Solutions for MAS277 Problems

Transcription

1 Solutions for MAS77 Problems Solutions for Chapter problems: Inner product spaces 1. No. The only part that fails, however, is the condition that f, f should mean that f. If f is a non-zero function that vanishes at x (e.g., f(x) x), then f, f but f.. We see that and x + y x + y, x + y x, x + x, y + y, x + y, y, x y x y, x y x, x x, y y, x + y, y, using our rules for inner products. Then as required.. (a) We have x + y + x y x, x + y, y x + y, C 1, C 1 6, C 1, C 1, C 1, C 6, C, C 1 1, C, C 7, C, C, C, C 1 6, C, C, C, C 8. (b) As, is a real inner product, A, B B, A. But B, A trace(ba T ) trace(ba) trace(ab) trace(ab T ) A, B, so A, B A, B, and so A, B. (Alternatively, one could expand out the products, but this is simpler.) 4. (a) We have x+1, x +x (b) We have (x+1)(x +x) dx x i, x j [ x x +x 4 +x dx 4 + x + x [ x x i+j i+j+1 dx i + j + 1 since i + j + 1 is even, meaning that () i+j+1 1 i+j+1. (c) We have 4f() 8f() + 4f(1) 4(a b + c) 8c + 4(a + b + c) 8a. 4. MAS77 1 Spring 1-14

2 Also, f, u (px + q)(ax + bx + c) dx apx 4 + bpx + (pc + qa)x + bqx + cq dx [ apx bpx4 4 + (pc + qa)x ap (pc + qa) + + cq 5 ( p a 5 + q ) ( ) p + c + q. + bqx + cqx 5. We have We require that this equal 8a, so we need p 5 + q 8 p + q, and solving this as usual gives p 45, q 5. x + y x + y, x + y x, x + x, y + y, x + y, y x + x, y + y, so the given formula holds if and only if x, y, as required. 6. (a) Notice that, for f, g C (R), f, g is a function. So, does not necessarily take real number values, and is not therefore an inner product. But restricting to V, things are better: f, g (fg + f g ) f g + fg + f g + f g f (g + g ) + g (f + f ), as both f and g lie in V, so that f + f and g + g. So f, g is a constant function, that is, a real number. (b) If f + f, then (f + f ), so f + f, i.e., f + (f ), and so f V. (c) Clearly, satisfies conditions IP1 IP of the definition of inner product, and is also valued in R by part (a). Also, a typical element in V is f a sin t+b cos t. Then f, f f + f (a sin t + b cos t) + (a cos t b sin t) a + b, which is clearly non-negative, and is only zero when a b, i.e., when f as required. (d) As D(sin) cos and D(cos) ( ) sin, the matrix representation of D relative to the basis {sin, cos} is. 1 MAS77 Spring 1-14

3 7. We use the Cauchy-Schwarz inequality on the space C[, 1] with its usual inner product. Put g(x) 1 + x. Then g, g g(x) dx (1 + x) dx [ x x + x + 1 dx + x + x The Cauchy-Schwarz inequality tells us that f(x)(1 + x) dx 1 + x, 1 + x f, f, 7. which is exactly the statement required. The Cauchy-Schwarz inequality is an equality if f(x) is a scalar multiple of 1 + x, so f(x) 1 + x is one function where we have equality. 8. We use the Cauchy-Schwarz inequality on the space C[, 1] with its usual inner product. Put g(x) 1 x. Then g, g g(x) dx 1 x dx [x x 4. The Cauchy-Schwarz inequality tells us that f(x) 1 x dx 1 x, 1 x f, f, which is exactly the statement required. The Cauchy-Schwarz inequality is an equality if f(x) is a scalar multiple of 1 x, so f(x) 1 x is one function where we have equality. 9. We use the Cauchy-Schwarz inequality on the space C[, 1] with its usual inner product. Given a function f(x), put g(x) f(x). Then g, g g(x) dx f(x) 4 dx. The Cauchy-Schwarz inequality tells us that f(x)g(x) dx g, g f, f, which is exactly the statement required when we substitute g(x) f(x). The Cauchy-Schwarz inequality is an equality if f(x) is a scalar multiple of f(x), so we need f(x) λf(x), i.e., f(x) 1/λ. Thus this is an equality precisely for constant functions. (One ought to be a little more careful about the possibility that f(x) at some points, but as f is continuous, one sees that either f(x) for all x, or f(x) 1/λ for all x.) MAS77 Spring 1-14

4 1. f(x) ax + bx + c is orthogonal to 1 and x if and only if f(x), 1 f(x), x. Using the definition of the inner product we require f(x), 1 (ax + bx + c)dx a + b + c and f(x), 1 (ax + bx + cx)dx a 4 + b + c. Setting a 1 in the above equations we find b and c 1. So the polynomial 6 f(x) x x + 1 is orthogonal to both 1 and x First, φ(a + b cos t + c sin t) b sin t + c cos t, so φ(a + b cos t + c sin t), α + β cos t + γ sin t b sin t + c cos t, α + β cos t + γ sin t If φ is the adjoint of φ, we need π(βc + γb). a + b cos t + c sin t, φ (α + β cos t + γ sin t) π(βc + γb) and so φ (α + β cos t + γ sin t) β sin t + γ cos t, showing that φ φ. 1. If v (x, y, z) T, we have ( ) ( ) x y a b φ(v), A, y z c d 1. Let On the other hand, if w (p, q, r) T, then v, w px + qy + rz. ax + by + cy + dz. These two agree when p a, q b + c and r d. So w (a, b + c, d) T. a 1 a a b 1 b b A a 4 a 5 a 6, B b 4 b 5 b 6. a 7 a 8 a 9 b 7 b 8 b 9 We need to find φ so that φ(a), B A, φ (B). But φ(a), B a 4 b + a 7 b + a 8 b 6. In order that this is equal to A, φ (B), we need b 1 b b φ b 4 b 5 b 6 b. b 7 b 8 b 9 b b 6 MAS77 4 Spring 1-14

5 ( ) ( ) ( ) a b a + b + c + d a + b + c + d α β 14. Explicitly, φ. Let B. Then c d a + b + c + d a + b + c + d γ δ φ(a), B (a + b + c + d)(α + β + γ + δ). We need this to equal A, φ (B), and as this is a(α + β + γ + δ) + b(α + β + γ + δ) + c(α + β + γ + δ) + d(α + β + γ + δ), ( ) α + β + γ + δ α + β + γ + δ we need φ (B), and so φ α + β + γ + δ α + β + γ + δ φ, as required. 15. We see χ(ax + bx + c) a. If u αx + βx + γ, then f, u 1/ / 1/ (ax + bx + c)(αx + βx + γ) dx aαx 4 + (aβ + αb)x + (aγ + bβ + cα)x + (bγ + βc)x + cγ dx / [ aαx (aβ + αb)x4 4 + (aγ + bβ + cα)x aα aγ + bβ + cα + + cγ 8 ( α a 8 1) + γ + b β ( α ) 1 + c 1 + γ. + bγ + βc)x / + cγx / We require that this always equal a. So β and α 1 γ. Then α 8 α 144, and this gives α 6, and so γ. Thus u 6x. We need χ(f), t f, χ (t) for t R. The left-hand side is so χ (t) tu t(6x ). 16. We simply observe that for i j, χ(f), t f (), t f ().t f, u t f, tu, φ(v i ), φ(v j ) v i, φ (φ(v j )) v i, v j, the first equality being the definition of the adjoint, the middle by the given property of φ, and the final one because v i and v j are orthogonal. Similarly, the same argument gives φ(v i ), φ(v i ) v i, v i 1, so {φ(v 1 ),..., φ(v n )} is an orthonormal set. 17. (a) Notice that f is just a constant. So α(α(f)) α((x 1)f ) f α(x 1) f (x 1)(x 1) 6f (x 1) 6α(f). (b) If α(f) λf, then α(α(f)) α(λf) λ.α(f) λ f. α(α(f)) 6α(f) 6λf. As f is nonzero, λ 6λ. But we also have MAS77 5 Spring 1-14

6 (c) The eigenvectors with eigenvalue are those polynomials where f, i.e., f(x) ax + b. The eigenvector with eigenvalue 6 must have α(f) 6f; but α(f) is always a scalar multiple of x 1, and indeed this is an eigenvector with eigenvalue 6. Thus we begin with the sequence 1, x, x 1. It is easy to see that 1 and x are already orthogonal. And x 1 is orthogonal to both 1 and x, since two eigenvectors of a self-adjoint operator with different eigenvalues are necessarily orthogonal. MAS77 6 Spring 1-14

7 Solutions for Chapter 4 problems: Gram-Schmidt and Fourier theory. 1. Integrating by parts, we have [ ] π cos kt t, sin kt t sin kt dt t + k [ cos kπ + () cos( kπ) ] k k cos kt k dt π k cos kπ ()k+1 π. k Notice that computing these inner products involves carrying out exactly the same integrals that you use when computing Fourier series.. (a) We use the angle formula: cos(a + B)t + cos(a B)t cos At cos Bt to see that cos t cos 4t (cos 5t + cos t)/. Then cos t, cos t cos 4t On the other hand, cos t, cos 5t + cos t, cos t π, cos t, cos t + π. and cos t cos 4t, cos t cos 4t cos 5t + cos t, cos 5t + cos t 4 cos 5t, cos 5t cos t, cos t + π 4 4. Combining these, we see that if θ is the angle between the functions, then (b) We use the angle formula: cos θ π/ 1. π π/ sin(a + B)t + sin(a B)t sin At cos Bt to see that cos t sin t (sin t sin t)/ and cos 5t sin t (sin 6t sin 4t)/. Then as sin mt, sin nt if m n, we see that cos t sin t, cos 5t sin t, as required.. We use the Cauchy-Schwarz inequality. Put g(x) sin x. Then g, g sin x dx π. The Cauchy-Schwarz inequality tells us that sin x f(x) dx sin x, sin x f, f, which is exactly the statement required. The Cauchy-Schwarz inequality is an equality if f(x) is a scalar multiple of sin x, so f(x) sin x is one function where we have equality. MAS77 7 Spring 1-14

8 4. We put f 1 1, f x, f x. Put g 1 f 1 1, and g f λg 1 x λ. We wish to choose λ so that g is orthogonal to f 1 : we know that λ f,g 1 g 1,g 1, but it is almost easier in this case just to work it out directly: f 1, g [ x (x λ) dx λx and so we must choose λ 1/. Thus g x 1. 1 λ, Next, we put g f λg 1 µg x λ µ(x 1 ), and we want to choose λ and µ so that g is orthogonal to f 1 and to f. We can work out λ and µ using the formulae, or just compute the inner products. Let s do that. f 1, g So λ 1/. Next, f, g x λ µ(x 1 ) dx [ x x λx µ(x 1 )x dx [ x 4 λx µx + µx 4 λx µx + µx 4 1 λ. 1 4 λ µ 1, and so 1 4 1/ µ 1, giving µ 1. Then g x 1 (x 1 ) x x We can take g 1 f 1 sin t. Notice that f 1 sin t is already orthogonal to g 1 as sin mt, sin nt if m n. So we can put g f, and g is already orthogonal to g 1. Now notice that f (t) (sin t sin t)/ is already orthogonal to g (t), again using the remark about sin mt, sin nt above. To make it orthogonal to g 1, put g f λg 1 sin t(cos t λ), and try to choose λ so that g, g 1. But g, g 1 sin t. sin t(cos t λ) dt (1 cos t)(cos t λ)/ dt λ π/, so λ /. Thus g sin t(cos t + 1/). ( ) ( ) ( ) Clearly a basis for V is B 1, B and B 1, and these 1 three matrices are clearly orthogonal to each other, so form an orthogonal basis for V. For A V, π(a) A, B 1 B 1, B 1 B 1 + A, B B, B B + A, B B, B B a 1 B 1 + b + c B + d 1 B ( ) a b+c b+c d (A + A T )/ MAS77 8 Spring 1-14

9 An alternative approach is to identify V and to write A in its ( unique ) form C + D, a b where C V and D V. Then π(a) C. A matrix A is orthogonal c d to B 1 if {( a ; is) orthogonal to B if b + c, and orthogonal to B if d. Then b V b R}. Given any matrix A, we want to write it as C + D, b where C V and D V. The unique way to do this is with ( ) ( ) a b+c C b c b+c and D c b. d Then as required. π(a) C ( ) a b+c b+c (A + A T )/, d 7. Let s take B 1 A 1. To find a matrix B A λb 1 orthogonal to B 1, it is easy to see that we can take λ 1 so that B 1 1, clearly orthogonal to B 1. Similarly, if B A λb 1 µb is orthogonal to B 1 and B, it is easy to see that we can take µ 1 and λ 1 to get 1 1 B Notice that B A A. The real reason that this is orthogonal to A 1 and A is that we ve arranged that there s no non-zero entry of B corresponding to a non-zero entry of A 1 or A (which have the same span as B 1 and B ) so it must be orthogonal to them. We could have done this by inspection. In the same way, we can see that 1 1 B 4 A 4 A 1 1 is orthogonal to all of B 1, B and B. Thus we have an orthogonal set B 1 1 1, B 1 1, B , B 4, MAS77 9 Spring 1-14

10 and we can make them normal by dividing by their norms: C , B 1 1 1, C , C 4 1, as required. 8. Let v 1 u 1. We need v u λv 1 to be orthogonal to v 1, and λ u,v 1 v 1,v 1 4 5, so that v (1/5, 1/5, 1/5, 1/5, 4/5) T. Next, we need v u λv 1 µv to be orthogonal to v 1 and v (or equivalently to u 1 and u, since they have the same span). But we know that we should take λ u,v 1 v 1,v 1 5, and µ u,v v,v /5 4/5 4. This gives v (1/4, 1/4, 1/4, /4, ) T. Next we need v 4 u 4 λv 1 µv νv to be orthogonal to v 1, v and v. But then λ u 4,v 1 v 1,v 1 5, µ u 4,v v,v /5 4/5 1, and ν u 4,v v,v 1/ /4, and so v 4 (1/, 1/, /,, ) T. I hope that you spotted the pattern before now, and worked out, just by staring at the vectors, why this had to be the case. So I won t bother to write down the final set of equations; if you do, you will find that v 5 (1/, /,,, ) T, but you should have been able to predict it anyway from how things have been working so far. Of course, these aren t yet orthonormal; we need to divide by the norms of the vectors, to get ˆv (1, 1, 1, 1, 1) T ˆv 1 (1, 1, 1, 1, 4) T ˆv 1 1 (1, 1, 1,, ) T ˆv (1, 1,,, ) T ˆv 5 1 (1,,,, ) T. MAS77 1 Spring 1-14