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

Linear Algebra Chapter 9: Complex Scalars Section 9.2. Matrices and Vector Spaces with Complex Scalars—Proofs of Theorems April 14, 2018 () Linear Algebra April 14, 2018 1 / 21

Table of contents Page 472 Number 8 1 Page 473 Number 10 2 Theorem 9.2. Properties of the Euclidean Inner Product 3 Page 473 Number 24 4 Page 473 Number 28 5 Theorem 9.3. Properties of the Conjugate Transpose 6 Page 471 Example 9 7 Page 474 Number 34 8 Page 474 Number 38 9 () Linear Algebra April 14, 2018 2 / 21

Page 472 Number 8 Page 472 Number 8   i 1 − i 1 + i Page 472 Number 8. Find A − 1 if A =  . 0 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”:  1 − i 1 + i 1 0 0  i [ A | I ] = 0 1 i 0 1 0   1 − i − i 1 − i 0 0 1   1 − 1 − i 1 − i − i 0 0 R 3 → R 3 − (1 − i ) R 1 R 1 →− iR 1 � � 0 1 0 1 0 i   1 − i − i 1 − i 0 0 1 () Linear Algebra April 14, 2018 3 / 21

Page 472 Number 8 Page 472 Number 8   i 1 − i 1 + i Page 472 Number 8. Find A − 1 if A =  . 0 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”:  1 − i 1 + i 1 0 0  i [ A | I ] = 0 1 i 0 1 0   1 − i − i 1 − i 0 0 1   1 − 1 − i 1 − i − i 0 0 R 3 → R 3 − (1 − i ) R 1 R 1 →− iR 1 � � 0 1 0 1 0 i   1 − i − i 1 − i 0 0 1 − i − (1 − i )( − 1 − i ) = − i − ( − 2) 2 3 1 − 1 − i 1 − i − i 0 0 = 2 − i 5 since 0 1 i 0 1 0 1 − i − (1 − i )(1 − i ) = 1 − i − ( − 2 i ) 4 0 2 − i 1 + i 1 + i 0 1 = 1 + i 0 − (1 − i )( − i ) = 1 + i () Linear Algebra April 14, 2018 3 / 21

Page 472 Number 8 Page 472 Number 8   i 1 − i 1 + i Page 472 Number 8. Find A − 1 if A =  . 0 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”:  1 − i 1 + i 1 0 0  i [ A | I ] = 0 1 i 0 1 0   1 − i − i 1 − i 0 0 1   1 − 1 − i 1 − i − i 0 0 R 3 → R 3 − (1 − i ) R 1 R 1 →− iR 1 � � 0 1 0 1 0 i   1 − i − i 1 − i 0 0 1 − i − (1 − i )( − 1 − i ) = − i − ( − 2) 2 3 1 − 1 − i 1 − i − i 0 0 = 2 − i 5 since 0 1 i 0 1 0 1 − i − (1 − i )(1 − i ) = 1 − i − ( − 2 i ) 4 0 2 − i 1 + i 1 + i 0 1 = 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 0 0  R 1 → R 1 − (1 − i ) R 2 � 0 1 i 0 1 0 R 3 → R 3 − (2 − i ) R 2   0 2 − i 1 + i 1 + i 0 1   1 0 0 − i 1 + i 0  since (1 + i ) − ( − 1 − i )( i ) = 0 0 1 i 0 1 0  (1 + i ) − (2 − i )( i ) = − i 0 0 − i 1 + i − 2 + i 1   1 0 0 − i 1 + i 0 R 2 → R 2 + R 3 � 0 1 0 2 + i − 1 + i 1   0 0 − i 1 + i − 2 + i 1   1 0 0 − i 1 + i 0 R 3 → iR 3  . � 0 1 0 2 + i − 1 + i 1  0 0 1 − 1 + i − 1 − 2 i 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 0 0  R 1 → R 1 − (1 − i ) R 2 � 0 1 i 0 1 0 R 3 → R 3 − (2 − i ) R 2   0 2 − i 1 + i 1 + i 0 1   1 0 0 − i 1 + i 0  since (1 + i ) − ( − 1 − i )( i ) = 0 0 1 i 0 1 0  (1 + i ) − (2 − i )( i ) = − i 0 0 − i 1 + i − 2 + i 1   1 0 0 − i 1 + i 0 R 2 → R 2 + R 3 � 0 1 0 2 + i − 1 + i 1   0 0 − i 1 + i − 2 + i 1   1 0 0 − i 1 + i 0 R 3 → iR 3  . � 0 1 0 2 + i − 1 + i 1  0 0 1 − 1 + i − 1 − 2 i i () Linear Algebra April 14, 2018 4 / 21

Page 472 Number 8 Page 472 Number 8 (continued 2)   i 1 − i 1 + i Page 472 Number 8. Find A − 1 if A =  . 0 1 i  1 − i − i 1 − i Solution (continued).   − i 1 + i 0 So A − 1 = 2 + i − 1 + i 1  . �  − 1 + i − 1 − 2 i i () Linear Algebra April 14, 2018 5 / 21

Page 473 Number 10 Page 473 Number 10   − 1 + i  where A is as Page 473 Number 10. Solve the system A � z = 2 + i  1 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  − 1 + i   − i 1 + i 0   − 1 + i  z = A − 1  = 2 + i 2 + i − 1 + i 1 2 + i      1 − 1 + i − 1 − 2 i 1 i () Linear Algebra April 14, 2018 6 / 21

Page 473 Number 10 Page 473 Number 10   − 1 + i  where A is as Page 473 Number 10. Solve the system A � z = 2 + i  1 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  − 1 + i   − i 1 + i 0   − 1 + i  z = A − 1  = 2 + i 2 + i − 1 + i 1 2 + i      1 − 1 + i − 1 − 2 i 1 i   ( − i )( − 1 + i ) + (1 + i )(2 + i ) + (0)(1) = (1 + i )( − 1 + i ) + ( − 1 + i )(2 + i ) + (1)(1)   ( − 1 + i )( − 1 + i ) + ( − 1 − 2 i )(2 + i ) + ( i )(1) () Linear Algebra April 14, 2018 6 / 21

Page 473 Number 10 Page 473 Number 10   − 1 + i  where A is as Page 473 Number 10. Solve the system A � z = 2 + i  1 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  − 1 + i   − i 1 + i 0   − 1 + i  z = A − 1  = 2 + i 2 + i − 1 + i 1 2 + i      1 − 1 + i − 1 − 2 i 1 i   ( − i )( − 1 + i ) + (1 + i )(2 + i ) + (0)(1) = (1 + i )( − 1 + i ) + ( − 1 + i )(2 + i ) + (1)(1)   ( − 1 + i )( − 1 + i ) + ( − 1 − 2 i )(2 + i ) + ( i )(1) () Linear Algebra April 14, 2018 6 / 21

Page 473 Number 10 Page 473 Number 10 (continued)   − 1 + i  where A is as Page 473 Number 10. Solve the system A � z = 2 + i  1 given in Exercise 8. Solution (continued). . . .  ( − i )( − 1 + i ) + (1 + i )(2 + i ) + (0)(1)  A − 1 = (1 + i )( − 1 + i ) + ( − 1 + i )(2 + i ) + (1)(1)   ( − 1 + i )( − 1 + i ) + ( − 1 − 2 i )(2 + i ) + ( i )(1)  (1 + i ) + (1 + 3 i ) + (0)   2 + 4 i   =  . = ( − 2) + ( − 3 + i ) + (1) − 4 + i   ( − 2 i ) + ( − 5 i ) + ( i ) − 6 i � () Linear Algebra April 14, 2018 7 / 21

Page 473 Number 10 Page 473 Number 10 (continued)   − 1 + i  where A is as Page 473 Number 10. Solve the system A � z = 2 + i  1 given in Exercise 8. Solution (continued). . . .  ( − i )( − 1 + i ) + (1 + i )(2 + i ) + (0)(1)  A − 1 = (1 + i )( − 1 + i ) + ( − 1 + i )(2 + i ) + (1)(1)   ( − 1 + i )( − 1 + i ) + ( − 1 − 2 i )(2 + i ) + ( i )(1)  (1 + i ) + (1 + 3 i ) + (0)   2 + 4 i   =  . = ( − 2) + ( − 3 + i ) + (1) − 4 + i   ( − 2 i ) + ( − 5 i ) + ( i ) − 6 i � () 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. w ∈ C n and let z be a complex scalar. Then: Let � u ,� v , � (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 � . u = [ u 1 , u 2 , . . . , u n ] ∈ C n we have Proof. (1) (Page 473 Number 16) For � u � = u 1 u 1 + u 2 u 2 + · · · + u n u n = | a 1 | 2 + | u 2 | 2 + · · · + | u n | 2 ≥ 0 since � � u ,� each | u k | 2 is a nonnegative real number. () Linear Algebra April 14, 2018 8 / 21