VS6-Basis and Dimension

7/28/2019 VS6-Basis and Dimension

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Lecture: Basis and Dimension

By

Dr. Mai Duc Thanh


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A basis for a vector space V is a set S of

vectors of V such that

a) S is linearly independent

b) S spans V

The plural of basis is bases

Definition


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Proof. First, we have to show that S is linearly

independent.

Example 1

1 2

1 2

In , let (1,0,...,0), (0,1,0,....,0), ... (0,0,...0,1)

The set { , ,..., } forms a basis in , called

standard basis for

n

n

n

n

n

R e e e

S e e e R

R

1 1 2 2

1 2

1 2 1 2

... 0

(1, 0,...0) (0,1, 0,...,0) ... (0, 0,..., 0,1) 0

( , ,..., ) 0 (0,0,..0) 0, 0,... 0

n n

n

n n

a e a e a e

a a a

a a a a a a


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Next, we have to show that S spans Rn .

This means that any vector in Rn can beexpressed as a linear combination of vectors

in S In fact, for any x in Rn

Thus, S spans Rn

Example 1

1 2

1 1 2 2

( , ,..., ) . Then we can express

x= ...

n

n

n n

x x x x R

x e x e x e


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Example 2

2The set {1, , , ..., } is a basis for ,

called the standard basis for

n

n

n

S x x x P

P


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Let M23 be the vector space of all 2x3matrices

Is a basis ofM23, called the standard basisforM

23

Example 3

1 2 6

1 2 3

4 5 6

{ , ,..., }, where

1 0 0 0 1 0 0 0 1, ,0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0, ,

1 0 0 0 1 0 0 0 1

S M M M

M M M

M M M


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From this point, we take the 2nd Textbook

for the course:

Kreyzig, Advanced Engineering

Mathematics (Part: Linear Algebra)

New textbook


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Let V be a vector space, S a set of vectors

of V, and W the subspace spanned by S. If

S is linearly independent, then S is a basis

for W

Theorem


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Theorem 1: Suppose that a vector space

V has a basis S with n vectors. If m>n,

then any set with m vectors is linearly

dependentUsing Theorem 1 one can easily prove:

Theorem 2: If a vector space V has a

basis S with n elements, then any otherbasis for V also has n elements

Useful Results


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Let V be a vector spacea) The dimension of V is n if V has a basis of

n (n>0, finite) elements

b) The dimension of zero vectorspace iszero

c) V is finite dimensional if the dimension ofV is a nonnegative integer

d) V is infinite dimensional if it is not finitedimensional

Notation: We write dim V for the dimension ofV

Dimension


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Theorem: Let V be a finite dimensional vector space andbe a set of vectors of V.

a) If S is linearly independent, then we can extend S toa basis for V

b) If S spans V and V not zero, then some subset of S isa basis for V

c) If dim V=n, S has n elements, and S spans V, then Smust also be linearly independent and hence S is abasis for V

d) If dim V=n, S has n elements, and S is linearlyindependent, then S must also span V and hence Sis a basis for V

Finite Dimensional Space

1 2{ , ,..., }kS v v v

1 2 1{ , ,..., , ,..., }k k mv v v v v


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Recall: If A is an mxn matrix, then the set ofall solutions of Ax=0, for x in Rn forms asubspace of Rn . This space is called the NullSpace of A, and is denoted by NS(A).

If U is a matrix in row echelon form, thendim NS(U) is equal to the number of freevariables in the equation Ux = 0

If U is of row echelon form obtained from anmxn matrix A by elementary row oprations,then NS(A)=NS(U)

The Null Space NS(A)

(Because ( ) ( )x NS A x NS U


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Find a basis of the null space of the matrix

Remark: Elementary row operations on theaugmented matrix (A|0) of the linear systemAx=0 is the same as the ones on A.Therefore, we will use row operations toreduce A to row echelon form instead of(A|0), for simplicity

Example

1 2 1

2 2 1

1 0 2

A


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Row operations on A:

Example

2 2 1 3 2

1 3

1 2 1 1 2 1 1 2 1

2 2 1 0 2 3 0 2 3

1 0 2 0 2 3 0 0 0

2 0 2, choose z=r as a free variable.

2 3 0 2 3

Then, we have

3 / 2, 3 2

R R R R

R RA

x y z x y z

y z y z

y r x r r r


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Example

Thus, the solution has the form:

( , , ) (2 , 3 / 2, ), for any

(2 , 3 / 2, ) (4, 3,2)2

So NS(A)=span(4,-3,2) and we can choose {(4,-3,2)}

as a basis for NS(A) and thusdim( ( )) 1

x y z r r r r R

rr r r

NS A


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1. Determine whether the given vectors forms a

basis for the given vector spaces

2. Find a basis and dimension of the null space of

the given matrix

CWA

2 4 2 02 1 1 1

, 3 1 2 54 2 2 1

2 2 4 6

A B

3

2 2 22

a) (3, 2,1), (2,3,1), and (5,0,1) in

b) ( ) 1 , ( ) 2 , and ( ) 4 for

u v w R

p x x x q x x x r x x P


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Deadline: 18th May

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VS6-Basis and Dimension