Assignment 9 solution

# Assignment 9 solution - MATH 223 Linear Algebra Winter 2008...

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MATH 223, Linear Algebra Winter, 2008 Solutions to Assignment 9 1. Let V be the vector space of real-valued functions deﬁned and continuous on [ - π 2 , π 2 ] . (a) Show that, for any f V , ( Z π 2 - π 2 f ( x )cos xdx ) 2 2 Z π 2 - π 2 f ( x ) 2 cos xdx. Identify those functions f for which equality holds. Solution: We have seen that < f,g > = R π 2 - π 2 f ( x ) g ( x )cos xdx de- ﬁnes an inner product on the space. So Cauchy-Schwarz applies. In particular, < f, 1 > 2 < f,f >< 1 , 1 > (with equality if and only if f is a constant multiple of 1; that is, f is a constant function). Now < f, 1 > = R π 2 - π 2 f ( x )cos xdx , < f,f > = R π 2 - π 2 f ( x ) 2 cos xdx and < 1 , 1 > = R π 2 - π 2 cos xdx = sin x ] π 2 - π 2 = 2. The given inequality follows, and is an equality if and only if f is a constant function. (b) Show that, for any f V , ( Z π 2 - π 2 f ( x )cos 2 xdx ) 2 4 3 Z π 2 - π 2 f ( x ) 2 cos xdx. Identify those function f for which equality holds. Solution: The same basic idea as above, except that now, instead of taking g ( x ) = 1, we take g ( x ) = cos x . So < f, cos x > 2 < f,f >< cos x, cos x > . < f, cos x > = R π 2 - π 2 f ( x )cos 2 xdx , < f,f > is still R π 2 - π 2 f ( x ) 2 cos xdx and < cos x, cos x > = R π 2 - π 2 cos 3 xdx = 4 3 . We have equality if and only if f ( x ) = r cos x for some constant r . [Incidentally, to do the integral R π 2 - π 2 cos 3 xdx , ﬁrst write cos 3 x = cos x (1 - sin 2 x ) = cos x - cos x sin 2 x . I presume you can integrate cos x ; for the other part, let u = sin x , so du = cos xdx and you’re oﬀ to the races.] 2. (a) Suppose that V is any inner product space, and that ~v and ~w are orthogonal vectors in V . Show that || ~v + ~w || 2 = || ~v || 2 + || ~w || 2 . Solution: || ~v + ~w || 2 = < ~v + ~w,~v + ~w > = < ~v,~v + ~w > + < ~w,~v + ~w > = < ~v,~v > + < ~v, ~w > + < ~w,~v > + < ~w, ~w > . As we are assuming that ~v and ~w are orthogonal, < ~v, ~w > = < ~w,~v > = 0. So || ~v + ~w || 2 = < ~v,~v > + < ~w, ~w > = || ~v || 2 + || ~w || 2 , as advertized. 1

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(b) Now suppose that V is ﬁnite-dimensional and W is a subspace of V . For any ~u V , show that Proj W ~u is the “closest” vector to ~u that’s in W ; that is, if ~x = Proj W ~u and ~ y 6 = ~x but ~ y W , then || ~u - ~x || < || ~u - ~ y || . Solution: We know that ~u - ~x is orthogonal to every vector in W . In particular, it is orthogonal to ~x - ~ y , which is in W since both ~x and ~ y are. We let ~v = ~u - ~x and ~w = ~x - ~ y , so that ~v and ~w are orthogonal, and ~v + ~w = ~u - ~ y ; note that ~w 6 = ~ 0, so that || ~w || > 0. Part (a) applies, so || ~v + ~w || 2 = || ~v || 2 + || ~w || 2 > || ~v || 2 ; taking square roots, we see that || ~v + ~w || > || ~v || . That is, || ~u - ~x || < || ~u - ~ y || . 3. Let
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Assignment 9 solution - MATH 223 Linear Algebra Winter 2008...

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