Linear Algebra Chapter 1. Vectors, Matrices, and Linear Systems - PowerPoint PPT Presentation

Linear Algebra Chapter 1. Vectors, Matrices, and Linear Systems Section 1.2. The Norm and Dot Product—Proofs of Theorems July 19, 2018 () Linear Algebra July 19, 2018 1 / 15

Table of contents Page 31 Number 8 1 Page 31 Number 12 2 Page 33 Number 42(b) 3 Page 31 Number 14 4 Page 31 Number 16 5 Page 26 Example 7, Parallelogram Law 6 Theorem 1.4, Schwarz’s Inequality 7 Page 31 Number 36 8 () Linear Algebra July 19, 2018 2 / 15

Page 31 Number 8 Page 31 Number 8 Page 31 Number 8. Find the unit vector parallel to � w = [ − 2 , − 1 , 3] which has the opposite direction. Solution. If we divide � w by the scalar � � w � > 0, we get a vector of length 1 (i.e., a unit vector; this process is called normalizing a vector). Such a vector is in the same direction as � w (by Definition 1.2 of “parallel and same direction”). () Linear Algebra July 19, 2018 3 / 15

Page 31 Number 8 Page 31 Number 8 Page 31 Number 8. Find the unit vector parallel to � w = [ − 2 , − 1 , 3] which has the opposite direction. Solution. If we divide � w by the scalar � � w � > 0, we get a vector of length 1 (i.e., a unit vector; this process is called normalizing a vector). Such a vector is in the same direction as � w (by Definition 1.2 of “parallel and same direction”). By Definition 1.5, “Vector Norm,” we have ( − 2) 2 + ( − 1) 2 + (3) 2 = √ 4 + 1 + 9 = √ � � � w � = 14, so � − 2 � � w 1 , − 1 3 w � = √ [ − 2 , − 1 , 3] = √ √ , √ is a unit vector in the same � � 14 14 14 14 direction as � w . () Linear Algebra July 19, 2018 3 / 15

Page 31 Number 8 Page 31 Number 8 Page 31 Number 8. Find the unit vector parallel to � w = [ − 2 , − 1 , 3] which has the opposite direction. Solution. If we divide � w by the scalar � � w � > 0, we get a vector of length 1 (i.e., a unit vector; this process is called normalizing a vector). Such a vector is in the same direction as � w (by Definition 1.2 of “parallel and same direction”). By Definition 1.5, “Vector Norm,” we have ( − 2) 2 + ( − 1) 2 + (3) 2 = √ 4 + 1 + 9 = √ � � � w � = 14, so � − 2 � � w 1 , − 1 3 w � = √ [ − 2 , − 1 , 3] = √ √ , √ is a unit vector in the same � � 14 14 14 14 direction as � w . To get a unit vector in the opposite direction, by Definition 1.2, we simply multiply by − 1 and take − � w / � � w � as the desired � − 2 � 2 vector: − � , − 1 3 � 1 , − 3 � w √ √ √ √ √ √ w � = − , = , . � � � 14 14 14 14 14 14 () Linear Algebra July 19, 2018 3 / 15

Page 31 Number 8 Page 31 Number 8 Page 31 Number 8. Find the unit vector parallel to � w = [ − 2 , − 1 , 3] which has the opposite direction. Solution. If we divide � w by the scalar � � w � > 0, we get a vector of length 1 (i.e., a unit vector; this process is called normalizing a vector). Such a vector is in the same direction as � w (by Definition 1.2 of “parallel and same direction”). By Definition 1.5, “Vector Norm,” we have ( − 2) 2 + ( − 1) 2 + (3) 2 = √ 4 + 1 + 9 = √ � � � w � = 14, so � − 2 � � w 1 , − 1 3 w � = √ [ − 2 , − 1 , 3] = √ √ , √ is a unit vector in the same � � 14 14 14 14 direction as � w . To get a unit vector in the opposite direction, by Definition 1.2, we simply multiply by − 1 and take − � w / � � w � as the desired � − 2 � 2 vector: − � , − 1 3 � 1 , − 3 � w √ √ √ √ √ √ w � = − , = , . � � � 14 14 14 14 14 14 () Linear Algebra July 19, 2018 3 / 15

Page 31 Number 12 Page 31 Number 12 Page 31 Number 12. Find the angle between � u = [ − 1 , 3 , 4] and � v = [2 , 1 , − 1]. u · � � v Solution. We have by definition that the desired angle is cos − 1 v � . � � u �� � () Linear Algebra July 19, 2018 4 / 15

Page 31 Number 12 Page 31 Number 12 Page 31 Number 12. Find the angle between � u = [ − 1 , 3 , 4] and � v = [2 , 1 , − 1]. � u · � v Solution. We have by definition that the desired angle is cos − 1 v � . � � u �� � Now by Definition 1.5, “Vector Norm,” ( − 1) 2 + (3) 2 + (4) 2 = √ 1 + 9 + 16 = √ � � � u � = 26 and √ (2) 2 + (1) 2 + ( − 1) 2 = √ 4 + 1 + 1 = � � � v � = 6. Also, by Definition 1.6, “Dot Product,” � u · � v = [ − 1 , 3 , 4] · [2 , 1 , − 1] = ( − 1)(2)+(3)(1)+(4)( − 1) = − 2+3 − 4 = − 3 . () Linear Algebra July 19, 2018 4 / 15

Page 31 Number 12 Page 31 Number 12 Page 31 Number 12. Find the angle between � u = [ − 1 , 3 , 4] and � v = [2 , 1 , − 1]. � u · � v Solution. We have by definition that the desired angle is cos − 1 v � . � � u �� � Now by Definition 1.5, “Vector Norm,” ( − 1) 2 + (3) 2 + (4) 2 = √ 1 + 9 + 16 = √ � � � u � = 26 and √ (2) 2 + (1) 2 + ( − 1) 2 = √ 4 + 1 + 1 = � � � v � = 6. Also, by Definition 1.6, “Dot Product,” � u · � v = [ − 1 , 3 , 4] · [2 , 1 , − 1] = ( − 1)(2)+(3)(1)+(4)( − 1) = − 2+3 − 4 = − 3 . � u · � − 3 v v is cos − 1 v � = cos − 1 So the angle between � u and � √ √ = � � u �� � 26 6 − 3 cos − 1 √ . 156 () Linear Algebra July 19, 2018 4 / 15

Page 31 Number 12 Page 31 Number 12 Page 31 Number 12. Find the angle between � u = [ − 1 , 3 , 4] and � v = [2 , 1 , − 1]. � u · � v Solution. We have by definition that the desired angle is cos − 1 v � . � � u �� � Now by Definition 1.5, “Vector Norm,” ( − 1) 2 + (3) 2 + (4) 2 = √ 1 + 9 + 16 = √ � � � u � = 26 and √ (2) 2 + (1) 2 + ( − 1) 2 = √ 4 + 1 + 1 = � � � v � = 6. Also, by Definition 1.6, “Dot Product,” � u · � v = [ − 1 , 3 , 4] · [2 , 1 , − 1] = ( − 1)(2)+(3)(1)+(4)( − 1) = − 2+3 − 4 = − 3 . � u · � − 3 v v is cos − 1 v � = cos − 1 So the angle between � u and � √ √ = � � u �� � 26 6 − 3 cos − 1 √ . We can use a calculator to approximate the true answer to 156 find that the angle is roughly 103 . 90 ◦ . () Linear Algebra July 19, 2018 4 / 15

Page 31 Number 12 Page 31 Number 12 Page 31 Number 12. Find the angle between � u = [ − 1 , 3 , 4] and � v = [2 , 1 , − 1]. � u · � v Solution. We have by definition that the desired angle is cos − 1 v � . � � u �� � Now by Definition 1.5, “Vector Norm,” ( − 1) 2 + (3) 2 + (4) 2 = √ 1 + 9 + 16 = √ � � � u � = 26 and √ (2) 2 + (1) 2 + ( − 1) 2 = √ 4 + 1 + 1 = � � � v � = 6. Also, by Definition 1.6, “Dot Product,” � u · � v = [ − 1 , 3 , 4] · [2 , 1 , − 1] = ( − 1)(2)+(3)(1)+(4)( − 1) = − 2+3 − 4 = − 3 . � u · � − 3 v v is cos − 1 v � = cos − 1 So the angle between � u and � √ √ = � � u �� � 26 6 − 3 cos − 1 √ . We can use a calculator to approximate the true answer to 156 find that the angle is roughly 103 . 90 ◦ . () Linear Algebra July 19, 2018 4 / 15

Page 33 Number 42(b) Page 33 Number 42(b) w ∈ R n . Prove the Distributive Law: Page 33 Number 42(b). Let � u ,� v , � � u · ( � v + � w ) = � u · � v + � u · � w . w ∈ R n , then by our first definition in Section 1.1, we Proof. Since � u ,� v , � have that � u = [ u 1 , u 2 , . . . , u n ], � v = [ v 1 , v 2 , . . . , v n ], and w = [ w 1 , w 2 , . . . , w n ] where all u i , v i , w i are real numbers. � () Linear Algebra July 19, 2018 5 / 15

Page 33 Number 42(b) Page 33 Number 42(b) w ∈ R n . Prove the Distributive Law: Page 33 Number 42(b). Let � u ,� v , � � u · ( � v + � w ) = � u · � v + � u · � w . w ∈ R n , then by our first definition in Section 1.1, we Proof. Since � u ,� v , � have that � u = [ u 1 , u 2 , . . . , u n ], � v = [ v 1 , v 2 , . . . , v n ], and w = [ w 1 , w 2 , . . . , w n ] where all u i , v i , w i are real numbers. Then � � u · ( � v + � w ) = [ u 1 , u 2 , . . . , u n ] · ([ v 1 , v 2 , . . . , v n ] + [ w 1 , w 2 , . . . , w n ]) = [ u 1 , u 2 , . . . , u n ] · [ v 1 + w 1 , v 2 + w 2 , . . . , v n + w n ] by Definition 1.1.(1), “Vector Addition” () Linear Algebra July 19, 2018 5 / 15

Page 33 Number 42(b) Page 33 Number 42(b) w ∈ R n . Prove the Distributive Law: Page 33 Number 42(b). Let � u ,� v , � � u · ( � v + � w ) = � u · � v + � u · � w . w ∈ R n , then by our first definition in Section 1.1, we Proof. Since � u ,� v , � have that � u = [ u 1 , u 2 , . . . , u n ], � v = [ v 1 , v 2 , . . . , v n ], and w = [ w 1 , w 2 , . . . , w n ] where all u i , v i , w i are real numbers. Then � � u · ( � v + � w ) = [ u 1 , u 2 , . . . , u n ] · ([ v 1 , v 2 , . . . , v n ] + [ w 1 , w 2 , . . . , w n ]) = [ u 1 , u 2 , . . . , u n ] · [ v 1 + w 1 , v 2 + w 2 , . . . , v n + w n ] by Definition 1.1.(1), “Vector Addition” = u 1 ( v 1 + w 1 ) + u 2 ( v 2 + w 2 ) + · · · + u n ( v n + w n ) by Definition 1.6, “Dot Product” () Linear Algebra July 19, 2018 5 / 15