Chap. 4 Vector Spaces 4.1 Vectors in R n 4.2 Vector Spaces 4.3 Subspaces of Vector Space 4.4 Spanning Sets & Linear Independent 4.5 Basis & Dimension 4.6

Chap. 4Chap. 4Vector SpacesVector Spaces

4.1 Vectors in Rn

4.2 Vector Spaces4.3 Subspaces of Vector Space4.4 Spanning Sets & Linear Independent4.5 Basis & Dimension4.6 Rank of a Matrix & Systems of Linear Equations4.7 Coordinates & Change of Basis4.8 Applications of Vector Spaces

Ming-Feng Yeh Chapter 4 4-2

Vectors in the plane (R2) is represented geometrically by a directed line segment whose initial point is the origin and whose terminal point is the point x = (x1, y1).x1, y1: the components of x

u = (u1, u2), v = (v1, v2), c is a scalarEqual: u = v iff u1 = v1& u2 = v2

Scalar multiplication: cv = c(v1, v2) = (cv1, cv2)

4.1 4.1 Vectors in RVectors in Rnn

x

y

Initial point

Terminal point (x1, y1)

x


Vector in the PlaneVector in the Plane Given u = (u1, u2), v = (v1, v2), and c is a scalar

Scalar multiplication:

Zero vector: 0 = (0, 0)Negative of v: v = (1)v

Section 4-1

x

y

u

cu, c > 0

x

y

u

cu, c < 0


Vector AdditionVector Addition Given u = (u1, u2) and v = (v1, v2)

Vector addition: u + v = (u1, u2) + (v1, v2) = (u1 + v1, u2 + v2)Difference of u and v: u v = u + (v)

u

v

u+v(u1, u2)

(v1, v2)

(u1 + v1, u2 + v2)

v1 u1

v2

u2

Section 4-1


Example 3Example 3 Given v = (2, 5) and u = (3, 4)

(a) ½v = （ ½(2), ½(5) ） = (1, 5/2)(b) u v = （ 3(2), 45 ） = (5, 1)(c) ½v + u = (1, 5/2) + (3, 4) = (2, 13/2)

x

y

)5,2(

v

u

)4,3(

),2( 213

0.5v + u

)1,5( v - u

),1( 25

v21

Section 4-1


Theorem 4.1Theorem 4.1

Vector Addition1. u + v is a vector in the plane

(closure under addition)

2. u + v = v + u(commutative property)

3. (u + v) + w = v + (u + w)(associative property)

4. u + 0 = u

5. u + (u) = 0

Scalar Multiplication6. cu is a vector in the plane

(closure under scalar multi.)

7. c(u + v) = cu + cv(left distributive property)

8. (c + d)u = cu + du(right distributive property)

9. c(du) = (cd)u

10. 1(u) = u

Let u, v, and w be vectors in the plane, andlet c and d be scalars.

Section 4-1


Proof of Theorem 4.1Proof of Theorem 4.14. (u + v) + w = [(u1, u2) + (v1, v2)] + (w1, w2)

= (u1 + v1, u2 + v2) + (w1, w2) = （ (u1 + v1) + w1, (u2 + v2) + w2 ）= （ u1 + (v1 + w1), u2 + (v2 + w2) ）= (u1, u2) + [(v1, v2) + (w1, w2)]= u + (v + w)

8. (c + d)u = (c + d) (u1, u2) = （ (c + d)u1, (c + d)u2 ）= (cu1 + du1, cu2 + du2)= (cu1, cu2) + (du1, du2)= c(u1, u2) + d(u1, u2)= cu + du

Section 4-1


Vectors in RVectors in Rnn

A vector in n-space is represented by an ordered n-tuple.x = (x1, x2, x3, …, xn)

The set of all n-tuples is called n-space (Rn)R1 = 1-space = set of all real numbersR2 = 2-space = set of all ordered pairs of real numbersR3 = 3-space = set of all ordered triples of real numbersR4 = 4-space = set of all ordered quadruples of real numbers::Rn = n-space = set of all ordered n-tuples of real numbers

Section 4-1


Vector Addition & Scalar Multi.Vector Addition & Scalar Multi. Let u = (u1, u2, u3, …, un) and v = (v1, v2, v3, …, vn) be

vectors in Rn and let c be a real number.Then the sum of u and v is defined to be the vector u + v = (u1+v1, u2+v2, u3+v3, …, un+vn),and the scalar multiple of u by c is defined to be cu = (cu1, cu2, cu3, …, cun).

The negative of u is defined asu = (u1, u2, u3, …, un)

The difference of u and v is defined as u v = (u1v1, u2v2, u3v3, …, unvn)

The Zero vector in Rn is given by 0 = (0, 0, …, 0)

Section 4-1



Vector Addition1. u + v is a vector in the plane

(closure under addition)

2. u + v = v + u(commutative property)

3. (u + v) + w = v + (u + w)(associative property)

4. u + 0 = u

5. u + (u) = 0

Scalar Multiplication6. cu is a vector in the plane

(closure under scalar multi.)

7. c(u + v) = cu + cv(left distributive property)

8. (c + d)u = cu + du(right distributive property)

9. c(du) = (cd)u

10. 1(u) = u

Let u, v, and w be vectors in the Rn, andlet c and d be scalars.

Section 4-1


Example 5Example 5 Let u = (2, 1, 5, 0), v = (4, 3, 1, 1), and w = (6, 2, 0, 3)

be vectors in R4. Solve for x in each of the following. x = 2u (v + 3w)

x = 2u v 3w = 2(2, 1, 5, 0) (4, 3, 1, 1) 3(6, 2, 0, 3) = (18, 11, 9, 8)

3(x + w) = 2u v + x 3x + 3w = 2u v + x x = ½(2u v 3w) = (9, 11/2, 9/2, 4)

Section 4-1


Additive Identity & Additive InverseAdditive Identity & Additive Inverse The zero vector 0 in Rn is called the additive identity in Rn. The vector v is called the additive inverse of v. Theorem 4.3: Let v be a vector in Rn, and let c be a scalar.

Then the following properties are true.1. The additive identity is unique. That is, if v + u = v, then u = 0.2. The additive inverse of v is unique. That is, if v + u = 0, then u = v.3. 0v = 0. 4. c0 = 0. 6. (v) = v5. If cv = 0, then c = 0 or v = 0.

Section 4-1


Linear Combination & Ex. 6Linear Combination & Ex. 6 The vector x is called a linear combination of the vectors

v1, v2, …, vn if x = c1v1 + c2v2 +…+ cnvn.

Example 6: Given x = (1, 2, 2), u = (0, 1, 4),v = (1, 1, 2), and w = (3, 1, 2) in R3, find scalarsa, b, and c such that x = au + bv + cw.

Sol: (1, 2, 2) = a(0, 1, 4) + b(1, 1, 2) + c(3, 1, 2)

(1, 2, 2) = (b + 3c, a + b + c, 4a + 2b +2c)

a = 1, b = 2, c = 1 x = u 2v w

Section 4-1


Row Vector & Column VectorRow Vector & Column Vector

Represent a vector u = (u1, u2, u3, …, un) in Rn as

a 1n row matrix (row vector)

or

an n1 column matrix (column vector)

Section 4-1

nuuu 21u

nu

u

u

2

1

u


4.2 4.2 Vector SpacesVector Spaces

Vector Addition1. u + v is in V.

(closure under addition)2. u + v = v + u

(commutative property)3. (u + v) + w = v + (u + w)

(associative property)4. V has a zero vector 0 s.t.for

every u in V, u + 0 = u(additive identity)

Scalar Multiplication6. cu is in V.

(closure under scalar multi.)7. c(u + v) = cu + cv

(left distributive property)8. (c + d)u = cu + du

(right distributive property)9. c(du) = (cd)u (associative

prop.)10. 1(u) = u (scalar property)

Let V be a set on which two operations (vector addition & scalar multiplication) are defined. If the following axioms are satisfied for every u, v, and w in V, and every scalar c and d, then V is called a vector space.

5. For every u in V, there is a vector in V denoted by u s.t. u + (u) = 0. (additive inverse)


Examples 1 ~ 3Examples 1 ~ 3 Example 1: R2 with the standard operations is a vector

space see Theorem 4.1 Example 2: Rn with the standard operations is a vector

space see Theorem 4.2 Example 3: Show that the set of all 23 matrices with the

operations of matrix addition and scalar multiplication is a vector space.

pf: If A and B are 23 matrices and c is a scalar, then A+B and cA are also 23 matrices. Hence, the set is closed under matrix addition and scalar multiplication. Moreover, the other eight vector space axioms follow from Theorems 2.1 and 2.2. Thus, the set is a vector space.

Section 4-2


Example 4Example 4The Vector Space of All Polynomials of Degree 2 or Less

Let P2 be the set of all polynomials of the formp(x) = a2x2 + a1x + a0; q(x) = b2x2 + b1x + b0

where a0(b0), a1(b1), and a2(b2) are real numbers.

The sum of two polynomials p(x) and q(x) is defined byp(x) + q(x) = (a2+b2)x2 + (a1+b1)x + (a0+b0),

and the scalar multiple of p(x) by the scalar c is defined bycp(x) = ca2x2 + ca1x + ca0

Show that P2 is a vector space.

proof: omitted

Section 4-2


Important Vector SpacesImportant Vector Spaces R = set of all real number R2 = set of all ordered pairs R3 = set of all ordered triples Rn = set of all n-tuples C(, ) = set of all continuous functions defined on

the real line C[a, b] = set of all continuous functions defined on

a closed interval [a, b] P = set of all polynomials Pn = set of all polynomials of degree n Mm,n = set of all mn matrices Mn,n = set of all nn square matrices

Section 4-2



Properties of Scalar Multiplication Let v be any element of a vector space V, and let c be any

scalar. Then the following properties are true.1. 0v = 0. 2. c0 = 0.3. If cv = 0, then c = 0 or v = 0. 4. (1)v = v.

pf: Suppose that cv = 0. To show that this implies either c = 0 or v = 0, assume that c 0.Because c 0, we can you the reciprocal 1/c to show thatv = 0 as follows

Section 4-2

00vvvv )1())(1())(1(1 ccccc


Example 6 ~ 7Example 6 ~ 7Example 6The set of integers is NOT a vector space The set of all integers (with the standard operations) does

not form a vector space because it is not closed under scalar multiplication. For example,

Example 7The set of second-degree polynomials is NOT a vector space The set of all 2nd-degree polynomials is not a vector space

because it is not closed under addition. For example,

.)1( 21

21

1)()(1)(

)(2

2

xxqxpxxxq

xxp

(noninteger)

(the 1st-degree poly.)

Section 4-2


Example 8Example 8 Let V = R2, the set of all ordered pairs of real number, with

the standard addition and the following nonstandard definition of scalar multiplication: c(x1, x2) = (cx1, 0).Show that V is not a vector space.

pf: This example satisfies the first nine axioms of the definition of a vector space. For example,let u = (1, 1), v = (3, 4), and c = 2, then we have c(u + v) = 2(1+3, 1+4) = (8, 0), cu = (2, 0), cv = (6, 0).Therefore, c(u + v) = cu + cv. However, when c =1, 1(1, 1) = (1, 0) (1,1). The tenth axiom is not verified. Hence, the set (together with the two given operations) is not a vector space.

Section 4-2


4.3 4.3 Subspaces of Vector SpacesSubspaces of Vector Spaces Definition of Subspace of a Vector SpaceDefinition of Subspace of a Vector Space

A nonempty subset W of a vector space V is called a subspace of V if W is itself a vector space under the operations of addition and scalar multiplication defined in V.

Remark: If W is a subspace of V, it must be closed under the operations inherited from V.


Example 1Example 1A subspace of R3

Show that the set W={(x1, 0, x3): x1, x3 R} is a subspace of R3 with the standard operations.

The set W can be interpreted as simply the xz-plane.

pf: The set W is nonempty because it contains the zero vector (0, 0, 0).The set W is closed under addition because the sum of any two vectors in the xz-plane must also lie in the xz-plane.That is, if (x1, 0, x3) and (y1, 0, y3) are in W, then their sum (x1+y1, 0, x3 +y3) is also in W.

Section 4-3


Example 1 (cont.)Example 1 (cont.)

To see that W is closed under scalar multiplication,let (x1, 0, x3) be in W and let c be a scalar.Then c(x1, 0, x3) = (cx1, 0, cx3) has zero as its second component and must therefore be in W.

The other eight vector space axioms can be verified as well, and these verifications as left to you.

Section 4-3


Theorem 4.5Theorem 4.5 Test for a Subspace

If W is a nonempty subset of a vector space V, then W is a subspace of V if and only if the following closure conditions hold.1. If u and v are in W, then u + v is in W.2. If u is in W and c is any scalar, then cu is in W.

To establish that a set W is a vector space, you must verify all ten vector space properties. However, if W is a subset of a larger vector space V (and the operations defined on W are the same as those defined on V), then most of the ten properties are inherited from the larger space and need no verification.

Section 4-3


Zero SubspaceZero Subspace The simplest subspace of a vector space is the one

containing of only the zero vector, W = {0}.This subspace is called the zero subspace.

If W is a subspace of a vector space V, then both W and V must have the same zero vector 0.

Another obvious subspace of V is V itself. Every vector space contains two trivial subspaces (the

zero subspace and the vector space itself), and subspaces other than these two are called proper (or nontrivial) subspace.

Section 4-3


Example 2Example 2A Subspace of M2,2 Let W be the set of all 22 symmetric matrices. Show

that W is a subspace of the vector space M2,2, with the standard matrix addition and scalar multiplication.

pf: Because M2,2 is a vector space, we need only show that W satisfies the conditions of Theorem 4.5.1. W is nonempty. why?2. if A1 and A2 are symmetric matrices of order 2, then so is A1+A2.3. if A is a symmetric matrix of order 2, then so is cA.

2121212211 )(, AAAAAAAAAA TTTTT

cAcAcAAA TTT )(

Section 4-3


Example 3Example 3The Set of Singular Matrices is NOT a Subspace of Mn,n Let W be the set of singular matrices of order 2. Show

that W is NOT a subspace of the vector space M2,2, with the standard operations.

pf: W is nonempty and closed under scalar multiplication, but it is NOT closed under addition. For example, let

A and B are both singular, but their sum A + B is nonsingular. Hence, W is not closed under addition, and it is not a subspace of M2,2.

10

01

10

00,

00

01BABA

Section 4-3


Example 4Example 4The Set of First-Quadrant Vectors is NOT a Subspace of R2

Show that W = {(x1, x2): x1 0 and x2 0}, with the standard operations, is NOT a subspace of R2.

pf: W is nonempty and closed under addition. It is NOT, however, closed under scalar multiplication.Note that (1, 1) is in W, but the scalar multiple(1)(1, 1) = (1, 1)is not in W. Therefore, W is not a subspace of R2.

Section 4-3


Intersection of Two SubspacesIntersection of Two Subspaces If U, V, and W are vector spaces such that

W is a subspace of V and V is a subspaceof U, then W is also a subspace of U.

Theorem 4.6If V and W are both subspacesof a vector space U, then theintersection of V and W(denoted by VW) is alsoa subspace of U.

W V U

U

V WWV

Section 4-3


Proof of Theorem 4.6Proof of Theorem 4.61. V and W are both subspaces of U.

Both contain the zero vector. VW is nonempty.

2. Let v1 and v2 be any vectors in VW. V and W are both subspaces of U. both are closed under addition. v1 and v2 are both in V. v1 + v2 must be in V.Similarly, v1 + v2 is in W ( v1 and v2 are both in W ). v1 + v2 is in VW, and it follows that VW is closed under addition.

3. VW is closed under scalar multiplication. (left to you)

Section 4-3


Subspace of RSubspace of R22

If W is a subspace of R2, then it is a subspace if and only if one of the following is true.1. W consists of the single point (0, 0).2. W consists of all points on a line that passes through the origin.3. W consists of all of R2.

x

y

x

y

x

y

Section 4-3


Example 6Example 6 Which of these two subsets is a subspace of R2?(a) The set of points on the line given by x + 2y = 0Sol: A point in R2 is on the line x + 2y = 0 if and only if

it has the form (2t, t), tR.1. The set is nonempty since it contains the origin (0, 0).2. Let v1 = (2t1, t1) and v2 = (2t2, t2) be any two points on the line. Then v1+v2 = (2(t1+t2), t1 +t2) = (2t3, t3). Thus v1+v2 is on the line, and the set is closed under addition.3. Similarly, you can show that the set is closed under scalar multiplication.Therefore, this set is a subspace of R2.

Section 4-3


Example 6 (cont.)Example 6 (cont.) Which of these two subsets is a subspace of R2?

(b) The set of points on the line given by x + 2y = 1

Sol: This subset of R2 is not a subspace of R2

because every subspace must contain the zero vector,and the zero vector (0, 0) is not on the line.

Section 4-3


Example 7Example 7 Show that the subset of R2 that consists of all points on the

unit circle x2 + y2 = 1 is not a subspace.

Sol: [Method I] This subset R2 is not a subspace of R2

because every subspace must containthe zero vector, and the zero vector(0, 0) is not on the circle.[Method II] This subset R2 is not a subspace of R2 because the points(1, 0) and (0, 1) are in the subset,but their sum (1, 1) is not. So thissubset is not closed under addition.

x

y

Section 4-3


Subspace of RSubspace of R33

If W is a subspace of R3, then it is a subspace if and only if one of the following is true.1. W consists of the single point (0, 0, 0).2. W consists of all points on a line that passes through the origin.3. W consists of all points on a plane that passes through the origin.4. W consists of all of R3.

Section 4-3


Example 8Example 8 Let W = {(x1, x1+x3, x3): x1 and x3 are real number}.

Show that W is a subspace of R3.

pf: Let v = (v1, v1+v3, v3) and u = (u1, u1+u3, u3) be two vectors in W, and let c be any real number.

1. W is nonempty because it contains the zero vector.2. v + u = (v1+ u1, (v1+ u1)+(v3+ u3), v3+ u3) = (x1, x1+x3, x3). Hence, v + u is in W (W is closed under addition).3. cv = (cv1, c(v1+v3), cv3) = (x1, x1+x3, x3).

Hence, cv is in W.We can conclude that W is a sunspace of R3.

Section 4-3


Linear Combination A vector v in a vector space V is called a linear combination of the vectors u1, u2, …, uk in V if v = c1u1 + c2u2 + … + ckuk,where c1, c2, …, ck are scalars.

Example 1.aS = { (1, 3, 1), (0, 1, 2), (1, 0, 5)} in R3, v1 v2 v3

v1 is a linear combination of v2 and v3 becausev1 = 3v2 + v3 = 3(0, 1, 2) + (1, 0, 5) = (1, 3, 1)

4.4 4.4 Spanning Sets & LinearSpanning Sets & Linear Independence Independence


Example 1.bExample 1.b For the set of vectors in M2,2,

v1 is a linear combination of v2, v3 and v4 because

Section 4-4

31

02,

21

31,

01

20,

12

804321

S

vvvv

4321 2 vvvv


Example 2Example 2 Write the vector w = (1, 1, 1) as a linear combination of

vectors in the set S. v1 v2 v3

S = { (1, 2, 3), (0, 1, 2), (1, 0, 1)}

Sol: w = c1v1 + c2v2 + c3v3

(1, 1, 1) = c1(1, 2, 3) + c2(0, 1, 2) + c3(1, 0, 1)

= (c1 c3, 2c1 + c2, 3c1 + 2c2 + c3)

)1let(32,

,21,1

123121

321

3

2

1

321

21

31

t

Rttc

tctc

ccccc

cc

vvvw

Section 4-4


Example 3Example 3 If possible, write the vector w = (1, 2, 2) as a linear

combination of vectors in the set S given in Example 2.

Sol:

The system of equations is inconsistent, and therefore there is no solution. Consequently, w CANNOT be written as a linear combination of v1, v2, and v3.

7000

4210

1101

22322

1

321

21

31

ccccc

cc

Section 4-4


Spanning SetsSpanning Sets Definition: Let S = {v1, v2, …, vk} be a subset of a vector

space V. The set S is called a spanning set of V if every vector in V can be written as a linear combination of vectors in S. In such cases it is said that S spans V.

Example 4: Standard Spanning Setsa. The set S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} spans R3

because any vector u = (u1, u2, u3) in R3 can be written as u = u1(1, 0, 0) + u2(0, 1, 0) + u3(0, 0, 1) = (u1, u2, u3) b. The set S = {1, x, x2} spans P2. p(x) = a(1) + b(x) + c(x2) = a + bx + cx2

Section 4-4


Example 5Example 5A Spanning Set for R3

Show that the set S = {(1, 2, 3), (0, 1, 2), (2, 0, 1)} spans R3.

Pf: Let u = (u1, u2, u3) be any vector in R3.

The coefficient matrix has a nonzero determinant. Hence, the system has a unique solution. Any vector in R3

can be written as a linear combination of the vectors in S, and we can conclude that the set S spans R3.

)23,2,2()1,0,2()2,1,0()3,2,1(),,(

3212131

321321

ccccccccccuuu

3321

221

131

232

ucccuccucc

Section 4-4


Example 6Example 6A Set That Does Not Spans R3

Show that the set S = {(1, 2, 3), (0, 1, 2), (1, 0, 1)} does not spans R3.

Pf: We can see form Example 3 that the vector w = (1, 2, 2)is in R3 and cannot be expressed as a linear combination of the vectors in S.Therefore, the set S does not span R3.

Section 4-4


Example 5 vs Example 6Example 5 vs Example 6 S1 = {(1, 2, 3), (0, 1, 2), (2, 0, 1)}

the vectors in S1 do not lie in a common plane

S2 = {(1, 2, 3), (0, 1, 2), (1, 0, 1)} the vectors in S2 lie in a common plane

x

y

z

x

y

z

Section 4-4


The Span of a SetThe Span of a Set Definition: If S = {v1, v2, …, vk} is a set of vectors in V,

then the span of S is the set of all linear combinations of the vectors in S,

span(S) = {c1v1+c2v2+…+ckvk: c1, c2, …, ck R}

The span of S is denoted by span(S) or span{v1, v2, …, vk}.

If span(S) = V, then V is spanned by {v1, v2, …, vk} orS spans V.

Section 4-4


Theorem 4.7Theorem 4.7 If S = {v1, v2, …, vk} is a set of vectors in V, then span(S) is

a subspace of V. Moreover, span(S) is the smallest subspace of V that contains S in the sense that every other subspace of V that contains S must contain span(S).

Pf: Suppose that u and v are any two vectors in span(S) u = c1v1+c2v2+…+ckvk; v = d1v1+d2v2+…+dkvk. Then, u + v = (c1+d1)v1+(c2+d2)v2+…+(ck+dk)vk

cu = cc1v1+ cc2v2 +…+ cckvk

which means that u + v and cu are also in span(S). Therefore, span(S) is a subspace of V.

Section 4-4


Linear Dependence &Linear Dependence &Linear IndependenceLinear Independence A set of vectors S = {v1, v2, …, vk} in a vector

space V is called linearly independent if the vector equationc1v1 + c2v2 + … + ckvk = 0 has only the trivial solution, c1= 0, c2= 0, …, ck= 0.

If there are also nontrivial solutions, then S is called linearly dependent.

Section 4-4


Example 7Example 7Linearly Dependent Sets The set S = {(1, 2), (3, 4)} in R2 is linearly dependent

2(1, 2) + 1(2, 4) = (0, 0) The set S = {(1, 0), (0, 1), (2, 5)} in R2 is linearly

dependent 2(1, 0) 5(0, 1) + 1(2, 5) = (0, 0) The set S = {(0, 0), (1, 2)} in R2 is linearly dependent

1(0, 0) + 0(1, 2) = (0, 0)

Section 4-4


Example 8Example 8 Testing for Linear Independence

Determine whether the set of vectors in R3 is linearly dependent or linearly independentS = { v1 = (1, 2, 3), v2 = (0, 1, 2), v3 = (2, 0, 1)}

Sol: c1v1 + c2v2 + c3v3 = 0

c1(1, 2, 3) + c2(0, 1, 2) + c3(2, 0, 1) = (0, 0, 0)

(c12c3, 2c1+c2, 3c1+2c2 +c3) = (0, 0, 0)

c1 = c2 = c3 = 0

Therefore, S is linearly independent.

Section 4-4



Determine whether the set of vectors in P2 is linearly dependent or linearly independentS = { 1 + x 2x2, 2 + 5x x2, x + x2}

Sol: c1v1 + c2v2 + c3v3 = 0

c1(1 + x 2x2) + c2(2 + 5x x2) + c3(x + x2) = 0 + 0x + 0x2

(c1+2c2) + (c1+5c2+c3)x + (2c1c2+c3) x2 = 0 + 0x + 0x2

The system has an infinite number of solutions. Therefore, the system must has nontrivial solutions, and we can conclude that the set S is linearly dependent.

c1 = 2t, c2 = t, c3 = 3t, tR.

Section 4-4



Determine whether the set of vectors in M2,2 is linearly dependent or linearly independent

Sol: c1v1 + c2v2 + c3v3 = 0

The system has only the trivial solution. Hence the set S is linear independently.

02

01,

12

03,

11

12S

00

00

02

01

12

03

11

12321 ccc

Section 4-4



Determine whether the set of vectors in M4,1 is linearly dependent or linearly independent

Sol: linear independence.

2

1

1

0

,

2

1

3

0

,

2

0

1

1

,

0

1

0

1

S

Section 4-4


Theorem 4.8Theorem 4.8A Property of Linearly Dependent Set A set S = {v1, v2, …, vk}, k2, is linear dependent if and

only if at least one of the vectors vj can be written as a linear combination of the other vectors in S.

Pf: 」 Assume that S is a linearly dependent set. Then there exist scalars c1, c2, …, ck (not all zero) such thatc1v1 + c2v2 + … + ckvk = 0Assume that c10. ThenThat is, v1 is a linear combination of the other vectors. 」 Suppose v1 in S is a linear combination of the other vectors, i.e., v1 = c2v2 + c3v3 +… + ckvk. Then the equationv1 + c2v2 + c3v3 +… + ckvk = 0 has at least one coefficient, 1, that is nonzero. Hence, S is linearly dependent.

kkcccc vvvv 332211

Section 4-4


Example 12Example 12 In Example 9, you determine set

S = { 1 + x 2x2, 2 + 5x x2, x + x2}is linearly dependent. Show that one of the vectors in this set can be written as a linear combination of the other two.

Sol: One nontrivial solution is c1 = 2, c2 = 1, and c3 =3, which yields2(1 + x 2x2) + (1)(2 + 5x x2) + 3(x + x2) = 0.Therefore, v2 = 2v1 + 3v3.

v2 can be written as a linear combination of v1 and v3.

Section 4-4


Corollary 4.8 & Example 13Corollary 4.8 & Example 13 Corollary 4.8

Two vectors u and v in a vector space V are linearly dependent if and only if one is a scalar multiple of the other.

Example 13(a) S = {(1, 2, 0), (2, 2, 1)} linear independent.(b) S = {(4, 4, 2), (2, 2, 1)} v1 = 2v2. linear dependent x

y

z

Section 4-4


4.5 4.5 Basis & DimensionBasis & Dimension A set of vectors S = {v1, v2, …, vn} in a vector

space V is called basis if the following conditions are true.1. S spans V. 2. S is linearly independent.

If a vector space V has a basis consisting of a finite number of vectors, then V is finite dimensional. Otherwise, V is called infinite dimensional.

The vector space V = {0} is finite dimensional.


Example 1Example 1 The Standard Basis for R3

Show that the set S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis for R3.

pf: 1. Example 4(a) in Section 4.4 showed that S spans R3. 2. S is linear independent. c1(1, 0, 0) + c2(0, 1, 0) + c3(0, 0, 1) = (0, 0, 0) has only the trivial solution c1 = c2 = c3 = 0. S is a basis for R3.

The Standard Basis for Rn

e1 = (1, 0, 0, …, 0), e2 = (0, 1, 0, …, 0), e3 = (0, 0, 1, …, 0),…, en = (0, 0, 0, …, 0, 1)

Section 4-5


Example 2Example 2 A Nonstandard Basis for R2

Show that the set S = {v1(1, 1), v2(1, 1)} is a basis for R2.

pf: 1. Let x = (x1, x2) represent an arbitrary vector in R2.Consider the linear combination c1v1 + c2v2 = x, (c1 + c2, c1 c2) = (x1, x2)

the coefficient matrix has a nonzero determinant, the system has a unique solution. S spans R2.2. S is linearly independent (verify it).

S is a basis for R2.

221

121

xccxcc

Section 4-5


Example 4Example 4 A Basis for Polynomials

Show that the vector space P3 has the following basis.S = {1, x, x2, x3}

pf: 1. the span of S consists of all polynomials of the form a0 + a1x + a2x2 + a3x3, a0, a1, a2, a3R S spans P3. 2. Consider the following linear combination a0 + a1x + a2x2 + a3x3 = 0(x) = 0, x the solutions are a0 = a1 = a2 = a3 = 0. That is, S is linear independent. S is a basis for P3.

Section 4-5


Some Standard BasesSome Standard Bases Standard basis for P3: {1, x, x2, x3}

Standard basis for Pn: {1, x, x2, x3, …, xn}

Standard basis for M2,2:

Standard basis for Mm,n:The set that consists of the mn distinct m n matrices having a single 1 and all other entries equal to zero.

10

00,

01

00,

00

10,

00

01

Section 4-5


Theorem 4.9Theorem 4.9Uniqueness of Basis Representation If S = {v1, v2, …, vn} is a basis for a vector space V, then

every vector in V can be written in one and only one way as a linear combination of vectors in S.

Pf: S spans V. an arbitrary vector u in V can be expressed as u = c1v1 + c2v2 + … + cnvn Suppose that u has another representation u = b1v1 + b2v2 + … + bnvn (c1 b1)v1 + (c2 b2)v2 + … + (cn bn)vn = 0 S is linear independent. The only solution to is

c1 b1 = 0, c2 b2 = 0, cn bn = 0 ci = bi for all i.Hence, u has only one representation for S.

Section 4-5


Example 6Example 6 Uniqueness of Basis Representation

Let u = (u1, u2, u3)R3. Show that u = c1v1 + c2v2 + c3v3 has a unique solution for the basis S = {v1, v2, v3} ={ (1, 2, 3), (0, 1, 2), (2, 0, 1) }

Sol: (u1, u2, u3) = c1(1, 2, 3) + c2(0, 1, 2) + c3(2, 0, 1)

that is, u has a unique solution for the basis S.

uc

A

u

u

u

c

c

c

3

2

1

3

2

1

123

012

201

uc 1,0)det( AA

Section 4-5


Theorem 4.10Theorem 4.10Bases & Linear Dependence If S = {v1, v2, …, vn} is a basis for a vector space V, then

every set containing morn than n vectors in V is linearly dependent.

Pf: Let S1 = {u1, u2, …, um} V, m > n.Consider: k1u1 + k2u2 + … + kmum = 0 If k1, k2 , …, km are not all zero, then S1 is linear dependent. S is a basis for V.each ui is a linear combination of vectors in S, i.e., Substituting each of these representations of ui into ,

nniiii ccc vvvu 2211

mimiiinn kckckcdddd 22112211 ,0vvv

Section 4-5


Theorem 4.10 (cont.)Theorem 4.10 (cont.)

vi’s form a linearly independent set. (S is a basis)di = 0

This homogeneous system has fewer equations than variables k1, k2 , …, km, (n < m). Hence it must have nontrivial solutions. That is, S1 is linearly dependent.

mimiiinn kckckcdddd 22112211 ,0vvv

0

00

2211

2222121

1212111

mnmnn

mm

mm

kckckc

kckckckckckc

Section 4-5



Number of Vectors in a Basis If a vector space V has one basis with n vectors,

then every basis for V has n vectors.Pf: Let S1 = {v1, v2, …, vn} be a given basis for V, and

let S2 = {u1, u2, …, um} be another basis for V.Because S1 is a basis and S2 is linearly independent,Theorem 4.10 implies that m n.Similarly, n m because S1 is linearly independentand S2 is a basis.Consequently, n = m.

Section 4-5


Example 8Example 8 Spanning Sets & Bases

Explain why each of the following statements is true.

(a) The set S1 = { (3, 2, 1), (7, 1, 4)} is NOT a basis for R3.The standard basis for R3 has 3 vectors, and S1 has only 2.Hence, S1 cannot be a basis for R3.

(b) The set S2 = { x + 2, x2, x3 1, 3x +1, x2 2x + 3} is NOT a basis for P3.The standard basis for P3 has 4 vectors.The set S2 has too many elements to be a basis for P3.

Section 4-5


Dimension of a Vector SpaceDimension of a Vector Space Definition: If a vector space V has a basis consisting of n

vectors, then the number n is called the dimension of V, denoted by dim(V) = n.If V consists of the zero vectors alone, the dimension of V is defined as zero, i.e., dim(0) = 0.

dim(Rn) = ndim(Pn) = n +1dim(Mm,n) = mn

If W is a subspace of an n-dimensional vector space, thendim(W) n.

Section 4-5


Example 9Example 9 Finding the Dimension of a Subspacefind a set of linearly independent vectors that spans the subspace

Determine the dimension of each subspace of R3.(a) W = { (d, cd, c): c, d R}Sol: (d, cd, c) = c(0, 1, 1) + d(1, 1, 0)

W is spanned by the set S = {(0, 1, 1), (1, 1, 0)}. Hence, dim(W) = 2.

(b) W = { (2b, b, 0): b R}Sol: W is spanned by the set S = {(1, 1, 0)}.

Hence, dim(W) = 1.

Section 4-5


Example 10Example 10 Finding the Dimension of a Subspace

Find the dimension of the subspace W of R4 spanned byS = { v1(1, 2, 5, 0), v2(3, 0, 1, 2), v3(5, 4, 9, 2)}

Sol: Although W is spanned by the set S, S is NOT a basis for W because S is a linear dependent set.v3 = 2v1 v2

This means that W is spanned by the set S1 = {v1, v2}.Moreover, S1 is linearly independent.Hence, dim(W) = 2.

Section 4-5


Example 11Example 11 Finding the Dimension of a Subspace

Let W be a subspace of all symmetric matrices M2,2.What is the dimension of W?

Sol:

S is linearly independent and S spans W.Hence, dim(W) = 3.

10

00,

11

10,

00

01

10

00

11

10

00

01

S

cbacb

ba

Section 4-5


Theorem 4.12 & Example 12Theorem 4.12 & Example 12Theorem 4.12: Basis Test in an n-dimensional Space

Let V be a vector space of dimension n.1. If S = {v1, v2, …, vn} is a linearly independent set of vectors in V, then S is a basis for V.2. If S = {v1, v2, …, vn} spans V, then S is a basis for V.

Example 12: Show the set S is a basis for M5,1.

S is a linearly independent set and dim(M5,1) = 5

S is a basis for M5,1.

2

0

0

0

0

,

3

2

0

0

0

,

5

1

2

0

0

,

3

2

3

1

0

,

4

3

1

2

1

S

Section 4-5


Definition Let A be an m n matrix.1. The row space of A is the subspace of Rn spanned by the row vectors of A.2. The column space of A is the subspace of Rm

spanned by the column vectors of A.

4.6 4.6 Rank of a Matrix & SystemsRank of a Matrix & Systems of Linear Equations of Linear Equations

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

Row vectors: 1 n

Column vectors: m 1


Theorems 4.13 & 4.14Theorems 4.13 & 4.14 Theorem 4.13

Row-Equivalent Matrices Have the Same Row SpaceIf an m n matrix A is row-equivalent to an m n matrix B, then the row space of A is equal to the row space of B.

Theorem 4.14Basis for the Row Space of a MatrixIf a matrix A is row-equivalent to a matrix B, then the nonzero row vectors of B form a basis for the row space of A.

Section 4-6


Example 2Example 2 Find a basis for a row space of A.

Sol:

w1 = (1, 3, 1, 3), w2 = (0, 1, 0, 1), and w3 = (0, 0, 0, 1)form a basis for the row space of A.

3

2

1

0000

0000

1000

0110

3131

2402

1243

1603

0110

3131

w

w

w

BA

Section 4-6


Example 3Example 3 Find a basis for the subspace of R3 spanned by

S = { (1, 2, 5), (3, 0, 3), (5, 1, 8) }

Sol:

w1 = (1, 2, 5) and w2 = (0, 1, 3) form a basis for the row space of A.That is, they form a basis for the subspace spanned by S.

2

1

3

2

1

000

310

521

815

303

521

w

w

v

v

v

BA

Section 4-6


Column Vectors ofColumn Vectors ofRow-Equivalent MatricesRow-Equivalent Matrices

b3 = 2b1 + b2 a3 = 2a1 + a2

The column vectors b1, b2, and b4 of matrix B are linearly independent, and so are the corresponding columns of A.

0000

0000

1000

0110

3131

2402

1243

1603

0110

3131

BA

4321 aaaa b1 b2 b3 b4

Section 4-6


Example 4Example 4 Find a basis for the column space of the matrix A.

Sol: [Method 1]

w1 = (1, 0, 3, 3, 2), w2 = (0, 1, 9, 5, 6), andw3 = (0, 0, 1, 1, 1) form a basis for the row space of AT.That is equivalent to saying that form a basis for the column space of A.

3

2

1

00000

11100

65910

23301

21103

42611

04013

23301

w

w

w

TA

TTT321 and,, www

Section 4-6


Example 4 (cont.)Example 4 (cont.)Sol: [Method 2]

a1, a2, and a4 form a basis for the column space of A.

0000

0000

1000

0110

3131

2402

1243

1603

0110

3131

BA

4321 aaaa

Section 4-6


Thm 4.15 & Rank of a MatrixThm 4.15 & Rank of a Matrix Theorem 4.15

Row and Column Spaces Have Equal DimensionsIf A is an m n matrix, then the row space and column space of A have the same dimension.

DefinitionThe dimension of the row (or column) space of a matrix A is called the rank of A and is denoted by rank(A).

Example 5

3)(rank

3100

1110

1021

5310

3512

1021

ABA

Section 4-6


Thm 4.16: Nullspace of a MatrixThm 4.16: Nullspace of a MatrixSolutions of a Homogeneous System If A is an m n matrix, then the set of all solutions of the

homogeneous system of linear equations Ax = 0is a subspace of Rn called the nullspace of A and is denoted by N(A). So, N(A) = {xRn: Ax = 0}.

The dimension of the nullspace of A is calledthe nullity of A. nullity(A) = dim( N(A) )

Pf: 1. Nonempty: A0 = 0 2. Addition: Ax1 = 0 and Ax2 = 0 A(x1+x2) = Ax1+ Ax2 = 0 3. Scalar multiplication: A(cx1) = c(Ax1) = 0

Section 4-6


Example 6Example 6 The nullspace of A is also called the solution space of the

system Ax = 0. Find the Solution Space of a Homogeneous System

Find the nullspace of A.Sol:

0000

1100

3021

3021

4563

1221

A

Rtsts

t

t

s

ts

,

1

1

0

3

0

0

1

232

x

},,:{)( 21 RtstsAN vvxx

Section 4-6


Remark of Example 6Remark of Example 6

A basis for N(A) is

because all solutions of Ax = 0 are linear combinations of these two vectors.

When homogeneous systems are solved from the reduced row-echelon form, the spanning set is linear independent.

A is a 3 4 matrix, rank(A) = 2, nullity(A) = 2 rank(A) + nullity(A) = 4

}

1

1

0

3

,

0

0

1

2

{

Section 4-6



Dimension of the Solution Space If A is an m n matrix of rank r, then the

dimension of the solution space of Ax = 0 is n r. That is,rank(A) + nullity(A) = n.

Section 4-6


Example 7Example 7Rank and Nullity of a Matrix Let the column vectors of the matrix A be denoted by

a1, a2, …, a5.(a) Find the rank and nullity of A.

rank(A) = 3, n = 5 nullity(A) = 5 3 = 2.

00000

11000

40310

10201

120930

31112

31310

01201

BA

Section 4-6


Example 7 (cont.)Example 7 (cont.)(b) Find a set of the column vectors of A that forms a basis

for the column space of A.

form a basis for the column space of A.

00000

11000

40310

10201

120930

31112

31310

01201

BA

,

0

2

0

1

1

a ,

3

1

1

0

2

a ,

0

1

1

1

4

a

Section 4-6


Example 7 (cont.)Example 7 (cont.)(c) If possible, write the third column of A as a linear

combination of the first two columns.

213213 3232

00000

11000

40310

10201

120930

31112

31310

01201

aaabbb

BA

Section 4-6


Solutions of Linear SystemsSolutions of Linear Systems The set of all solution vectors of Ax = 0 is a

subspace. Is the set of all solution vectors of Ax = b (b 0)

also a subspace?The answer is “no,” because the zero vector is never a solution of Ax = b.

Section 4-6


Theorem 4.18Theorem 4.18 Solutions of a Nonhomogeneous Linear System

If xp is a particular solution of Ax = b (b 0), then every solution of this system can be written in the formx = xp + xh,where xh is a solution of Ax = 0.

Pf: Let x be any solution of Ax = b.Then (x xp) is a solution of Ax = 0,because A(x xp) = Ax Axp = b b = 0.Let xh = x xp x = xp + xh

Section 4-6


Example 8Example 8 Find the set of all solution vectors of the following system

952

853

52

421

321

431

xxx

xxx

xxx

Rtsts

t

s

ts

ts

x

x

x

x

,,

0

0

1

5

1

0

3

1

0

1

1

2

73

52

00000

73110

51201

95021

80513

51201

4

3

2

1

x

xh xp

Section 4-6


Theorem 4.19Theorem 4.19 Solution of a System of Linear Equations

The system of linear equation Ax = b is consistent if and only if b is in the column space of A.

Ax = b iff b is a linear combination of the columns of A

Section 4-6


Example 9Example 9

Consider

The rank of A is equal to the rank of [A b]. Hence, b is in the column space of A, and the system is consistent.

1

3

1

123

101

111

3

2

1

x

x

x

2)(rank.

000

210

101

123

101

111

AA

2])([rank.

0000

4210

3101

1123

3101

1111

][

bb AA

Section 4-6


Equivalent ConditionsEquivalent ConditionsIf A is an n n matrix, then the following conditions are

equivalent. A is invertible. Ax = b has a unique solution for any n 1 matrix b. Ax = 0 has only trivial solution. A is row-equivalent to In. Rank(A) = n The n row (column) vectors of A are linearly independent.

Section 4-6

0A


4.4.7 Coordinates and7 Coordinates and Change of Basis Change of Basis

Coordinate Representation Relative to a Basis Let B = {v1, v2, …, vn} is an ordered basis for a vector

space V and let x be a vector in V s.t.

x = c1v1 + c2v2 + … + cnvn

The scalars c1, c2, …, cn are called the coordinates of x relative to the basis B.

The coordinate matrix (or coordinate vector)of x relative to B is the column matrix in Rn

whose components are the coordinate of x.

n

B

c

c

c

2

1

][x


Example 2Example 2 Finding a Coordinate Matrix Relative to a Standard Basis

The coordinate matrix of x relative to the (nonstandard)

basis B = {v1, v2} = {(1, 0), (1, 2)} is

Find the coordinates of x relative to the standard basisB = {u1, u2} = {(1, 0), (0, 1)}.

Sol: x = 3v1 + 2v2 = 3(1, 0) + 2(1, 2) = (5, 4) = 5(1, 0) + 4(0, 1) = 5u1 + 4u2

Section 4-7

2

3][ Bx

4

5][ Bx

1v

2v

1u

2u


Example 3Example 3 Finding a Coordinate Matrix Relative to a Nonstandard Basis

Find the coordinate matrix of x = (1, 2, 1) in R3 relative to the (nonstandard) basis B = {u1, u2 , u3} = {(1, 0, 1),(0, 1, 2), (2, 3, 5)}.

Sol: x = c1u1 + c2u2 + c3u3

(1, 2, 1) = c1(1, 0, 1) + c2(0, 1, 2) + c3(2, 3, 5)

1

2

1

521

310

201

3

2

1

c

c

c

321 uuu B][x B][x

2

8

5

][

3

2

1

c

c

c

Bx


Change of Basis in RChange of Basis in Rnn

Transition matrix from B to B:

Change of basis from B to B: Change of basis from B to B:

transition matrix from B to B:

1

2

1

521

310

201

3

2

1

c

c

c

321 uuu B][x B][x

321 uuuP

2

8

5

1

2

1

121

373

241

3

2

1

c

c

c

B][xB][x 1P

BBP ][][ xx BB P ][][ 1 xx

1P


Theorems 4.20 & 4.21Theorems 4.20 & 4.21 Theorem 4.20

The Inverse of a Transition MatrixIf P is the transition matrix from a basis B to a basis B in Rn, then P is invertible and the transition matrix from B to B is given by P1.

Theorem 4.21Transition Matrix from B to BLet B = {v1, v2, …, vn} and B = {u1, u2, …, un} be two bases for Rn. Then the transition matrix P1 from B to B can be found by using Gauss-Jordan elimination on the n 2n matrix [B | B], as follows ][][ 1 PIBB n


Example 4Example 4 Finding a Transition Matrix

Find the transition matrix from B to B for the following bases in R3. B = {(1,0,0), (0,1,0), (0,0,1)} and B = {(1,0,1), (0,1,2), (2,3,5)}.

Sol:

521

310

201

,

100

010

001

BB

121100

373010

241001

100521

010310

001201

][

BB

3I 1P


Example 5Example 5 Finding a Transition Matrix

Find the transition matrix from B to B for the following bases in R2. B = {(3,2), (4, 2)} and B = {(1,2), (2,2)}.

Sol:

2222

4321

BB

123210

2101

PI

32

211P

PIBB

21210

2301

2222

2143


Examples 6 & 7Examples 6 & 7Ex. 6: Coordinate Representation in P3

Find the coordinate matrix of p = 3x3 2x2 + 4 relative to the standard basis in P3. S = {1, x, x2, x3}

Sol: p = 4(1) + 0(x) 2(x2) + 3(x3)

Ex. 7: Coordinate Representation in M3,1

Find the coordinate matrix ofrelative to the standard basis in M3,1.

Sol:

3

2

0

4

][ Sp

TX 341

3

4

1

][

1

0

0

3

0

1

0

4

0

0

1

)1(

3

4

1

SXX

Documents

Chap. 4 Vector Spaces 4.1 Vectors in R n 4.2 Vector Spaces 4.3 Subspaces of Vector Space 4.4 Spanning Sets & Linear Independent 4.5 Basis & Dimension 4.6