Download Solutions to Homework 8 - Linear Algebra | MATH 115A and more Assignments Linear Algebra in PDF only on Docsity! MATH 115 SOLUTION SET 8-10 ANSWERS TO SELECTED PROBLEMS 1. Set 8 §5.2 8. We always have dimEλ2 ≥ 1. On the other hand, since dimEλ1 = n− 1, the algebraic multiplicity of λ1 is at least n − 1. So the multiplicity mλ2 of λ2 is at most 1 (the two multiplicities must sum to n). We have 1 ≥ mλ2 ≥ dimEλ2 ≥ 1, from which we conclude dimEλ2 = m and A is diagonalizable. 2. Set 9 §6.1 16. Linearity is easy. For the property 〈f, g〉 = 〈g, f〉, we check: 〈g, f〉 = 1 2π ∫ 2π 0 g(t)f(t)dt = 1 2π ∫ 2π 0 f(t)g(t)dt = 〈f, g〉 For the last property, suppose f ∈ H is nonzero. We claim 〈f, f〉 > 0. For suppose x0 ∈ [0, 2π] is a point with f(x0) 6= 0; say |f(x0)| = δ. Then since f is continuous, there must be some subinterval I ⊂ [0, 2π] for which |f(t)| ≥ δ/2 whenever t ∈ I. We have 2π 〈f, f〉 = ∫ 2π 0 |f(t)|2 dt ≥ ∫ I |f(t)|2 dt ≥ ∫ I ( δ 2 )2 dt > 0. 21. a) We have A∗1 = 1 2 (A ∗+A∗∗) = 12 (A ∗+A) = A1. The case of A2 is similar. If A1 and A2 were instead defined to be the real and imaginary parts of A, then we would have A = A1 + iA2, but then there is no reason to expect A∗1 = A1 or A∗2 = A2. b) For if A = B1 + iB2, then A∗ = B∗1 − iB∗2 = B1 − iB2 (remember: (cM)∗ = cM∗), so A+ A∗ = 2B1 and A− A∗ = 2iB2. Solving, we find that B1 and B2 are the same as the matrices A1 and A2 from part a. 1 2 ANSWERS TO SELECTED PROBLEMS 3. Set 10 §6.2 4. S⊥ is the span of the single vector (2i,−1− i, 2). 5. S⊥0 is a plane, and S ⊥ is a line. 7. The rightward direction is easy: If z ∈ W⊥, then z is orthogonal to every vector in W , including every vector in β. For the leftward direction, suppose 〈z, v〉 = 0 for every v ∈ β. Now suppose w ∈ W . Write w = ∑ i civi as a linear combination of vectors vi ∈ W . Then 〈z, w〉 = ∑ i ci 〈z, vi〉 = 0 because each 〈z, vi〉 = 0. 11. Let the rows of A be the vectors v1, . . . , vn, so that A = v1... vn . Then AA∗ = v1... vn (vt1 . . . vtn) = 〈v1, v1〉 . . . 〈v1, vn〉... . . . ... 〈vn, v1〉 . . . 〈vn, vn〉 . This matrix is the identity if and only if 〈vi, vj〉 = δij , i.e. if {v1, . . . , vn} is an orthonormal basis for Cn. §6.3 8. If T is invertible, suppose U is such that TU = I and UT = I. Taking adjoints of everything we find (TU)∗ = U∗T ∗ = I and (UT )∗ = T ∗U∗ = I. Thus T ∗ is invertible with inverse U∗ = (T−1)∗. 10. One direction is easy: If 〈T (x), T (y)〉 = 〈x, y〉 for all x and y, apply to x = y to see that ||T (x)|| = ||x||. Now assume ||T (x)|| = ||x|| for all x. The polar identity shows that 〈x, y〉 can be written in terms of ∣∣∣∣x+ iky∣∣∣∣ for k = 0, 1, 2, 3. We have∣∣∣∣T (x+ iky)∣∣∣∣ = ∣∣∣∣x+ iky∣∣∣∣, so we must have 〈T (x), T (y)〉 = 〈x, y〉 also.
