Math 221: LINEAR ALGEBRA 5-3. Vector Space R n - Orthogonality Le - PowerPoint PPT Presentation

Math 221: LINEAR ALGEBRA §5-3. Vector Space R n - Orthogonality Le Chen 1 Emory University, 2020 Fall (last updated on 08/18/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.

Definitions  x 1   y 1  x 2 y 2      be vectors in R n . Let � x =  and � y =  .   .  . .     . .   x n y n

Definitions  x 1   y 1  x 2 y 2      be vectors in R n . Let � x =  and � y =  .   .  . .     . .   x n y n 1. The dot product of � x and � y is x T � � x · � y = x 1 y 1 + x 2 y 2 + · · · x n y n = � y . Note: � x · � y is a scalar, but we also treat it as a 1 × 1 matrix.

Definitions  x 1   y 1  x 2 y 2      be vectors in R n . Let � x =  and � y =  .   .  . .     . .   x n y n 1. The dot product of � x and � y is x T � � x · � y = x 1 y 1 + x 2 y 2 + · · · x n y n = � y . Note: � x · � y is a scalar, but we also treat it as a 1 × 1 matrix. 2. The length of � x, denoted || � x || is √ � x 2 1 + x 2 || � x || = 2 · · · + x 2 n = � x · � x .

Definitions  x 1   y 1  x 2 y 2      be vectors in R n . Let � x =  and � y =  .   .  . .     . .   x n y n 1. The dot product of � x and � y is x T � � x · � y = x 1 y 1 + x 2 y 2 + · · · x n y n = � y . Note: � x · � y is a scalar, but we also treat it as a 1 × 1 matrix. 2. The length of � x, denoted || � x || is √ � x 2 1 + x 2 || � x || = 2 · · · + x 2 n = � x · � x . 3. � x is called a unit vector if || � x || = 1 .

Theorem (Properties of length and the dot product) z ∈ R n , and let a ∈ R . Then Let � x ,� y ,�

Theorem (Properties of length and the dot product) z ∈ R n , and let a ∈ R . Then Let � x ,� y ,� 1. � x · � y = � y · � x (the dot product is commutative)

Theorem (Properties of length and the dot product) z ∈ R n , and let a ∈ R . Then Let � x ,� y ,� 1. � x · � y = � y · � x (the dot product is commutative) 2. � x · ( � y + � z ) = � x · � y + � x · � z (the dot product distributes over addition)

Theorem (Properties of length and the dot product) z ∈ R n , and let a ∈ R . Then Let � x ,� y ,� 1. � x · � y = � y · � x (the dot product is commutative) 2. � x · ( � y + � z ) = � x · � y + � x · � z (the dot product distributes over addition) 3. ( a � x ) · � y = a ( � x · � y ) = � x · ( a � y )

Theorem (Properties of length and the dot product) z ∈ R n , and let a ∈ R . Then Let � x ,� y ,� 1. � x · � y = � y · � x (the dot product is commutative) 2. � x · ( � y + � z ) = � x · � y + � x · � z (the dot product distributes over addition) 3. ( a � x ) · � y = a ( � x · � y ) = � x · ( a � y ) x || 2 = � 4. || � x · � x.

Theorem (Properties of length and the dot product) z ∈ R n , and let a ∈ R . Then Let � x ,� y ,� 1. � x · � y = � y · � x (the dot product is commutative) 2. � x · ( � y + � z ) = � x · � y + � x · � z (the dot product distributes over addition) 3. ( a � x ) · � y = a ( � x · � y ) = � x · ( a � y ) x || 2 = � 4. || � x · � x. x = � 5. || � x || ≥ 0 with equality if and only if � 0 n .

Theorem (Properties of length and the dot product) z ∈ R n , and let a ∈ R . Then Let � x ,� y ,� 1. � x · � y = � y · � x (the dot product is commutative) 2. � x · ( � y + � z ) = � x · � y + � x · � z (the dot product distributes over addition) 3. ( a � x ) · � y = a ( � x · � y ) = � x · ( a � y ) x || 2 = � 4. || � x · � x. x = � 5. || � x || ≥ 0 with equality if and only if � 0 n . 6. || a � x || = | a | || � x || .

Example y ∈ R n . Then Let � x ,� y || 2 || � x + � = ( � x + � y ) · ( � x + � y ) = � x · � x + � x · � y + � y · � x + � y · � y = � x · � x + 2( � x · � y ) + � y · � y x || 2 + 2( � y || 2 . = || � x · � y ) + || �

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k.

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k. x = t 1 � f 1 + t 2 � f 2 + · · · + t k � Write � f k for some t 1 , t 2 , . . . , t k ∈ R (this is possible because � f 1 ,� f 2 , . . . ,� f k span R n ).

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k. x = t 1 � f 1 + t 2 � f 2 + · · · + t k � Write � f k for some t 1 , t 2 , . . . , t k ∈ R (this is possible because � f 1 ,� f 2 , . . . ,� f k span R n ). Then x || 2 || � = � x · � x

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k. x = t 1 � f 1 + t 2 � f 2 + · · · + t k � Write � f k for some t 1 , t 2 , . . . , t k ∈ R (this is possible because � f 1 ,� f 2 , . . . ,� f k span R n ). Then x || 2 || � = � x · � x x · ( t 1 � f 1 + t 2 � f 2 + · · · + t k � = � f k )

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k. x = t 1 � f 1 + t 2 � f 2 + · · · + t k � Write � f k for some t 1 , t 2 , . . . , t k ∈ R (this is possible because � f 1 ,� f 2 , . . . ,� f k span R n ). Then x || 2 || � = � x · � x x · ( t 1 � f 1 + t 2 � f 2 + · · · + t k � = � f k ) x · ( t 1 � x · ( t 2 � x · ( t k � = � f 1 ) + � f 2 ) + · · · + � f k )

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k. x = t 1 � f 1 + t 2 � f 2 + · · · + t k � Write � f k for some t 1 , t 2 , . . . , t k ∈ R (this is possible because � f 1 ,� f 2 , . . . ,� f k span R n ). Then x || 2 || � = � x · � x x · ( t 1 � f 1 + t 2 � f 2 + · · · + t k � = � f k ) x · ( t 1 � x · ( t 2 � x · ( t k � = � f 1 ) + � f 2 ) + · · · + � f k ) x · � x · � x · � = t 1 ( � f 1 ) + t 2 ( � f 2 ) + · · · + t k ( � f k )

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k. x = t 1 � f 1 + t 2 � f 2 + · · · + t k � Write � f k for some t 1 , t 2 , . . . , t k ∈ R (this is possible because � f 1 ,� f 2 , . . . ,� f k span R n ). Then x || 2 || � = � x · � x x · ( t 1 � f 1 + t 2 � f 2 + · · · + t k � = � f k ) x · ( t 1 � x · ( t 2 � x · ( t k � = � f 1 ) + � f 2 ) + · · · + � f k ) x · � x · � x · � = t 1 ( � f 1 ) + t 2 ( � f 2 ) + · · · + t k ( � f k ) = t 1 (0) + t 2 (0) + · · · + t k (0) = 0 . x || 2 = 0 , || � Since || � x || = 0 . By the previous theorem, || � x || = 0 if and only if x = � x = � � 0 n . Therefore, � 0 n .

Example f k } ∈ R n and suppose R n = span { � Let { � f 1 ,� f 2 , . . . ,� f 1 ,� f 2 , . . . ,� f k } . Furthermore, x ∈ R n for which � x · � suppose that there exists a vector � f j = 0 for all j, 1 ≤ j ≤ k. x = t 1 � f 1 + t 2 � f 2 + · · · + t k � Write � f k for some t 1 , t 2 , . . . , t k ∈ R (this is possible because � f 1 ,� f 2 , . . . ,� f k span R n ). Then x || 2 || � = � x · � x x · ( t 1 � f 1 + t 2 � f 2 + · · · + t k � = � f k ) x · ( t 1 � x · ( t 2 � x · ( t k � = � f 1 ) + � f 2 ) + · · · + � f k ) x · � x · � x · � = t 1 ( � f 1 ) + t 2 ( � f 2 ) + · · · + t k ( � f k ) = t 1 (0) + t 2 (0) + · · · + t k (0) = 0 . x || 2 = 0 , || � Since || � x || = 0 . By the previous theorem, || � x || = 0 if and only if x = � x = � � 0 n . Therefore, � 0 n .

Theorem (Cauchy Inequality) y ∈ R n , then | � If � x ,� x · � y | ≤ || � x || || � y || with equality if and only if { � x ,� y } is linearly dependent.