Linear Algebra Chapter 9: Complex Scalars Section 9.2. Matrices and - - PowerPoint PPT Presentation

Linear Algebra

April 14, 2018 Chapter 9: Complex Scalars Section 9.2. Matrices and Vector Spaces with Complex Scalars—Proofs of Theorems

() Linear Algebra April 14, 2018 1 / 21

Table of contents

1

Page 472 Number 8

2

Page 473 Number 10

3

Theorem 9.2. Properties of the Euclidean Inner Product

4

Page 473 Number 24

5

Page 473 Number 28

6

Theorem 9.3. Properties of the Conjugate Transpose

7

Page 471 Example 9

8

Page 474 Number 34

9

Page 474 Number 38

() Linear Algebra April 14, 2018 2 / 21

Page 472 Number 8

Page 472 Number 8

Page 472 Number 8. Find A−1 if A =   i 1 − i 1 + i 1 i 1 − i −i 1 − i   .

• Solution. We augment A with I and use the same technique introduced

in Section 1.5, “Inverses of Square Matrices”: [A | I] =   i 1 − i 1 + i 1 1 i 1 1 − i −i 1 − i 1  

R1→−iR1

• 

 1 −1 − i 1 − i −i 1 i 1 1 − i −i 1 − i 1  

R3→R3−(1−i)R1

• ()

Linear Algebra April 14, 2018 3 / 21

Page 472 Number 8

Page 472 Number 8

Page 472 Number 8. Find A−1 if A =   i 1 − i 1 + i 1 i 1 − i −i 1 − i   .

• Solution. We augment A with I and use the same technique introduced

in Section 1.5, “Inverses of Square Matrices”: [A | I] =   i 1 − i 1 + i 1 1 i 1 1 − i −i 1 − i 1  

R1→−iR1

• 

 1 −1 − i 1 − i −i 1 i 1 1 − i −i 1 − i 1  

R3→R3−(1−i)R1

• 2

4 1 −1 − i 1 − i −i 1 i 1 2 − i 1 + i 1 + i 1 3 5 since −i − (1 − i)(−1 − i) = −i − (−2) = 2 − i 1 − i − (1 − i)(1 − i) = 1 − i − (−2i) = 1 + i 0 − (1 − i)(−i) = 1 + i

() Linear Algebra April 14, 2018 3 / 21

Page 472 Number 8

Page 472 Number 8

Page 472 Number 8. Find A−1 if A =   i 1 − i 1 + i 1 i 1 − i −i 1 − i   .

• Solution. We augment A with I and use the same technique introduced

in Section 1.5, “Inverses of Square Matrices”: [A | I] =   i 1 − i 1 + i 1 1 i 1 1 − i −i 1 − i 1  

R1→−iR1

• 

 1 −1 − i 1 − i −i 1 i 1 1 − i −i 1 − i 1  

R3→R3−(1−i)R1

• 2

4 1 −1 − i 1 − i −i 1 i 1 2 − i 1 + i 1 + i 1 3 5 since −i − (1 − i)(−1 − i) = −i − (−2) = 2 − i 1 − i − (1 − i)(1 − i) = 1 − i − (−2i) = 1 + i 0 − (1 − i)(−i) = 1 + i

() Linear Algebra April 14, 2018 3 / 21

Page 472 Number 8

Page 472 Number 8 (continued 1)

Solution (continued).   1 −1 − i 1 − i −i 1 i 1 2 − i 1 + i 1 + i 1  

R1→R1−(1−i)R2

• R3 → R3 − (2 − i)R2

  1 −i 1 + i 1 i 1 −i 1 + i −2 + i 1   since (1 + i) − (−1 − i)(i) = 0 (1 + i) − (2 − i)(i) = −i

R2→R2+R3

• 

 1 −i 1 + i 1 2 + i −1 + i 1 −i 1 + i −2 + i 1  

R3→iR3

• 

 1 −i 1 + i 1 2 + i −1 + i 1 1 −1 + i −1 − 2i i   .

() Linear Algebra April 14, 2018 4 / 21

Page 472 Number 8

Page 472 Number 8 (continued 1)

Solution (continued).   1 −1 − i 1 − i −i 1 i 1 2 − i 1 + i 1 + i 1  

R1→R1−(1−i)R2

• R3 → R3 − (2 − i)R2

  1 −i 1 + i 1 i 1 −i 1 + i −2 + i 1   since (1 + i) − (−1 − i)(i) = 0 (1 + i) − (2 − i)(i) = −i

R2→R2+R3

• 

 1 −i 1 + i 1 2 + i −1 + i 1 −i 1 + i −2 + i 1  

R3→iR3

• 

 1 −i 1 + i 1 2 + i −1 + i 1 1 −1 + i −1 − 2i i   .

() Linear Algebra April 14, 2018 4 / 21

Page 472 Number 8

Page 472 Number 8 (continued 2)

Page 472 Number 8. Find A−1 if A =   i 1 − i 1 + i 1 i 1 − i −i 1 − i   . Solution (continued). So A−1 =   −i 1 + i 2 + i −1 + i 1 −1 + i −1 − 2i i  .

() Linear Algebra April 14, 2018 5 / 21

Page 473 Number 10

Page 473 Number 10

Page 473 Number 10. Solve the system A z =   −1 + i 2 + i 1   where A is as given in Exercise 8.

• Solution. Since A−1 exists as seen in Exercise 8 so, as in Theorem 1.12(iii)

(from the real setting), the unique solution to the system of equations is z = A−1   −1 + i 2 + i 1   =   −i 1 + i 2 + i −1 + i 1 −1 + i −1 − 2i i     −1 + i 2 + i 1  

() Linear Algebra April 14, 2018 6 / 21

Page 473 Number 10

Page 473 Number 10

Page 473 Number 10. Solve the system A z =   −1 + i 2 + i 1   where A is as given in Exercise 8.

• Solution. Since A−1 exists as seen in Exercise 8 so, as in Theorem 1.12(iii)

(from the real setting), the unique solution to the system of equations is z = A−1   −1 + i 2 + i 1   =   −i 1 + i 2 + i −1 + i 1 −1 + i −1 − 2i i     −1 + i 2 + i 1   =   (−i)(−1 + i) + (1 + i)(2 + i) + (0)(1) (1 + i)(−1 + i) + (−1 + i)(2 + i) + (1)(1) (−1 + i)(−1 + i) + (−1 − 2i)(2 + i) + (i)(1)  

() Linear Algebra April 14, 2018 6 / 21

Page 473 Number 10

Page 473 Number 10

Page 473 Number 10. Solve the system A z =   −1 + i 2 + i 1   where A is as given in Exercise 8.

• Solution. Since A−1 exists as seen in Exercise 8 so, as in Theorem 1.12(iii)

(from the real setting), the unique solution to the system of equations is z = A−1   −1 + i 2 + i 1   =   −i 1 + i 2 + i −1 + i 1 −1 + i −1 − 2i i     −1 + i 2 + i 1   =   (−i)(−1 + i) + (1 + i)(2 + i) + (0)(1) (1 + i)(−1 + i) + (−1 + i)(2 + i) + (1)(1) (−1 + i)(−1 + i) + (−1 − 2i)(2 + i) + (i)(1)  

() Linear Algebra April 14, 2018 6 / 21

Page 473 Number 10

Page 473 Number 10 (continued)

Page 473 Number 10. Solve the system A z =   −1 + i 2 + i 1   where A is as given in Exercise 8. Solution (continued). . . . A−1 =   (−i)(−1 + i) + (1 + i)(2 + i) + (0)(1) (1 + i)(−1 + i) + (−1 + i)(2 + i) + (1)(1) (−1 + i)(−1 + i) + (−1 − 2i)(2 + i) + (i)(1)   =   (1 + i) + (1 + 3i) + (0) (−2) + (−3 + i) + (1) (−2i) + (−5i) + (i)   =   2 + 4i −4 + i −6i   .

• ()

Linear Algebra April 14, 2018 7 / 21

Page 473 Number 10

Page 473 Number 10 (continued)

Page 473 Number 10. Solve the system A z =   −1 + i 2 + i 1   where A is as given in Exercise 8. Solution (continued). . . . A−1 =   (−i)(−1 + i) + (1 + i)(2 + i) + (0)(1) (1 + i)(−1 + i) + (−1 + i)(2 + i) + (1)(1) (−1 + i)(−1 + i) + (−1 − 2i)(2 + i) + (i)(1)   =   (1 + i) + (1 + 3i) + (0) (−2) + (−3 + i) + (1) (−2i) + (−5i) + (i)   =   2 + 4i −4 + i −6i   .

• ()

Linear Algebra April 14, 2018 7 / 21

Theorem 9.2. Properties of the Euclidean Inner Product

Theorem 9.2

Theorem 9.2. Properties of the Euclidean Inner Product. Let u, v, w ∈ Cn and let z be a complex scalar. Then: (1) u, u ≥ 0 and u, u = 0 if and only if u = 0, (2) u, v = v, u, (3) ( u + v), w = u, w + v, w, (4) w, ( u + v) = w, u + w, v, (5) z u, v = z u, v and u, z v = z u, v.

• Proof. (1) (Page 473 Number 16) For

u = [u1, u2, . . . , un] ∈ Cn we have

• u,

u = u1u1 + u2u2 + · · · + unun = |a1|2 + |u2|2 + · · · + |un|2 ≥ 0 since each |uk|2 is a nonnegative real number.

() Linear Algebra April 14, 2018 8 / 21

Theorem 9.2. Properties of the Euclidean Inner Product

Theorem 9.2

Theorem 9.2. Properties of the Euclidean Inner Product. Let u, v, w ∈ Cn and let z be a complex scalar. Then: (1) u, u ≥ 0 and u, u = 0 if and only if u = 0, (2) u, v = v, u, (3) ( u + v), w = u, w + v, w, (4) w, ( u + v) = w, u + w, v, (5) z u, v = z u, v and u, z v = z u, v.

• Proof. (1) (Page 473 Number 16) For

u = [u1, u2, . . . , un] ∈ Cn we have

• u,

u = u1u1 + u2u2 + · · · + unun = |a1|2 + |u2|2 + · · · + |un|2 ≥ 0 since each |uk|2 is a nonnegative real number. In addition, |u1|2 + |u2|2 + · · · + |un|2 = 0 if and only if |u1|2 = |u2|2 = · · · = |un|2 = 0; that is u, u = 0 if and only if u1 = u2 = · · · − un, or u = 0.

() Linear Algebra April 14, 2018 8 / 21

Theorem 9.2. Properties of the Euclidean Inner Product

Theorem 9.2

Theorem 9.2. Properties of the Euclidean Inner Product. Let u, v, w ∈ Cn and let z be a complex scalar. Then: (1) u, u ≥ 0 and u, u = 0 if and only if u = 0, (2) u, v = v, u, (3) ( u + v), w = u, w + v, w, (4) w, ( u + v) = w, u + w, v, (5) z u, v = z u, v and u, z v = z u, v.

• Proof. (1) (Page 473 Number 16) For

u = [u1, u2, . . . , un] ∈ Cn we have

• u,

u = u1u1 + u2u2 + · · · + unun = |a1|2 + |u2|2 + · · · + |un|2 ≥ 0 since each |uk|2 is a nonnegative real number. In addition, |u1|2 + |u2|2 + · · · + |un|2 = 0 if and only if |u1|2 = |u2|2 = · · · = |un|2 = 0; that is u, u = 0 if and only if u1 = u2 = · · · − un, or u = 0.

() Linear Algebra April 14, 2018 8 / 21

Theorem 9.2. Properties of the Euclidean Inner Product

Theorem 9.2 (continued 1)

Theorem 9.2. Properties of the Euclidean Inner Product. Let u, v, w ∈ Cn and let z be a complex scalar. Then: (2) u, v = v, u, (3) ( u + v), w = u, w + v, w, Proof (continued). (3 and 4) (Page 473 Number 18) For

• u = [u1, u2, . . . , un]

v = [v1, v2, . . . , vn], and w = [w1, w2, . . . , wn]. (3) We have ( u + v), w = ([u1, u2, . . . , un] + [v1, v2, . . . , vn]), [w1, w2, . . . , wn] = [u1 + v1, u2 + v2, . . . , un + vn], [w1, w2, . . . , wn] = (u1 + v1)w1 + (u2 + v2)w2 + · · · + (un + vn)wn = (u1 + v1)w1 + (u2 + v2)w2 + · · · + (un + vn)wn by Theorem 9.1(1) = ((u1w1+u2w2+· · ·+unwn)+(v1w1+v2w2+· · ·+vnwn) = u, w+ v, w.

() Linear Algebra April 14, 2018 9 / 21

Theorem 9.2. Properties of the Euclidean Inner Product

Theorem 9.2 (continued 1)

Theorem 9.2. Properties of the Euclidean Inner Product. Let u, v, w ∈ Cn and let z be a complex scalar. Then: (2) u, v = v, u, (3) ( u + v), w = u, w + v, w, Proof (continued). (3 and 4) (Page 473 Number 18) For

• u = [u1, u2, . . . , un]

v = [v1, v2, . . . , vn], and w = [w1, w2, . . . , wn]. (3) We have ( u + v), w = ([u1, u2, . . . , un] + [v1, v2, . . . , vn]), [w1, w2, . . . , wn] = [u1 + v1, u2 + v2, . . . , un + vn], [w1, w2, . . . , wn] = (u1 + v1)w1 + (u2 + v2)w2 + · · · + (un + vn)wn = (u1 + v1)w1 + (u2 + v2)w2 + · · · + (un + vn)wn by Theorem 9.1(1) = ((u1w1+u2w2+· · ·+unwn)+(v1w1+v2w2+· · ·+vnwn) = u, w+ v, w.

() Linear Algebra April 14, 2018 9 / 21

Theorem 9.2. Properties of the Euclidean Inner Product

Theorem 9.2 (continued 2)

Theorem 9.2. Properties of the Euclidean Inner Product. Let u, v, w ∈ Cn and let z be a complex scalar. Then: (4) w, ( u + v) = w, u + w, v, Proof (continued). (4) We have

• w, (

u + v) = ( u + v), w by Part (2) =

• u,

w + v, w by Part (3) =

• u,

w + v, w by Theorem 9.1(1) =

• w,

u+ w, v by Part (2).

() Linear Algebra April 14, 2018 10 / 21

Page 473 Number 24

Page 473 Number 24

Page 473 Number 24. Find a unit vector perpendicular to [2 − i, 1 + i].

• Solution. Let

v = [v1, v2] ∈ C2 be a vector perpendicular to the given

• vector. Then

[2 − i, 1 + i], v = [2 − i, 1 + i], [v1, v2] = (2 − i)v1 + (1 + i)v2 = (2 + i)v1 + (1 − i)v2 = 0. So we need v1 = −1−i

2+i v2.

() Linear Algebra April 14, 2018 11 / 21

Page 473 Number 24

Page 473 Number 24

Page 473 Number 24. Find a unit vector perpendicular to [2 − i, 1 + i].

• Solution. Let

v = [v1, v2] ∈ C2 be a vector perpendicular to the given

• vector. Then

[2 − i, 1 + i], v = [2 − i, 1 + i], [v1, v2] = (2 − i)v1 + (1 + i)v2 = (2 + i)v1 + (1 − i)v2 = 0. So we need v1 = −1−i

2+i v2. Let’s take v2 = 2 + i so that v1 = −1 + i. So

• v = [−1 + i, 2 + i] is perpendicular to the given vector. So one possible

answer is v/

• v. Notice that

v =

• | − 1 + i|2 + |2 + i|2 =

√

• 7. So we

can take vector

1 √ 7[−1 + i, 2 + i]. We can also take the negative of this

vector, −1

√ 7[−1 + i, 2 + i].

() Linear Algebra April 14, 2018 11 / 21

Page 473 Number 24

Page 473 Number 24

Page 473 Number 24. Find a unit vector perpendicular to [2 − i, 1 + i].

• Solution. Let

v = [v1, v2] ∈ C2 be a vector perpendicular to the given

• vector. Then

[2 − i, 1 + i], v = [2 − i, 1 + i], [v1, v2] = (2 − i)v1 + (1 + i)v2 = (2 + i)v1 + (1 − i)v2 = 0. So we need v1 = −1−i

2+i v2. Let’s take v2 = 2 + i so that v1 = −1 + i. So

• v = [−1 + i, 2 + i] is perpendicular to the given vector. So one possible

answer is v/

• v. Notice that

v =

• | − 1 + i|2 + |2 + i|2 =

√

• 7. So we

can take vector

1 √ 7[−1 + i, 2 + i]. We can also take the negative of this

vector, −1

√ 7[−1 + i, 2 + i].

() Linear Algebra April 14, 2018 11 / 21

Page 473 Number 28

Page 473 Number 28

Page 473 Number 28. Use the Gram-Schmidt Process to convert the basis { a1, a2, a3} = {[1 − i, 1i, 1 + i], [1, 1, −1 − i], [1, i, −i]} of C3 into an

• rthogonal basis {

v1, v2, v3} of C3.

• Solution. First, we take

v1 = a1 = [1 − i, 1 + i, 1 + i]. Next,

• v2 =

a2 − v1, a2

• v1,

v1 v1 = [1, 1, −1−i]− [1 − i, 1 + i, 1 + i], [1, 1, −1 − i] [1 − i, 1 + i, 1 + i], [1 − i, 1 + i, 1 + i][1−i, 1+i, 1+i] =

() Linear Algebra April 14, 2018 12 / 21

Page 473 Number 28

Page 473 Number 28

Page 473 Number 28. Use the Gram-Schmidt Process to convert the basis { a1, a2, a3} = {[1 − i, 1i, 1 + i], [1, 1, −1 − i], [1, i, −i]} of C3 into an

• rthogonal basis {

v1, v2, v3} of C3.

• Solution. First, we take

v1 = a1 = [1 − i, 1 + i, 1 + i]. Next,

• v2 =

a2 − v1, a2

• v1,

v1 v1 = [1, 1, −1−i]− [1 − i, 1 + i, 1 + i], [1, 1, −1 − i] [1 − i, 1 + i, 1 + i], [1 − i, 1 + i, 1 + i][1−i, 1+i, 1+i] = [1, 1, −1−i]− (1 − i)(1) + (1 + i)(1) + (1 + i)(−1 − i) (1 − i)(1 − i) + (1 + i)(1 + i) + (1 + i)(1 + i) [1−i, 1+i, 1+i] = [1, 1, −1 − i] − (1 + i) + (1 − i) + (−2) 2 + 2 + 2 [1 − i, 1 + i, 1 + i] = [1, 1, −1 − i] − 0[1 − i, 1 + i, 1 + i] = [1, 1, −1 − i].

() Linear Algebra April 14, 2018 12 / 21

Page 473 Number 28

Page 473 Number 28

Page 473 Number 28. Use the Gram-Schmidt Process to convert the basis { a1, a2, a3} = {[1 − i, 1i, 1 + i], [1, 1, −1 − i], [1, i, −i]} of C3 into an

• rthogonal basis {

v1, v2, v3} of C3.

• Solution. First, we take

v1 = a1 = [1 − i, 1 + i, 1 + i]. Next,

• v2 =

a2 − v1, a2

• v1,

v1 v1 = [1, 1, −1−i]− [1 − i, 1 + i, 1 + i], [1, 1, −1 − i] [1 − i, 1 + i, 1 + i], [1 − i, 1 + i, 1 + i][1−i, 1+i, 1+i] = [1, 1, −1−i]− (1 − i)(1) + (1 + i)(1) + (1 + i)(−1 − i) (1 − i)(1 − i) + (1 + i)(1 + i) + (1 + i)(1 + i) [1−i, 1+i, 1+i] = [1, 1, −1 − i] − (1 + i) + (1 − i) + (−2) 2 + 2 + 2 [1 − i, 1 + i, 1 + i] = [1, 1, −1 − i] − 0[1 − i, 1 + i, 1 + i] = [1, 1, −1 − i].

() Linear Algebra April 14, 2018 12 / 21

Page 473 Number 28

Page 473 Number 28 (continued 1)

Page 473 Number 28. Use the Gram-Schmidt Process to convert the basis { a1, a2, a3} = {[1 − i, 1i, 1 + i], [1, 1, −1 − i], [1, i, −i]} of C3 into an

• rthogonal basis {

v1, v2, v3} of C3. Solution (continued). Finally,

• v3 =

a3 −

• v1,

a3

• v1,

v1 v1 + v2, a3

• v2,

v2 v2

• = [1, i, −i] −
• [1 − i, 1 + i, 1 + i], [1, i, −i]

[1 − i, 1 + i, 1 + i], [1 − i, 1 + i, 1 + i][1 − i, 1 + i, 1 + i] + [1, 1, −1 − i], [1, i, −i] [1, 1, −1 − i], [1, 1, −1 − i][1, 1, −1 − i]

• = [1, i, −i] − (1 − i)(1) + (1 + i)(i) + (1 + i)(−i)

2 + 2 + 2 [1 − i, 1 + i, 1 + i] − (1)(1) + (−1)(i) + (−1 − i)(−i) (1)(1) + (1)(1) + (−1 − i)(−1 − i) [1, 1, −1 − i]

() Linear Algebra April 14, 2018 13 / 21

Page 473 Number 28

Page 473 Number 28 (continued 1)

Page 473 Number 28. Use the Gram-Schmidt Process to convert the basis { a1, a2, a3} = {[1 − i, 1i, 1 + i], [1, 1, −1 − i], [1, i, −i]} of C3 into an

• rthogonal basis {

v1, v2, v3} of C3. Solution (continued). Finally,

• v3 =

a3 −

• v1,

a3

• v1,

v1 v1 + v2, a3

• v2,

v2 v2

• = [1, i, −i] −
• [1 − i, 1 + i, 1 + i], [1, i, −i]

[1 − i, 1 + i, 1 + i], [1 − i, 1 + i, 1 + i][1 − i, 1 + i, 1 + i] + [1, 1, −1 − i], [1, i, −i] [1, 1, −1 − i], [1, 1, −1 − i][1, 1, −1 − i]

• = [1, i, −i] − (1 − i)(1) + (1 + i)(i) + (1 + i)(−i)

2 + 2 + 2 [1 − i, 1 + i, 1 + i] − (1)(1) + (−1)(i) + (−1 − i)(−i) (1)(1) + (1)(1) + (−1 − i)(−1 − i) [1, 1, −1 − i]

() Linear Algebra April 14, 2018 13 / 21

Page 473 Number 28

Page 473 Number 28 (continued 2)

Solution (continued). = [1, i, −i] − (1 + i) + (1 + i) + (−1 − i) 6 [1 − i, 1 + i, 1 + i] −(1) + (i) + (1 + i) 1 + 1 + 2 [1, −1, −1 − i] = [1, i, −i] − 1 + i 6 [1 − i, 1 + i, 1 + i] − 1 + i 2 [1, 1, −1 − i] = [1, i, −i] − 1 6[1, 2i, 2i] − 1 2[1 + i, 1 + i, −2i] 1 6[6−2−3(2+i), 6i −2i −3(1+i), −6i −2i +6i] = 1 6[−1−3i, −3+i, −2i]. So an orthogonal basis is { v1, v2, v3} = {[1 − i, 1 + i, 1 + i], [1, 1, −1 − i], 1

6[−1 − 3i, −3 + 2i, −2i]}.

• ()

Linear Algebra April 14, 2018 14 / 21

Page 473 Number 28

Page 473 Number 28 (continued 2)

Solution (continued). = [1, i, −i] − (1 + i) + (1 + i) + (−1 − i) 6 [1 − i, 1 + i, 1 + i] −(1) + (i) + (1 + i) 1 + 1 + 2 [1, −1, −1 − i] = [1, i, −i] − 1 + i 6 [1 − i, 1 + i, 1 + i] − 1 + i 2 [1, 1, −1 − i] = [1, i, −i] − 1 6[1, 2i, 2i] − 1 2[1 + i, 1 + i, −2i] 1 6[6−2−3(2+i), 6i −2i −3(1+i), −6i −2i +6i] = 1 6[−1−3i, −3+i, −2i]. So an orthogonal basis is { v1, v2, v3} = {[1 − i, 1 + i, 1 + i], [1, 1, −1 − i], 1

6[−1 − 3i, −3 + 2i, −2i]}.

• ()

Linear Algebra April 14, 2018 14 / 21

Theorem 9.3. Properties of the Conjugate Transpose

Theorem 9.3

Theorem 9.3. Properties of the Conjugate Transpose. Let A and B be m × n matrices. Then (1) (A∗)∗ = A, (2) (A + B)∗ = A∗ + B∗, (3) (zA)∗ = zA∗ for an scalar z ∈ C, (4) If A and B are square matrices, then (AB)∗ = B∗A∗.

• Proof. (Page 473 Number 32)

(1) Since A∗ = [aij]T, the (i, j) entry of A∗ is aji. Let B = [bij] = A∗ so that bij = aji. Then (A∗)∗ = B∗ =

• (aji)

T = [aji]T by Theorem 9.1(5), “Properties of Conjugates in Cn.” So the (i, j) entry of (A∗)∗ = B∗ is aij. That is, (A∗)∗ = A.

() Linear Algebra April 14, 2018 15 / 21

Theorem 9.3. Properties of the Conjugate Transpose

Theorem 9.3

Theorem 9.3. Properties of the Conjugate Transpose. Let A and B be m × n matrices. Then (1) (A∗)∗ = A, (2) (A + B)∗ = A∗ + B∗, (3) (zA)∗ = zA∗ for an scalar z ∈ C, (4) If A and B are square matrices, then (AB)∗ = B∗A∗.

• Proof. (Page 473 Number 32)

(1) Since A∗ = [aij]T, the (i, j) entry of A∗ is aji. Let B = [bij] = A∗ so that bij = aji. Then (A∗)∗ = B∗ =

• (aji)

T = [aji]T by Theorem 9.1(5), “Properties of Conjugates in Cn.” So the (i, j) entry of (A∗)∗ = B∗ is aij. That is, (A∗)∗ = A.

() Linear Algebra April 14, 2018 15 / 21

Theorem 9.3. Properties of the Conjugate Transpose

Theorem 9.3 (continued)

Proof (continued). (2) Let A = [aij] and B = [bij]. Then A + B = [aij + bij] and (A + B)∗ =

• aij + bij

T = [aji + bji]T by Theorem 9.1(1) =

• [aji] +
• bji

T = [aji]T +

• bji

T by Note 1.3.B (which is stated for real matrices but also holds for complex matrices) = A∗ + B∗. (3) Let A = [aij] and B = [bij] be n × n matrices. Then (AB)∗ = ([aij[bij])∗ = n

• k=1

aikbkj ∗ = n

• k=1

aikbkj T = n

• k=1

bT

jkaT ki

• where bT

kj = bjk and aT ik = aki

() Linear Algebra April 14, 2018 16 / 21

Theorem 9.3. Properties of the Conjugate Transpose

Theorem 9.3 (continued)

Proof (continued). (2) Let A = [aij] and B = [bij]. Then A + B = [aij + bij] and (A + B)∗ =

• aij + bij

T = [aji + bji]T by Theorem 9.1(1) =

• [aji] +
• bji

T = [aji]T +

• bji

T by Note 1.3.B (which is stated for real matrices but also holds for complex matrices) = A∗ + B∗. (3) Let A = [aij] and B = [bij] be n × n matrices. Then (AB)∗ = ([aij[bij])∗ = n

• k=1

aikbkj ∗ = n

• k=1

aikbkj T = n

• k=1

bT

jkaT ki

• where bT

kj = bjk and aT ik = aki

() Linear Algebra April 14, 2018 16 / 21

Theorem 9.3. Properties of the Conjugate Transpose

Theorem 9.3 (continued)

Proof (continued). (AB)∗ = n

• k=1

bT

jkaT ki

• where bT

kj = bjk and aT ik = aki

= n

• k=1

b

T jkaT ki

• by Theorem 9.1(1) and (3)

= n

• k=1

b

T ikaT kj

• =
• b

T ij

aT

ij

• =
• bji
• [aji] =
• bij

T [aij]T = [bij]∗[aij]∗ = B∗A∗.

() Linear Algebra April 14, 2018 17 / 21

Page 471 Example 9

Page 471 Example 9

Page 471 Example 9. Let T : Cn → Cn be a linear transformation having a unitary matrix U as its matrix representation with respect to the standard basis. Prove that T( z) = z for all z ∈ Cn.

• Proof. Since U is the standard matrix representation of T then for

z ∈ Cn we have T( z) = U

• z. Then

T( z)2 = U z2 = U z, U z by the definition of norm = (U z)∗(U z) by Note 9.2.A = ( z∗U∗)U z by Theorem 9.3(4) =

• z∗(U∗U)

z = z∗I z = z∗ z =

• z,

z by Note 9.2.A =

• z2.

Hence T( z = z since norms are nonnegative.

() Linear Algebra April 14, 2018 18 / 21

Page 471 Example 9

Page 471 Example 9

Page 471 Example 9. Let T : Cn → Cn be a linear transformation having a unitary matrix U as its matrix representation with respect to the standard basis. Prove that T( z) = z for all z ∈ Cn.

• Proof. Since U is the standard matrix representation of T then for

z ∈ Cn we have T( z) = U

• z. Then

T( z)2 = U z2 = U z, U z by the definition of norm = (U z)∗(U z) by Note 9.2.A = ( z∗U∗)U z by Theorem 9.3(4) =

• z∗(U∗U)

z = z∗I z = z∗ z =

• z,

z by Note 9.2.A =

• z2.

Hence T( z = z since norms are nonnegative.

() Linear Algebra April 14, 2018 18 / 21

Page 474 Number 34

Page 474 Number 34

Page 474 Number 34. Prove that, for vectors v1, v2, . . . , vn in Cn, { v1, v2, . . . , vn} is a basis for Cn if and only if { v1, v2, . . . , vn} is a basis for Cn.

• Proof. Since the standard basis of Cn, {ˆ

e1, ˆ e2, . . . , ˆ en}, has n vectors then the dimension of Cn is, of course, n by Definition 3.7, “Dimension of a Vector Space.” Consider the equation a1 v1 + a2 v2 + · · · + an vn = 0.

() Linear Algebra April 14, 2018 19 / 21

Page 474 Number 34

Page 474 Number 34

Page 474 Number 34. Prove that, for vectors v1, v2, . . . , vn in Cn, { v1, v2, . . . , vn} is a basis for Cn if and only if { v1, v2, . . . , vn} is a basis for Cn.

• Proof. Since the standard basis of Cn, {ˆ

e1, ˆ e2, . . . , ˆ en}, has n vectors then the dimension of Cn is, of course, n by Definition 3.7, “Dimension of a Vector Space.” Consider the equation a1 v1 + a2 v2 + · · · + an vn =

• 0. Then

a1 v1 + a2 v2 + · · · + an vn = 0 of, by Theorem 9.1, a1 v1 + a2 v2 + · · · + an vn =

• 0. Since {

v1, v2, . . . , vn} is a basis for Cn by hypothesis then this set of vectors is linearly independent by the definition

• f basis (Definition 3.6) and so a1 = a2 = · · · = an = 0.

() Linear Algebra April 14, 2018 19 / 21

Page 474 Number 34

Page 474 Number 34

Page 474 Number 34. Prove that, for vectors v1, v2, . . . , vn in Cn, { v1, v2, . . . , vn} is a basis for Cn if and only if { v1, v2, . . . , vn} is a basis for Cn.

• Proof. Since the standard basis of Cn, {ˆ

e1, ˆ e2, . . . , ˆ en}, has n vectors then the dimension of Cn is, of course, n by Definition 3.7, “Dimension of a Vector Space.” Consider the equation a1 v1 + a2 v2 + · · · + an vn =

• 0. Then

a1 v1 + a2 v2 + · · · + an vn = 0 of, by Theorem 9.1, a1 v1 + a2 v2 + · · · + an vn =

• 0. Since {

v1, v2, . . . , vn} is a basis for Cn by hypothesis then this set of vectors is linearly independent by the definition

• f basis (Definition 3.6) and so a1 = a2 = · · · = an = 0. Hence

a1 = a2 = · · · = an = 0 is necessary and so { v1, v2, . . . , vn} is a linearly independent set of n vectors in Cn. By Corollary 3.2.B, { v1, v2, . . . , vn} is a basis for Cn.

() Linear Algebra April 14, 2018 19 / 21

Page 474 Number 34

Page 474 Number 34

Page 474 Number 34. Prove that, for vectors v1, v2, . . . , vn in Cn, { v1, v2, . . . , vn} is a basis for Cn if and only if { v1, v2, . . . , vn} is a basis for Cn.

• Proof. Since the standard basis of Cn, {ˆ

e1, ˆ e2, . . . , ˆ en}, has n vectors then the dimension of Cn is, of course, n by Definition 3.7, “Dimension of a Vector Space.” Consider the equation a1 v1 + a2 v2 + · · · + an vn =

• 0. Then

a1 v1 + a2 v2 + · · · + an vn = 0 of, by Theorem 9.1, a1 v1 + a2 v2 + · · · + an vn =

• 0. Since {

v1, v2, . . . , vn} is a basis for Cn by hypothesis then this set of vectors is linearly independent by the definition

• f basis (Definition 3.6) and so a1 = a2 = · · · = an = 0. Hence

a1 = a2 = · · · = an = 0 is necessary and so { v1, v2, . . . , vn} is a linearly independent set of n vectors in Cn. By Corollary 3.2.B, { v1, v2, . . . , vn} is a basis for Cn.

() Linear Algebra April 14, 2018 19 / 21

Page 474 Number 38

Page 474 Number 38

Page 474 Number 38. Prove that the product of two n × n unitary matrices is also a unitary matrix. What about the sum?

• Proof. Suppose A and B are unitary n × n matrices. Then, by definition,

A∗A = B∗B = I. Now (AB)∗(AB) = B∗A∗AB by Theorem 9.3(4), “Properties of Conjugate Transpose” = B∗(A∗A)B = B∗IB = B∗B = I. So AB is, by definition, unitary.

() Linear Algebra April 14, 2018 20 / 21

Page 474 Number 38

Page 474 Number 38

Page 474 Number 38. Prove that the product of two n × n unitary matrices is also a unitary matrix. What about the sum?

• Proof. Suppose A and B are unitary n × n matrices. Then, by definition,

A∗A = B∗B = I. Now (AB)∗(AB) = B∗A∗AB by Theorem 9.3(4), “Properties of Conjugate Transpose” = B∗(A∗A)B = B∗IB = B∗B = I. So AB is, by definition, unitary.

() Linear Algebra April 14, 2018 20 / 21

Page 474 Number 38

Page 474 Number 38 (continued)

Page 474 Number 38. Prove that the product of two n × n unitary matrices is also a unitary matrix. What about the sum? Solution (continued). For a sum, consider the 2 × 2 real matrices A = 1 1

• and B =

1 −1

• . Then A = A∗, B = B∗, A∗A = I,

and B∗B = I so that A and B are unitary (also “orthogonal” as real matrices).

() Linear Algebra April 14, 2018 21 / 21

Page 474 Number 38

Page 474 Number 38 (continued)

Page 474 Number 38. Prove that the product of two n × n unitary matrices is also a unitary matrix. What about the sum? Solution (continued). For a sum, consider the 2 × 2 real matrices A = 1 1

• and B =

1 −1

• . Then A = A∗, B = B∗, A∗A = I,

and B∗B = I so that A and B are unitary (also “orthogonal” as real matrices). But A + B = 1

• and (A + B)∗ =

1

• so that

(A + B)∗(A + B) = 1 1

• =

1

• =

1 1

• = I.

So A + B is NOT unitary.

() Linear Algebra April 14, 2018 21 / 21

Page 474 Number 38

Page 474 Number 38 (continued)

Page 474 Number 38. Prove that the product of two n × n unitary matrices is also a unitary matrix. What about the sum? Solution (continued). For a sum, consider the 2 × 2 real matrices A = 1 1

• and B =

1 −1

• . Then A = A∗, B = B∗, A∗A = I,

and B∗B = I so that A and B are unitary (also “orthogonal” as real matrices). But A + B = 1

• and (A + B)∗ =

1

• so that

(A + B)∗(A + B) = 1 1

• =

1

• =

1 1

• = I.

So A + B is NOT unitary.

() Linear Algebra April 14, 2018 21 / 21