notes 4.3 and 4.4

7/31/2019 Notes 4.3 and 4.4

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MATH 2250 Linear Algebra and Differential Equations

1

2/3u

7/6v

u

v

w

Chapter 4: Vector Spaces

This set of notes combines sections 4.3 and 4.4.

Key Terminology and Concepts for these sections:

Vector SpaceNullspace of a matrix A

Linear Combination of a set of vectors

Subspace of a Vector Space

Linear Independence

Linear Dependence

Span of a set of vectors

4.3 Linear Combinations and Independence of Vectors

The following examples illustrate the concepts and process of this section.

Example 1: Determine ifw = (3,-1) is a linear combination ofu = (1,2) and v = (2,-2).

We seek scalars 21 andcc such that wvu 21 cc . That is,

1

3

2

2

2

121 cc OR

122

32

21

21

cc

cc

Either way, we can view this as bAx and find the solution as bAx 1 .

Thus, we have (see the picture at right)

673

2

1

3

12

22

6

1

2

1

c

c

so

wvu1

3

2

2

6

7

2

1

3

2

6

7

3

2

This is a fundamental representation of the idea of a linear combination. If two vectors form aplane and a third vector is a linear combination of the first two, then it lies in the same

plane. This is true in all dimensional spaces.

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Example 2: Determine if the vector w = (-3,2,1) is a linear combination of the vectors

u = (1,1,3) and v = (1,-1,2).

This question is essentially asking if w lies in the same plane as the vectors u and v.

Solution: We seek scalars 21 and cc such that wvu 21 cc . That is,

1

2

3

2

1

1

3

1

1

21 cc

or in augmented matrix form,

123

211

311

.

This matrix can be row reduced to the echelon form

1500

520

311

.

In this way, we see that the system has no solution. We conclude that w is nota linear

combination ofu and v. Hence, w does not lie in the plane formed by u and v.

NOTE:We could have used the determinant of the above matrix to answer this question. Why?

Example3: Can we find scalars 21 and cc such that w = (2,5,-26,11) is a linear combination of

u = (1,3,-15,7) and v = (1,4,-19,10)? Look at the vector (or matrix form) of this question:

11

26

5

2

10

19

4

1

7

15

3

1

21 cc OR

11

26

5

2

107

1915

43

11

2

1

c

c.

What do you expect? WHY?

Try solving the system of equations. What do you find. What does this mean?

Take the first two rows and solve the 2x2 system

5

2

43

11

2

1

c

c

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Definition: The span of a set of vectors kvvv ,,, 21 is the set of all linear combinations of

kvvv ,,, 21 :

span{v1, v2, v3, , vk} = {x | x = c1v1 + c2v2 + c3v3+ + ckvk}

Theorem 1: The Span of a Set of Vectors

Let kvvv ,,, 21 be a vectors in a vector space V. (There is no restriction on this set.) Then the

set Wof all linear combinations of kvvv ,,, 21 is a subspace ofV.

Proof: We must show that Wis closed under scalar multiplication and vector addition. Let u and

w denote vectors in W. Since u and w are members ofW, there are scalars kaaa ,,, 21 and

kbbb ,,, 21 such that

kkaaa vvvu 2211 and kkbbb vvvw 2211 .

For any scalar t, we have

kk

kk

tatata

aaatt

vvv

vvvu

2211

2211 )(

which is clearly a linear combination of kvvv ,,, 21 so it is in W. Also,

kkk

kkkk

bababa

bbbaaa

vvv

vvvvvvwu

)()()(

)()(

222111

22112211

which is in W. We conclude that the set Wof linear combinations of kvvv ,,, 21 is a subspace

ofV.

As a reviewDefinition We say that the vectors nvvvv ,,,, 321 are linearly INDEPENDENT if the ONLY

solution to

0vvvv nncccc 332211

is 021 nccc . That is, the only solution is the trivial solution.

Fundamental Example: The standard unit vectors

)1,0,,0,0(

)0,,0,1,0(

)0,,0,0,1(

2

1

ne

e

e

in nR are linearly independent. This is easily seen because the only solution to

)0,,0,0(),,,( 212211 nnn cccccc eee

Is the trivial solution 021 nccc . It should be apparent that they also span Rn. We

refer to these vectors as the standard basis for Rn.

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Comment: It should be apparent that a subset of any set of linearly independent set of vectors is

linearly independent.

Question: How many vectors in nR can be linearly independent?

We will answer this question using the shape of corresponding matrices. By the previousexample, it should be apparent that there can be up to n linearly independent vectors in nR .

What if there are n + 1 vectors from nR ? Consider the matrix equation Ax = 0 resulting from

forming a matrix whose columns are the n + 1 vectors and ),,,( 21 nccc x is the vector of

unknown scalars nccc ,,, 21 . Thus, A is an nx (n + 1) matrix with more columns than rows

and the corresponding augmented matrix has n rows and n + 2 columns. The point is that thesystem cannot have a unique solution so the set must be linearly dependent.

Therefore, the maximum number of linearly independent vectors in nR is n.

For k< n vectors inn

R , a set ofkvectors inn

R is linearly independent if and only if the n xkmatrix of vectors contains some kxksubmatrix with nonzero determinant.

Example 4: To take this idea further, can we find scalars 4321 andc,c, cc such that w = (3,-1) is a

linear combination ofu = (1,2), v = (2,-2), x = (2,-5), and y = (-3,7)?

Write out the matrix form of the resulting system of equations and explain your answer in terms

of the number of equations and number of unknowns.

This results in a system of 2 equations with 4 unknowns. Since this problem is an extension of

Example 1, we know that the system has a solution. Therefore, we can conclude that this system

has infinitely many solutions. In fact, there are 2 free variables (c3 and c4).

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4.4 Bases and Dimension for Vector Space

Definition: BASISA finite set of vectors Sin a vector space Vis called a basis for Vprovided that

(a) the vectors in Sare linearly independent(b) the vectors in SspanV.

Furthermore, the number of vectors in any such set Sfor Vis the dimension ofV.

The standard basis for nR consists of the n vectors

e1=(1,0,,0), e2=(0,1,0,,0),, en=(0,,0,1).

A basis has the following additional characterizations:

(i) a basis is a maximal set of linearly independent vectors in V

(ii) a basis is a minimalspanning set for V

Eventually, our work will be directed at describing the set of solutions of a lineardifferentialequation or system of linear differential equations.

The following example contrasts the difference between representing the subspace of all

solutions to a homogeneous system of linear equations and identifying a basis for the

subspace.

Example 5: Find a basis for the subspace of all solutions of the homogeneous system

0112652

010194

07153

wzyx

wzyx

wzyx

Viewing the system as Ax = 0, the coefficient matrix A is row equivalent to the RREF

0000

3410

2301

We see that the variablesz and w are free variables so

x = 3z + 2w and y = 4z3w

Therefore, lettingz = s and w = t, the set of all solutions is given by

solutionsallofsubspacetheisThis

1

0

3

2

0

1

4

3

34

23

ts

t

s

ts

ts

w

z

y

x

x

or x = su + tv where u =(3,4,1,0) and v = (2,-3,0,1).

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We will refer to the set of all solutions to Ax = 0 as the nullspace ofA. Therefore, a basis for

the nullspace ofA is the set )1,0,3,2(),0,1,4,3( . In this case, a basis for the nullspace (solution

space ofAx = 0) is a set with two vectors. The solution space is a two dimensional plane

spanned by u and v.

The nullspace ofA is the subspace of 4R for which multiplication by A results in 0 (Ax = 0).We say that Aannihilates the vectors x in the nullspace.

Algorithm for Finding the Nullspace Ax = 0

1. Find the RREF for the n x n matrix A

2. Identify the rleading variables and nrfree variables3. Solve for the leading variables in terms of the free variables.

4. Rename (relabel) the free variables using parameters (s, t, etc. OR t1, t2, t3, etc.)

5. Write the parameterization of the nullspace in vector form with the parameters s, t, etc.

factored out as scalars (like su + tv)

Note: The set of vectors u and v form a basis for the nullspace.

Example: Find a basis for the plane 3x5y + 2z = 0.

Example: Is the plane 3x5y + 2z = 5 a subspace ofR3? Explain your reasoning.

Example: Are the vectors u = (1, 0, -1, 1), v = (1, 0, 1, 1), and w = (2, 3, 0, -1) linearlyindependent? Justify your answer. If yes, do they form a basis for the corresponding vectorspace ( )?

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Example: Find a basis for R4

containing the previous vectors u, v, and w.

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Algorithm for Finding the Nullspace Ax = 0

1. Find the RREF for the n x n matrix A

2. Identify the rleading variables and nrfree variables

3. Solve for the leading variables in terms of the free variables.

4. Rename (relabel) the free variables using parameters (s, t, etc. OR t1, t2, t3, etc.)5. Write the parameterization of the nullspace in vector form with the parameters s, t, etc.

factored out as scalars (like su + tv)

Note: The set of vectors u and v form a basis for the nullspace.

Comments: Keep in mind that the role of a basis is a representation of a vector space (or

subpace). We do not always want to use the standard basis because in most applications we use

specific linear transformations in the vector space to stretch, reflect, rotate, shear, etc. the space.

As such, we want to represent the vector space using vectors whose directions correspond to thestretches, reflections, etc.

Given a basis {v1, v2, v3, , vk} for a vector space (or subspace), for each vector x in the space,

there is a unique set of coefficients c1, c2, c3, , ckfor which x = c1v1 + c2v2 + c3v3+ + ckvk.

Finally, note the following inequality:

For a vector space V with dimension n:

Size of a linearly independent Size of a set of vectors

set of vectors that span V

notes 4.3 and 4.4

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