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examples152-ch10

# examples152-ch10 - MATH 152 Fall 2006-07 Applied Linear...

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MATH 152 Fall 2006-07 Applied Linear Algebra & Differential Equations Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology September 1, 2006

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Contents Table of Contents iii 1 Introduction 1 2 First-Order Differential Equations 5 3 Second-Order Linear Equations 39 4 Laplace Transform 83 5 Matrix 127 6 Systems of Linear Equations 135 7 Euclidean Vector 155 8 Eigenvalue and Eigenvector 177 9 Systems of Differential Equations 209 10 Orthogonality 231 iii

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Chapter 10 Orthogonality ¥ Example 10.1 (Dot product) Find the dot product of x = (2 , 2 , 1) and y = (2 , 5 , - 3) . Solution The dot product (or inner product or scalar product) of x and y is given by x · y = (2 , 2 , 1) · (2 , 5 , - 3) = 2 · 2 + 2 · 5 + 1 · ( - 3) = 11. That is, x · y is obtained by multiplying corresponding components and adding the resulting products. The vectors x and y are said to be orthogonal (or perpendicular) if their dot product is zero, that is, if x · y = 0. Therefore, for this example, the two given vectors x and y are not orthogonal. 2 ¥ Example 10.2 (Norm) For x = (1 , - 2 , 3 , - 4 , 5) , find the norm k x k . Solution The norm of x is given by k x k = x · x = p 1 2 + ( - 2) 2 + 3 2 + ( - 4) 2 + 5 2 = 55. The norm (or length) of a vector x in R n , denoted by k x k , is defined to be the nonnegative square root of x · x . In particular, if x = ( x 1 , x 2 , · · · , x n ), then k x k = p x 2 1 + x 2 2 + · · · + x 2 n . That is, k x k is the square root of the sum of the squares of the components of x . Thus, k x k > 0, and k x k = 0 if and only if x = 0 . 2 ¥ Example 10.3 (Normalize a vector) Find the rescaled vector x k x k , where x = (2 , 2 , 1) . Solution By finding the norm k x k = 2 2 + 2 2 + 1 2 = 3, we could normalize x to the following unit vector x k x k = 1 3 (2 , 2 , 1) = ( 2 3 , 2 3 , 1 3 ). Verify that the norm of the above rescaled vector is p (2 / 3) 2 + (2 / 3) 2 + (1 / 3) 2 = 1. In general, a vector x is called a unit vector if k x k = 1 or, equivalently, if x · x = 1. For any nonzero vector x in R n , the vector ˆ x = (1 / k x k ) x = x / k x k is the unique unit vector in the same direction of x . The process of finding ˆ x from x is called normalizing x . 2 231

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10. Orthogonality ¥ Example 10.4 (Schwarz inequality) Prove that | x · y | 6 k x k k y k . Solution For any real number t , we have 0 6 ( t x + y ) · ( t x + y ) = t 2 ( x · x ) + 2 t ( x · y ) + ( y · y ) = k x k 2 t 2 + 2( x · y ) t + k y k 2 . Let a = k x k 2 , b = 2( x · y ), c = k y k 2 . Then, for every value of t , at 2 + bt + c > 0. This means that the quadratic polynomial cannot have two distinct real roots. This implies that the discriminant D = b 2 - 4 ac 6 0. Equivalently, b 2 6 4 ac . Thus, 4( x · y ) 2 6 4 k x k 2 k y k 2 . Dividing by 4 and taking the square root of both sides gives the inequality. 2 Remark . The angle θ between nonzero vectors x and y in R n is defined by cos θ = x · y k x k k y k . This definition is well-defined, since, by the Schwarz inequality, - 1 6 x · y k x k k y k 6 1. Thus, - 1 6 cos θ 6 1, and so the angle exists and is unique. Note that if x · y = 0, then θ = π/ 2. This then agrees with our previous definition of orthogonality.
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