Math 1180 Solutions to Homework 7 April 10, 2013 LADW Chapter 5, Number 1.8 (a) Taking v = x gives kxk2 = hx, xi = 0, so x = 0. (b) Let v ∈ V . Since P {v1 , . . . , vn } is a spanning set, there are scalars c1 , . . . , cn such that v = cj vj . Then D X E X ck hx, vk i = 0 hx, vi = x, ck v k = so, by part (a), x = 0. (c) We have hx − y, vk i = hx, vk i − hy, vk i = 0 so, by part (b), x − y = 0. LADW Chapter 5 Number 3.1 Let v1 , v2 , v3 be the three given vectors. We have   1 w1 = v1 =  2  −2       1 1 2 hv2 , w1 i −9  −1  −  2 = 1  w = w2 = v2 − 1 kw1 k2 9 4 −2 2 hv3 , w1 i hv3 , w2 i w1 − w2 w3 = v3 − kw1 k2 kw2 k2           1 2 2 2 2/9 7 1 2 =  1  −  2  −  1  =  −2/9  =  −2  9 9 9 1 −2 2 −1/9 −1

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Finally, divide each wj by its norm to get an orthonormal set:   1 1 2  u1 = 3 −2   2 1 1  u2 = 3 2   2 1 −2  . u3 = 3 −1 LADW Chapter 5 Number 5.3 Let x ∈ ker A. Then (A∗ A)x = A∗ (Ax) = A∗ 0 = 0, so x ∈ ker(A∗ A). Therefore ker A ⊂ ker(A∗ A). For the reverse inclusion, let x ∈ ker(A∗ A). Then kAxk2 = hAx, Axi = hx, A∗ Axi = hx, 0i = 0, so Ax = 0. Therefore x ∈ ker A, so ker(A∗ A) ⊂ ker A. SE Number 1 (a) Letting Q = I − P . Then Q2 = (I − P )2 = I 2 − IP − P I + P 2 = I − P − P + P = I − P = Q. (b) Let λ be an eigenvalue for P with eigenvector v. Then λ2 is an eigenvalue for P 2 with the same eigenvector v. Since P 2 = P , it follows that λ2 = λ, so λ = 0 or λ = 1. If P is not the zero operator, then there is a v ∈ V with P v 6= 0. Then P (P v) = P 2 v = P v, so P v is an eigenvector with eigenvalue 1. If P is not the identity operator, then there is a v ∈ V with P v 6= v. Then v − P v 6= 0 and P (v − P v) = 0, so v − P v is an eigenvector with eigenvalue 0. (c) The eigenspace for the eigenvalue 1 is the range of P , and the eigenspace for the eigenvalue 0 is the kernel of P . By the Rank-Nullity Theorem, the sum of the dimensions of these two subspaces is the dimension of V , so combining bases for these two subspaces gives a basis for V consisting of eigenvectors for P . Alternatively, for any v ∈ V , the vector w = v − P v satisfies P w = P v − P 2 v = 0, so w is in the kernel of P . Therefore v = w + P v, with w and P v both eigenvectors for P , with eigenvalues 0 and 1, respectively. Therefore, the eigenvectors for P span V . 2

(d) Suppose S −1 P S = D where D is a diagonal matrix with zeroes and ones along the diagonal. Then D2 = D, so P 2 = (SDS −1 )2 = SDS −1 SDS −1 = SD2 S −1 = SDS −1 = P. Alternatively, if v is any eigenvector for P , then P v = λv with λ equal to either 0 or 1. It follows that P 2 v = λ2 v = λv = P v. Since V has a basis consisting of eigenvectors for P , the operators P and P 2 agree on a basis, so they are equal. SE Number 2 Let v and w be vectors in a real inner product space such that kv + wk2 = kvk2 + kwk2 . We have kv + wk2 = hv + w, v + wi = hv, vi + hv, wi + hw, vi + hw, wi = kvk2 + 2 hv, wi + kwk2 By hypothesis, this equals kvk2 + kwk2 , so hv, wi = 0. In the complex case, the above argument only gives < (hv, wi) = 0, which does not imply that v and w are orthogonal. Indeed, if v is any non-zero vector in a complex inner product space, and if w = iv, then kv + wk2 = k(1 + i)vk2 = |1 + iv|2 kvk2 = 2kvk2 = kvk2 + kwk2 , but hv, wi = −ikwk2 6= 0. SE Number 3 kv + wk2 = hv + w, v + wi = kvk2 + hv, wi + hw, vi + kwk2 and kv − wk2 = hv − w, v − wi = kvk2 − hv, wi − hw, vi + kwk2 . Adding gives the result. SE Number 4 (a) We must check that hh·, ·ii satisfies the three defining properties of a real inner product. Positivity: For any v ∈ V , we have hv, vi ≥ 0, so hhv, vii = hv, vi, so positivity of hh·, ·ii follows from that of h·, ·i. Symmetry: By the Hermitian symmetry of h·, ·i, we have hu, vi = hv, ui. Since conjugation of a complex number does not affect its real part, we obtain hhu, vii = hhv, uii 3

Linearity: By linearity of the complex inner product, we have for every u, v, w ∈ V and every s, t ∈ R, hsu + tv, wi = s hu, wi + t hv, wi , and taking real parts gives hhsu + tv, wii = shhu, wii + thhv, wii. (c) If u and v are orthogonal with respect to the complex inner product, then hu, vi = 0, and so its real part hhu, vii is also 0. On the other hand, for any non-zero v ∈ V , we have hv, ivi = −ikvk2 , which is non-zero, but has 0 real part, so u and v are orthogonal with respect to the real inner product, but not with respect to the complex inner product. Since the two inner products define the same norm, the conclusion ku + vk2 = kuk2 + kvk2 follows from the hypotheses