MATH 205 HOMEWORK #3 OFFICIAL SOLUTION. Problem 1: Find all eigenvalues and eigenvectors of the following linear transformations. (a) F = R, V = R 3,
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1 MATH 205 HOMEWORK #3 OFFICIAL SOLUTION Problem 1: Find all eigenvalues and eigenvectors of the following linear transformations. a F = R, V = R 3, b F = R or C, V = F 2, T = T = There should be two different answers. c F = R, V = C R, R, T = d dx. d F has characteristic 0, V is the vector space of polynomials of degree at most 3 in x, Solution:. T px = d2 dx 2 x 2 1px. Explain how you would solve this problem if V = F [x]. a Suppose that λ is an eigenvalue of T ; then T λi is not invertible. Therefore if we do row reduction on T λi we will get a condition for λ to be an eigenvalue of T. First, suppose that λ = 9. Then when we row-reduce T + 9I we get R3 2R2 R R which has 3 pivots and thus is invertible. Therefore λ 9. Then we have 9 λ λ 4 R3 2R2 9 λ λ λ λ 1 λ R2 8 9+λ R1 9 λ λ5+λ 41+λ 9+λ 9+λ λ 1 + λ Note that the bottom two rows are divisible by 1 + λ; thus λ = 1 is a possibility. When λ = 1 the top two rows of the matrix are equal and the bottom one is twice the top one; thus the kernel of T + I has dimension 2. A simple computation shows that the vectors 1, 2, 0 and 1, 0, 2 span this kernel and are, in fact, eigenvectors of T with eigenvalue 1. Since eigenvectors with different eigenvalues are linearly independent, we can have at most one more eigenvalue; it will have a one-dimensional subspace of eigenvectors. 1
2 2 MATH 205 HOMEWORK #3 OFFICIAL SOLUTION If λ 1 we can divide the bottom two rows by 1 + λ and continue. We get 9 λ λ 4 R λ 9 λ λ R3 9+λ 9+λ 3 λ λ In order for T λ to not be invertible we must have 3 λ = 0, or λ = 3. Then T λ is equal to = By inspection, 1, 1, 2 is in the kernel of this matrix. Thus 1, 1, 2 is an eigenvector with eigenvalue 3. b Note that this matrix has no kernel, so we can assume λ 0. Doing row reduction, we get λ 1 1 λ R1+λR2 0 λ λ Thus this matrix is not invertible if and only if λ = 0. If F = R then this has no eigenvalues. If F = C then it has two possibilities: λ = ±i. By inspection, the two eigenvectors are 1, i for λ = i and 1, i for λ = i. c We need to find all smooth functions f such that Multiplying both sides by dx f Integrating both sides we get Exponetiating we then conclude that df dx = λf. we get df f = λdx. log f = λx + C. f = e C e λx. Since all functions e λx work, we see that these are exactly the eigenvectors of T. d We will show that there is an eigenvector of each degree and that they all have different eigenvalues. Write px = a nx n. Then T px = d2 dx 2 a n a n+2 x n+2 = n + 2n + 1a n a n+2 x n. If px is an eigenvector with eigenvalue λ then we know that for all n, or in other words n + 2n + 1a n a n+2 = λa n a n+2 = λ 1 a n. n + 2n + 1 If we assume that deg p = m then we know that a m+2 = 0, so that λ = m + 2m + 1; in addition, we conclude that a m 1 = 0. Then we can use this recursion to construct the polynomial of degree m that will be an eigenvector of this eigenvalue. Thus we see that we ve found a basis of eigenvectors for F [x], and thus we have found all of the eigenvectors. Using this formula we can find the polynomials up to degree 3:
3 MATH 205 HOMEWORK #3 OFFICIAL SOLUTION 3 degree λ polynomial x x x 3 3x Problem 2: Look at the notes on the website showing that every linear operator over an algebraically closed field has an eigenvalue. Give an example of a vector space V over C, a linear transformation T : V V and two vectors v 1, v 2 such that the polynomials constructed in the proof are different for v 1 and v 2. Solution: Let V = C 2 and let T take z 1, z 2 to z 1, 0. Then if we select the vector 1, 0 then the sequence v, T v, T 2 v is just 1, 0, 1, 0, 1, 0, and we can take the polynomial x 1. On the other hand, if we take the vector 0, 1 then the sequence is 0, 1, 0, 0, 0, 0, so we can take the polynomial x instead. Problem 3: Let F be an algebraically closed field. Give an example of a vector space over F and a linear operator T LV such that T has no eigenvalues. Solution: Let V = F and let T be the right-shift operator which takes x 1, x 2,... to 0, x 1, x 2,.... Then T has no eigenvalues. Indeed, suppose it did, so that 0, x 1, x 2,... = λx 1, x 2,.... We must have λ = 0 in this case. Thus the only possible eigenvalue is 0. But T is injective, so the only vector in the kernel is 0. Thus T has no eigenvalues. Problem 4: Let LV be the set of linear transformations V V. For two linear transformations T, T LV, we define T + T v = T v + T v and T T v = T T v. Prove that LV is a ring but not a field. If V is finite dimensional, what is LV? Solution: The only nontrivial axiom to check is that composition distributes over addition. Thus we need to check that for transformations S, T, T we have ST + T = ST + ST and T + T S = T S + T S. We will check the first of these; the second follows analogously. Note that for any v V, ST + T v = ST v + T v = ST v + ST v = ST v + ST v, where the second equality follows because S is a linear transformation. Thus LV is a ring. However, in general it is not a field. The inverse in the ring is function inversion, so all we need to check is that there are non-invertible functions in LV. If V is one-dimensional the only non-invertible function will be 0, so in fact LV will be a field; however, in general it is possible to construct a non-invertible function by choosing any basis and taking one of the vectors to 0 and the others to themselves for example. Thus non-invertible functions exist, and LV is not a field. If V is finite dimensional then LV is the group of invertible n n matrices, which is GL n F. Problem 5: Suppose that S, T are in LV. Show that ST and T S have the same eigenvalues. Give an example of some F, V, S and T such that ST and T S have different eigenvectors. Solution: Suppose that λ is a nonzero eigenvalue of ST with eigenvector v. Then T ST v = T ST v = T λv = λt v.
4 4 MATH 205 HOMEWORK #3 OFFICIAL SOLUTION Thus if T v is nonzero it is an eigenvector of T S with eigenvalue λ. The reverse version works as well. Thus the only worry may be that T v = 0. But if T v = 0 then ST v = ST v = S0 = 0, so λ = 0, a contradiction. Thus T v is nonzero, and λ is an eigenvalue of T S. Thus the only place where the eigenvalues of ST and T S may differ is if one has a nontrivial kernel and the other does not. Suppose that ST is injective; we claim that T S is also injective. Note: this will only work if the vector space is finite dimensional; if the space is infinite dimensional then the right-shift and left-shift operators are a counterexample. Clearly, T must be injective. But S must also be injective, as dim im ST dim im S n. Thus if ST is injective then so are both S and T, and therefore T S must be injective as well. Thus 0 is an eigenvalue of ST if and only if it is an eigenvalue of T S, and we see that their eigenvalues are the same. Let F = R, V = R 2. Let S take x, y to 0, x and T take x, y to 0, y. Then ST x, y = S0, y = 0, 0 and T Sx, y = T 0, x = 0, x. In this case, everything is an eigenvector of ST with eigenvalue 0, but only the y-axis is eigenvectors of T S with eigenvalue 0. Problem 6: Let F = C. Suppose that T LV and that p C[x]. Show that a is an eigenvalue of pt if and only if a = pλ for some eigenvalue λ of T. Give an example to show that this is not true if F = R. Write px = a nx n. If λ is an eigenvalue of T with eigenvector v then pt v = a n T n v = a n λ n v = pλv. So pλ is an eigenvalue of pt with eigenvector v. Suppose that a is an eigenvalue of pt. Then pt ai has a kernel. Write px a = x λ 1 x λ m. If v is an eigenvector with eigenvalue a then we know that T λ 1 I T λ m Iv = 0. Thus for some j, T λ j has a kernel, and thus λ j is an eigenvalue of T. Then so a = pλ j, as desired. Let pλ j a = λ j λ 1 λ j λ m = 0, T = and let px = x 2. Then T 2 = 1, so 1 is an eigenvalue of T, but T has no eigenvalues over R. Problem 7: Let LV, W be the set of linear transformations V W. a Show that LV, W is a vector space. If dim V = n and dim W = m, what is dim LV, W? b Let {v 1,..., v n } be a basis for V and {w 1,..., w m } be a basis for W. Describe a basis for LV, W in terms of v s and w s. Solution: a Inside LV, W we define addition and scalar multiplication pointwise, which is possible since it is the set of functions into a vector space. Then it is a vector space simply because W is a vector space. dim LV, W = mn; the proof is in the next part.
5 MATH 205 HOMEWORK #3 OFFICIAL SOLUTION 5 b Let f ij be the linear transformation that takes v i to w j and all other v i to 0. We claim that the f ij are a basis for LV, W so that dim LV, W = mn. Indeed, suppose T is any linear transformation. If we write note that we can write T v i = a i1 w a im w m T = i,j a ij f ij. Thus the f ij span LV, W. To check that they are linearly independent, note that if there exist b ij such that i,j b ijf ij = 0 then by applying both sides to v i we get m b ij w j = 0. j=1 Thus b ij = 0 for all j. Since this holds for any i, we conclude that all b ij = 0 and the f ij are linearly independent, as desired. Problem 8: Let f : [0, 1] R/Z and let f : [0, 1] R be its lift. Recall that if f0 = f1 then the winding number of f is defined to be f1. a Look at the construction of f on the notes on the website. Prove that f is well-defined. b Check that if f0 = f1 then f1 f0 Z. c Let f be defined by fa = r exp2πia n for n Z. Show that f1 f0 = n. Show that if f is constant then so is f. d We define the path sum of two functions f, g : R/Z R/Z to be the function { f2a a 1 f + p ga = 2 g2a 1 a 1 2. Show that f + p g1 = f1 + g1. Note: this should make sense. Remember that f1 is supposed to count the number of times we ve gone around the origin. If you go around a circle n times and then m times then you ve gone around the circle m + n times. e Let f, g : R/Z C. Show that if fa > fa ga for all a R/Z then the winding numbers of 1 2π arg f and 1 arg g 2π are the same. Use this to show that if p is a polynomial of degree n then the winding number of a 1 2π argpre2πia is n for sufficiently large r. Solution: a Note that the definition of f chooses small enough intervals inside [0, 1] such that the value of f does not change by more than 1/4. Thus if f1/m = α then we think of f as taking values inside α 1/4, α + 1/4 while we are in the interval [1/m, 2/m]. In this way we consider f to be a continuous function [1/m, 2/m] R, and thus f will be a continuous function within this interval. Since f is defined by splicing together different continuous functions it is continuous if it is continuous at each boundary point. The fact that it is continuous at each boundary point follows from the formulas.
6 6 MATH 205 HOMEWORK #3 OFFICIAL SOLUTION b If f0 = f1 then f1 f0 f1 f0 = 0 mod R/Z so f1 f0 Z, as desired. c Choose m = 4n. Then fk/m + α = fk/m + nα; in particular, fk + 1/m = fk/m + 1/4, so by induction fx = nx for all x [0, 1]. Thus f1 = n, as desired. If f is constant then fk/m + α = fk/m, so by induction f is constant. d Let m f be the value chosen for m when constructing f, and m g be the value chosen for g when constructing g. Then m = 2 maxm f, m g works for constructing f + p g. Then from the definition of f + p g we know that f + p g1/2 = f1 f + p g1 f + p g1/2 = g1. Thus f + p g1 = f1 + g1, as desired. e Let φ = 1 2π arg f and γ = 1 2π arg g. We claim that if fa > fa ga then φ γ < 1/2 for all x. If we can show this then we are done, since φ1 γ1 = 0 as it must be an integer. Consider C as R 2. Then we want to show that if fa > fa ga then the angle between the vector fa and the vector ga is less than π/2. But the condition that fa > fa ga says that ga lies in the circle centered at fa with radius fa, the angle between fa and all such vectors is less than π/2, as desired. Suppose that p is a polynomial of degree n > 0; we assume without loss of generality that it is monic for simplicity. Then set fα = re 2πiα n and gα = pre 2πiα. Then f g is a function of degree n 1, and thus for large enough r we will have fα > fα gα. For a more precise statement, take r > 4 max a n 1 + a n a 0, 2. Thus the winding number of the given function is equal to the winding number of re 2πiα n, which we showed in part c is n.
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