Fall TMA4145 Linear Methods. Exercise set 10

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1 Norwegian University of Science and Technology Department of Mathematical Sciences TMA445 Linear Methods Fall 207 Exercise set 0 Please justify your answers! The most important part is how you arrive at an answer, not the answer itself. Let M be the plane of R 3 given by x + x 2 + x 3 = 0. Find the linear mapping that is the orthogonal projection of R 3 onto this plane. Solution. We defined the projection P : R 3 R 3 by decomposing each x R 3 into x = y +z using the projection theorem, where y M and z M, and defining P by P x = y. A natural starting point is therefore to find M. Since R 3 is 3-dimensional, M is 2-dimensional and R 3 = M M by the projection theorem, M must be one-dimensional. It is not difficult to see that the vector a = (,, ) belongs to M, since x, a = x + x 2 + x 3 = 0 if x M. Since M is one-dimensional and contains a, it follows that M = {λa : λ R}. Now let x = (x, x 2, x 3 ) R 3. We want to write decompose x = y + z with y M and z M. Since M = {λa : λ R}, we must have that z = λa for some λ C: so in terms of coordinates we have x = y + λa, (x, x 2, x 3 ) = (y, y 2, y 3 ) + (λ, λ, λ). We are looking for (y, y 2, y 3 ), and need to eliminate λ. To achieve this we have one unused assumption: y M, so y + y 2 + y 3 = 0. To use this, we solve for (y, y 2, y 3 ): (y, y 2, y 3 ) = (x λ, x 2 λ, x 3 λ). Since (y, y 2, y 3 ) M we must have that 0 = y + y 2 + y 3 = x λ + x 2 λ + x 3 λ = x + x 2 + x 3 3λ. We may solve this for λ to get λ = x + x 2 + x 3, 3 Remember that the inner product on R 3 is just the usual dot product. October 27, 207 Page of 5

2 Exercise set 0 and inserting this back into our expression for (y, y 2, y 3 ) we find that (y, y 2, y 3 ) = (x λ, x 2 λ, x 3 λ) = 3 (2x x 2 x 3, 2x 2 x x 3, 2x 3 x x 2 ). Since P x = y, this means that we have shown that P (x, x 2, x 3 ) = 3 (2x x 2 x 3, 2x 2 x x 3, 2x 3 x x 2 ). 2 Let A R be a set. Prove that if A is bounded from below, then there is a sequence (a n ) A such that a n inf A as n. (In other words, prove that inf A Ā.) Similarly, if A is bounded from above, prove that there is a sequence (a n ) A such that a n sup A as n. (In other words, prove that sup A Ā.) Solution. Let A R be bounded from below. First off, because of the least upper bound property, the infimum of A exists (as a real number), that is, inf A R. 2 Let ɛ > 0. Since inf A is the greatest lower bound of A, we have that inf A + ɛ cannot be a lower bound for A, so there is some element a ɛ A such that a ɛ < inf A + ɛ. Furthermore, since inf A is a lower bound of A, and since a ɛ A, we must have that inf A a ɛ. Thus inf A a ɛ < inf A + ɛ. For every n we may choose ɛ =. Using the above, there is some element n a n A such that inf A a n < inf A + n. We have obtained a sequence (a n ) A such that inf A a n < inf A + n Subtracting inf A from all sides of this inequality, we get that for all n. 0 a n inf A < n, which implies a n inf A n 0 as n, so a n inf A. 2 I m afraid that we did not show that the least upper bound property implies the existence of the infinum, and I would not require this to be part of the student s solution. The student should, however, be made aware that the existence of the infinum of a set bounded from below follows from the least upper bound property. October 27, 207 Page 2 of 5

3 Exercise set 0 The second part, regarding sup A is very similar. Let A R be bounded from above. By the least upper bound property, the supremum of A exists (as a real number), that is, sup A R. Let ɛ > 0. Since sup A is the least upper bound of A, we have that sup A ɛ cannot be an upper bound for A, so there is some element a ɛ A such that a ɛ > sup A ɛ. Furthermore, since sup A is an upper bound of A, and since a ɛ A, we must have that a ɛ sup A. Thus sup A ɛ < a ɛ sup A. For every n we may choose ɛ =. Using the above, there is some element n a n A such that sup A n < a n sup A. We have obtained a sequence (a n ) A such that sup A n < a n sup A for all n. Subtracting sup A from all sides of this inequality, we get that which implies n < a n sup A 0, a n sup A n 0 as n, so a n sup A. 3 Let T be a bounded linear operator on a Hilbert spae X. Show that the operator norm of T can be expressed in terms of the innerproduct of X: T = sup{ T x, y : x, y X with x = y = }. Solution. We will first show that sup{ x, y : y X with y = } = x for all x X. () By the Cauchy-Schwarz inequality we have x, y x, y x y = x, for all x, y X with y =. It follows that sup{ x, y : y X with y = } x for all x X. It remains to show the inequality sup{ x, y : y X with y = } x for all x X. October 27, 207 Page 3 of 5

4 Exercise set 0 This clearly holds when x = 0, since 0, y = 0 for all y X. Now suppose x 0. Let y = x and notice that y =. We have x The inequality follows. x, y = x, x x = x, x = x x x 2 = x. We will now show that the norm of T can be expressed in terms of the innerproduct. We have T = sup{ T x, x X with x = } = sup{ sup{ T x, y : y X, y = } : x X, x = } (equation ()) = sup{ T x, y : x, y X with x = y = } 4 Let M = {x l 2 : x = (x, 0, x 3, 0, x 5,...)} be the subspace of odd sequences in l 2. Determine the orthogonal complement M. You must prove that the space you find really is the orthogonal complement of M. Solution. We claim that M = M e, where M e = {x l 2 : x = (0, x 2, 0, x 4, 0, x 6...)} is the set of even sequences in l 2. First assume that x M e. If y = (y, 0, y 3, 0, y 5,...) M, then x, y = x i y i = x i y i = 0. i= i even i odd Hence x M, and we have shown that M e M. Now assume that x M. We can decompose x into the sum x = (x, 0, x 3, 0, x 5,...) + (0, x 2, 0, x 4, 0, x 6,...) := x o + x e, where x o M and x e M e. Since we assume that x M, we know that 0 = x, x o = x o + x e, x o = x o, x o + x e, x o = x o 2, where we have used that x e, x o = 0, as is clear by inspection. But this means that x o = 0, so x = x e M e. Thus M M e. 5 Let c f be the subspace of l 2 that consists of all sequences with finitely many non-zero terms. a) Show that best approximation fails for c f. b) Why does this not contradict the best approximation theorem from class? October 27, 207 Page 4 of 5

5 Exercise set 0 Solution. a) Let x = (, /2, /3,...). We start by showing that inf{ x m : m M} = 0. Since the norm is always non-negative, we have that inf{ x m : m M} 0. We will show equality by constructing a sequence {m n } n N in M such that x m n 0. Let m n = (, /2, /3,..., /n, 0, 0,...). We have lim x m n = lim n n (0, 0,..., /(n + ), /(n + 2), /(n + 3),...) = lim n k=n+ ( = lim n = k= k= k lim 2 n = x x = 0 n k= n k= ) which was what we needed to prove. Now, suppose there exist a sequence m M such that x m = 0. Then we have m = x, but x is not in M since it has no non-zero terms. b) This does not contradict the best approximation theorem because the conditions for the theorem are not satisfied. Indeed, the subspace M is not closed. To see this, observe that that the sequence {m n } n N converges to x, but x is not in M. October 27, 207 Page 5 of 5

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