Ch04_1

4.8 Rank

DefinitionLet A be an m n matrix. The rows of A may be viewed as row vectors r1, …, rm, and the columns as column vectors c1, …, cn. Each row vector will have n components, and each column vector will have m components, The row vectors will span a subspace of Rn called the row space of A, and the column vectors will span a subspace of Rm called the column space of A.

Rank enables one to relate matrices to vectors, and vice versa.

Ch04_2

Example 1

Consider the matrix

(1) The row vectors of A are

014561432121

)0 ,1 ,4 ,5(),6 ,1 ,4 ,3( ),2 ,1 ,2 ,1( 321 rrr

(2) The column vectors of A are

These vectors span a subspace of R3 called the column space of A.

062

111

442

531

4321 cccc

These vectors span a subspace of R4 called the row space of A.

Ch04_3

Theorem 4.16The row space and the column space of a matrix A have the same dimension.

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

Ch04_4

Example 2Determine the rank of the matrix

852210321

A

Solution

The third row of A is a linear combination of the first two rows:(2, 5, 8) = 2(1, 2, 3) + (0, 1, 2)

Hence the three rows of A are linearly dependent. The rank of A must be less than 3. Since (1, 2, 3) is not a scalar multiple of (0, 1, 2), these two vectors are linearly independent. These vectors form a basis for the row space of A. Thus rank(A) = 2.

Ch04_5

Theorem 4.17The nonzero row vectors of a matrix A that is in reduced echelon form are a basis for the row space of A. The rank of A is the number of nonzero row vectors.

Ch04_6

Example 3

Find the rank of the matrix

This matrix is in reduced echelon form. There are three nonzero row vectors, namely (1, 2, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1). According to the previous theorem, these three vectors form a basis for the row space of A. Rank(A) = 3.

0000100001000021

A

Ch04_7

Theorem 4.18Let A and B be row equivalent matrices. Then A and B have the same the row space. rank(A) = rank(B).

Theorem 4.19

Let E be a reduced echelon form of a matrix A. The nonzero row vectors of E form a basis for the row space of A. The rank of A is the number of nonzero row vectors in E.

Ch04_8

Example 4Find a basis for the row space of the following matrix A, and determine its rank.

511452321

A

Solution

Use elementary row operations to find a reduced echelon form of the matrix A. We get

000210701

210210321

511452321

The two vectors (1, 0, 7), (0, 1, 2) form a basis for the row space of A. Rank(A) = 2.

Ch04_9

4.9 Orthonormal Vectors and Projections

DefinitionA set of vectors in a vector space V is said to be an orthogonal set if every pair of vectors in the set is orthogonal. The set is said to be an orthonormal set if it is orthogonal and each vector is a unit vector.

Ch04_10

Example 1

5

3 ,

5

4 0, ,

5

4 ,

5

3 0,),0 ,0 ,1(

Solution

(1) orthogonal: ;0,,0)0,0,1( 54

53

;0,,0)0,0,1( 53

54

;0,,0,,0 53

54

54

53

(2) unit vector:

10 ,0,

10 , 0,

1001)0,0,1(

222

222

222

53

54

53

54

54

53

54

53

Thus the set is thus an orthonormal set.

Show that the set is an orthonormal set.

Ch04_11

Theorem 4.20An orthogonal set of nonzero vectors in a vector space is linearly independent.

Proof Let {v1, …, vm} be an orthogonal set of nonzero vectors in a vector space V. Let us examine the identity

c1v1 + c2v2 + … + cmvm = 0Let vi be the ith vector of the orthogonal set. Take the dot product of each side of the equation with vi and use the properties of the dot product. We get

Since the vectors v1, …, v2 are mutually orthogonal, vj‧vi = 0 unless j = i. Thus

Since vi is a nonzero, then vi‧vi 0. Thus ci = 0. Letting i = 1, …, m, we get c1 = 0, cm = 0, proving that the vectors are linearly independent.

0 iiic vv

0)(

2211

2211

immii

iimm

cccccc

vvvvvvv0vvvv

Ch04_12

DefinitionA basis that is an orthogonal set is said to be an orthogonal basis. A basis that is an orthonormal set is said to be an orthonormal basis.Standard Bases

• R2: {(1, 0), (0, 1)}• R3: {(1, 0, 0), (0, 1, 0), (0, 0, 1)} orthonormal bases• Rn: {(1, …, 0), …, (0, …, 1)}

Theorem 4.21Let {u1, …, un} be an orthonormal basis for a vector space V. Let v be a vector in V. v can be written as a linearly combination of these basis vectors as follows:

nn uuvuuvuuvv )()()( 2211

Ch04_13

Example 2The following vectors u1, u2, and u3 form an orthonormal basis for R3. Express the vector v = (7, 5, 10) as a linear combination of these vectors.

5

3 ,

5

4 0, ,

5

4 ,

5

3 0, 0), 0, 1,( 321 uuu

Solution

105

3 ,

5

4 0,10) ,5 ,7(

55

4 ,

5

3 0,10) ,5 ,7(

70) 0, ,1(10) ,5 ,7(

3

2

1

uv

uv

uv

Thus

321 1057 uuuv

Ch04_14

Projection of One vector onto Another Vector

Figure 4.17

Let v and u be vectors in Rn with angle (0 ) between them.

||||

|||| |||| ||||

cos |||| cos

u

uv

uv

uvv

v

OBOA

OA : the projection of v onto u

uuu

uv

u

u

u

uv

)||||

)(||||

( OA

.0 then 2 / If : Note

uu

uv

.proj uuu

uvvu

So we define

Ch04_15

DefinitionThe projection of a vector v onto a nonzero vector u in Rn is denoted projuv and is defined by

uuuuv

vu proj

Figure 4.18O

Ch04_16

Example 3Determine the projection of the vector v = (6, 7) onto the vector u = (1, 4).

Solution

171614) (1,4) (1,342864) (1,7) (6,

uuuv

Thus

The projection of v onto u is (2, 8).

8) (2,4) ,1(1734

proj uuuuv

vu

Ch04_17

Theorem 4.22The Gram-Schmidt Orthogonalization ProcessLet {v1, …, vn} be a basis for a vector space V. The set of vectors {u1, …, un} defined as follows is orthogonal. To obtain an orthonormal basis for V, normalize each of the vectors u1, …, un .

nnnn nvvvu

vvvu

vvuvu

uu

uu

u

11

21

1

projproj

projproj

proj

3333

222

11

Figure 4.19

Ch04_18

Example 4The set {(1, 2, 0, 3), (4, 0, 5, 8), (8, 1, 5, 6)} is linearly independent in R4. The vectors form a basis for a three-dimensional subspace V of R4. Construct an orthonormal basis for V.

Solution

Let v1 = (1, 2, 0, 3), v2 = (4, 0, 5, 8), v3 = (8, 1, 5, 6)}. Use the Gram-Schmidt process to construct an orthogonal set {u1, u2, u3} from these vectors.

2) 0, ,1 ,4()(

)(

)(

)(

projprojLet

2) 5, 4, 2,()()(

projLet

3) 0, 2, 1,(Let

222

231

11

133

3333

111

222222

11

21

1

uuu

uvu

uu

uvv

vvvu

uuuuv

vvvu

vu

uu

u

Ch04_19

The set {(1, 2, 0, 3), (2, 4, 5, 2), (4, 1, 0, 2)} is an orthogonal basis for V. Normalize them to get an orthonormal basis:

21)2(0142) 0, 1, (4,

725)4(22) 5, 4, (2,

1430213) 0, 2, (1,

2222

2222

2222

orthonormal basis for V:

212

0, ,211

,214

,72

,75

,74

,72

,143

0, ,142

,141