that

1 1 2 2 3 3 0 (3)k v k v k v

has only a trivial solution, i.e. 1 2 3 0.k k k Comparing equation (3) with equation (1), we

observe that the coefficient matrix of the equations (2) and (3) is same.

Since determinant of the coefficient matrix of equation (3) is non-zero, the system has only

trivial solution.

Hence, S is linearly independent and spans 3R and so is a basis for 3R

0 2, 0,,0 0, 2,S

Ex: Check the following set of vectors form a basis of R3

or not.

0vv 2211 cc

Sol: (i)

solution trivialonly the 021 cc

t.independenlinearly is S

(a) For LI Set,

)0,0,0(0,2,00,0,2 21 cc

)0,0,0(0,2,2 21 cc

)( 2211

3vvuu ccRb

(2,2,2).ufor solution no has bxA

3)( RSspan

From (a) & (b) S is not a basis of R3.

)2,2,2(0,2,00,0,2 21 cc(2,2,2)uLet,

Ex: Check the following set of vectors form a basis of R3

or not.

Sol: (a) Its not a LI Set.

S = {(1,0,0), (0,1,0), (0,0,1), (1,1,1)}

(b) Its a Span set of R3.

From (a) & (b) S is not a basis of R3.

1 2, 3,,1- 6, 5,,3 2,- 1,)(

3 3, 3,,0 2, 2,,0 0, 1,)(

Sii

Si

Ex: Which of the following set of vectors form basis of R3

?

0vvv 332211 ccc

Sol: (i)

0300

0320

0321

solution trivialonly the 0321 ccc

t.independenlinearly is S

(a) For LI Set,

0

300

320

321

A

)( 332211

3vvvuu cccRb

33

232

1321

3

3 2

32

uc

ucc

uccc

0

300

320

321

A

u.every for solution oneexactly has bx A

3)( RSspan

From (a) & (b) S is a basis of R3.

0vvv 332211 ccc

(ii)

0113

0262

0351

solution lnon trivia the

dependent.linearly is S

(a) For LI Set,

0

113

262

351

A

From (a) & (b) S is not a basis of R3.

222

22

x,7xxx,4x3x-2)(

4x3x9x,3,2x2x1)(

Sii

Si

Ex: Which of the following set of vectors form basis of P2 ?

0vvv 332211 ccc

Sol: (i)

0401

0392

0321

solution trivialonly the 0321 ccc

t.independenlinearly is S

(a) For LI Set,

0

401

392

321

A

)( 332211

3vvvuu cccRb

331

2321

1321

4

3 9 2

32

ucc

uccc

uccc

u.every for solution oneexactly has bx A

3)( RSspan

From (a) & (b) S is a basis of P2.

0

401

392

321

A

0vvv 332211 ccc

(ii)

0111

0713

0042

solution lnon trivia the

dependent.linearly is S

(a) For LI Set,

0

111

713

042

A

From (a) S is not a basis of P2.

Next Lecture : Extension and Reduction to Basis