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8 Pages

### HW8

Course: MATH 430, Fall 2011
School: Penn State
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Word Count: 660

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1 Page of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 3.1.3. Show that v, w = v1 w1 + v1 w2 + v2 w1 + v2 w2 does not dene an inner product on R2 . By Inner Product Axioms, the inner product has to be positive denite i.e. v , v &gt; 0; Assume, v = v = 0 1 . Then using the denition of inner product provided, 1 v , v = v1 v1 + v1 v2 + v2 v1 + v2 v2 v , v = (1)(1) + (1)(1) + (1)(1)...

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1 Page of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 3.1.3. Show that v, w = v1 w1 + v1 w2 + v2 w1 + v2 w2 does not dene an inner product on R2 . By Inner Product Axioms, the inner product has to be positive denite i.e. v , v > 0; Assume, v = v = 0 1 . Then using the denition of inner product provided, 1 v , v = v1 v1 + v1 v2 + v2 v1 + v2 v2 v , v = (1)(1) + (1)(1) + (1)(1) + (1)(1) v, v = 1 1 1 + 1 = 0 The inner product is zero when v is not zero. That is, it is not positive denite. Hence, v, w = v1 w1 + v1 w2 + v2 w1 + v2 w2 does not dene an inner product on R2 . Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 Page 2 of 8 3.1.5(b). The unit circle for an inner product on R2 is dened as the set of all vectors of unit length: v = 1. w1 v1 . ,w = Graph the unit of circles for the weighted inner product v, w = 2v1 w1 + 5v2 w2 , v = w2 v2 2 2 v , v = 2v1 + 5v2 = 1 This gives us an eclipse of the dual radii, 1/ 2 and 1/ 5 Page 3 of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 3.1.10. Let V be an inner product space. (a) Prove that x, v = 0 for all v V if and only if x = 0. Let v = x. Then n n x2 i xi xi = x, x = i=1 i=1 By denition of norm, n 2 x x2 i = i=1 x, x = x 2 =0 x =0 x=0 (b) Prove that x, v = y, v for all v V if and only if x = y. Rewriting the equation, we get, x, v y , v = 0 By denition of inner products, we get, x y, v = 0 Assume, v = x y , then n n x y, x y = i=1 (x y )i (x y )i = By denition of norm, i=1 (x y )2 i n (x y ) 2 = i=1 (x y )2 i (x y ), (x y ) = (x y ) (x y ) = 0 (x y ) = 0 x=y 2 =0 Page 4 of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 (c) Let v1 , . . . , vn be a basis for V . Prove that x, vi = y, v i , i = 1, . . . , n, if and only if x = y. Since, v1 , . . . , vn be a basis for V , then, v = v1 c1 + + cn vn . By bilinearity, x, v = c1 x, v1 + + cn x, vn = 0 Now, this holds true for v V . Then, x y, v = 0 And, (x y ), v = c1 (x y ), v1 + + cn (x y ), vn = 0 n n i=1 ci (x y )vi = (x y ) ci vi = 0 i=1 However, ci = 0 and vi = 0, therefore, xy =0x=y Page 5 of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 3.1.20. Let f (x) = x, g(x) = 1 + x2 . Compute f , g , f , and g for (a) the L2 inner product 1 f, g = f (x)g(x)dx, 0 1 x(x2 + 1)dx = f, g = 3 4 0 1 f= f, f = 0 1 xxdx = 3 1 g= 28 15 (1 + x2 )2 dx = g, g = 0 (b) the L2 inner product 1 f, g = f (x)g(x)dx, 1 1 x(x2 + 1)dx = 0 f, g = 1 f= f, f = xxdx = 2 3 1 1 g= (1 + x2 )2 dx = g, g = 1 56 15 Page 6 of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 (c) the weighted inner product 1 f, g = f (x)g(x)xdx, 0 1 x(x2 + 1)xdx = f, g = 8 15 0 1 f= (x)2 xdx = x f, f = 1 2 0 1 g= (1 + x2 )2 xdx = x g, g = 0 7 6 Page 7 of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 3.2.2(a). Find the Euclidean angle between the vectors (1, 1, 1, 1)T and (1, 1, 1, 1)T in R4 . Let v1 = (1, 1, 1, 1)T and v2 = (1, 1, 1, 1) Euclidean Angle is: cos = v1 v2 v1 v2 1 (1)(1) + (1)(1) + (1)(1) + (1)(1) 2 cos = == 4 2 ( 12 + 12 + 12 + 12 )( 12 + 12 + 12 + (1)2 ) = 3 Page 8 of 8 Math 430 Siddharth Dahiya syd5144@psu.edu October 30, 2011 Homework # 8 3.2.7. Prove that (a1 + a2 + + an )2 n(a2 + a2 + + a2 ) 1 2 n for any real numbers a1 , . . . , an . When does equality hold? Proof. By Cauchy-Schwarz Inequality, |v w| v w Assume, v = (a1 , a2 , . . . , an )T and w = (1, 1, . . . , 1)T . Then, |v w| = |a1 + a2 + + an | w v= n a2 + a2 + a2 n 1 2 By Cauchy-Schwarz Inequality, |a1 + a2 + + an | n a2 + a2 + a2 n 1 2 Squaring both sides, |a1 + a2 + + an |2 n(a2 + a2 + a2 ) 1 2 n
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