Solutions to the Multivariable Calculus and Linear Algebra Problems on the Comprehensive Examination

Multivariable Calculus and Linear Algebra Solutions January 2017 Solutions to the Multivariable Calculus and Linear Algebra problems on the Comprehensive Examination of January 27, 2017 1. (a) [15 points] Find an equation of form ax + by + cz = d for the plane passing through ( 2, 1, 4) that is perpendicular to the line with parametric equations x = 2t, y = 3t 1, z = 5 t. Solution: The plane should have the same normal vector as the line parameterized as (x, y, z) = (2t, 3t 1, 5 t) = t(2, 3, 1) + (0, 1, 5). Therefore, the plane has normal vector (2, 3, 1). It must additionally pass through ( 2, 1, 4), hence, its equation is given by 2(x ( 2)) + 3(y ( 1)) + ( 1)(z , 2x + 3y z = 11. 4) = 0 (b) [15 points] Find an equation for the tangent plane to the sphere x2 + y 2 + z 2 = 3 at the point (1, 1, 1). Solution: The sphere x2 + y 2 + z 2 = 3 is a level surface of the function F (x, y, z) = x2 + y 2 + z 2 . We have Fx (x, y, z) = 2x, Fy (x, y, z) = 2y, Fz (x, y, z) = 2z, and hence Fx (1, 1, 1) = 2, Fy (1, 1, 1) = 2, Fz (1, 1, 1) = 2. The tangent plane at (1, 1, 1) is thus given by Fx (1, 1, 1)(x 1) + Fy (1, 1, 1)(y ( 1)) + Fz (1, 1, 1)(z , 2(x 1) + 2(y + 1) + 2(z 1) = 0 , 2x 2y + 2z = 6 , x y + z = 3. 8 3 5y 3 < 3x 2. Let f (x, y) = x2 + y 2 : 0 1) = 0 if (x, y) 6= (0, 0) if (x, y) = (0, 0). (a) [15 points] Compute fx (0, 0) and fy (0, 0). Solution: First observe that when h 6= 0, we have f (h, 0) = 3h3 0 = 3h, h2 + 0 f (0, h) = 0 5h3 = 0 + h2 5h. Then we use the definition of partial derivative to obtain fx (0, 0) = lim f (h, 0) h h!0 and fy (0, 0) = lim h!0 f (0, 0) f (0, h) f (0, 0) h = lim 3h h!0 = lim Page 1 of 5 h!0 5h h 0 h = lim 3 = 3 h!0 0 = lim h!0 5= 5. Multivariable Calculus and Linear Algebra Solutions January 2017 (b) [10 points] Is f is continuous at (0, 0)? Justify your answer. Solution: Recall that f is continuous at (0, 0) i↵ lim(x,y)!(0,0) f (x, y) = f (0, 0). Here, f (0, 0) = 0, so it suffices to show that lim(x,y)!(0,0) f (x, y) = 0. In polar coordinates, x = r cos ✓, y = r sin ✓, and (x, y) ! (0, 0) becomes r ! 0. Then 3x3 5y 3 3r3 cos3 ✓ 5r3 sin3 ✓ = lim r!0 r 2 cos2 ✓ + r 2 sin2 ✓ (x,y)!(0,0) x2 + y 2 r3 (3 cos3 ✓ 5 sin3 ✓) = lim 2 r!0 r (sin2 ✓ + cos2 ✓) = lim r(3 cos3 ✓ 5 sin3 ✓) = 0 lim r!0 as desired, where we have used sin2 ✓ + cos2 ✓ = 1, and that 3 cos3 ✓ bounded. 5 sin3 ✓ is 3. [20 points] Find the volume of the solid region bounded by the surfaces y = 0, y = 3, z = x2 , and z = 2 x2 . Solution: The curves z = x2 and z = 2 x2 intersect when x2 = 2 x2 , 2x2 = 2 , x = ±1. For 1  x  1, the curve z = 2 x2 sits above the curve z = x2 . Therefore, the volume is given by Z 3 Z 1 Z 3 Z 1 2 2 V = (2 x ) x dx dy = 2 2x2 dx dy y=0 3 x= 1 y=0 x= 1 ✓ Z 1 ◆ Z 3  2 = 2 2 2x dx dy = 2 2x y=0 x=0 y=0 ✓ ◆Z 3 ✓ ◆ 2 2 =2 2 dy = 2 2 3 = 8. 3 3 y=0 Z 2 3 x 3 x=1 dy x=0 4. [25 points] Show that the line integral Z z 2 dx + 2y dy + 2xz dz C depends only on the endpoints of the path C and not on the path taken between those endpoints. Solution: Let x = (x, y, z), and let F (x) = (z 2 , 2y, 2xz). If we let f (x) = z 2 x + y 2 , then the gradient of f satisfies rf (x) = (fx , fy , fz ) = (z 2 , 2y, 2xz) = F (x). That is, F is a conservative vector field with potential function f . If C is given by r(t), where a  t  b, then Z Z Z 2 z dx + 2y dy + 2xz dz = F · dr = rf · dr = f (r(b)) f (r(a)) C C C by the Fundamental Theorem for Line Integrals, and is therefore independent of path, and depends only on the end points r(a) and r(b). Page 2 of 5 Multivariable Calculus and Linear Algebra Solutions January 2017 5. Let V denote the vector space of polynomials of degree less than or equal to 2 with real coefficients. Let T : V ! R be a linear transformation. (a) [5 points] Explain what is meant by the kernel, or null space, of T . Solution: The set N (T ) = {x 2 V | T (x) = 0}. (b) [15 points] Prove that the kernel of T is a subspace of V . Solution: Given x1 , x2 2 N (T ) arbitrary: T (x1 + x2 ) = T (x1 ) + T (x2 ) = 0 + 0 = 0. So x1 + x2 2 N (T ) and so N (T ) is closed under addition. Given x 2 N (T ) and c 2 R arbitrary: T (cx) = cT (x) = c0 = 0. So cx 2 N (T ) and so N (T ) is closed under scalar multiplication. Since T (0) = 0 we have 0 2 N (T ). Therefore N (T ) is a subspace of V . (c) [10 points] What are the possible values of the nullity (that is, the dimension of the kernel) of T ? Justify your answer. Solution: By the Rank-Nullity Theorem, the nullity of T is equal to dim(V ) rank(T ). Since the target vector space is one-dimensional, the rank is either 0 or 1. The dimension of V is 3. Therefore, the nullity of T is equal to either 2 or 3. 6. [20 points] Suppose that {u1 , u2 } is a basis for a vector space U . Is the set {u1 u2 , u1 + 2u2 } also a basis for U ? Justify your answer. Solution: To decide if the set is linearly independent, suppose that a(u1 u2 ) + b(u1 + 2u2 ) = 0. Rearranging we have (a + b)u1 + ( a + 2b)u2 = 0. Since {u1 , u2 } is a basis, it is linearly independent and so a + b = 0, a + 2b = 0. Adding these two equations, we get 3b = 0 and so b = 0. Therefore, also a = 0. So the set {u1 u2 u1 + 2u2 } is linearly independent. We know that dim(U ) = 2 since {u1 , u2 } is a basis. Therefore, the linearly independent set {u1 u2 u1 + 2u2 } must also be a basis. Page 3 of 5 Multivariable Calculus and Linear Algebra Solutions January 2017 7. (a) [5 points] Explain what it means to say that a real number is an eigenvalue of an n ⇥ n matrix A. Solution: It means that there is some x 2 Rn with x 6= 0 such that Ax = x. (b) [10 points] Calculate the eigenvalues of the matrix 2 3 2 1 0 1 5. A = 40 1 0 3 1 Solution: The characteristic polynomial is 2 2 1 4 0 1 det 0 3 0 1 1 which gives (2 )((1 )( 1 ) 3) = (2 Therefore, the eigenvalues of A are multiplicity 1). )(( 2 3 5 4) = ( 2)2 ( + 2). = 2 (with multiplicity 2) and = 2 (with (c) [15 points] Decide if A is diagonalizable or not. Justify your answer. Solution: To find eigenvectors for = 2, we solve (A which gives 2 0 40 0 2I)x = 0 1 1 3 In reduced row echelon form, this linear 2 0 1 40 0 0 0 which has solutions 3 0 1 5 x = 0. 3 system is 3 0 15 x = 0 0 82 3 9 < t = 4 5 0 |t2R . x= : ; 0 This eigenspace is 1-dimensional. Since the multiplicity of the eigenvalue 2 is 2, this means that A is not diagonalizable. Page 4 of 5 Multivariable Calculus and Linear Algebra Solutions January 2017 8. (a) [5 points] Let U and W be vector spaces. Explain what it means to say that a linear transformation S : U ! W is invertible. Solution: It means that there is some linear transformation S 1 : W ! U such that SS 1 = IW , the identity on W , and S 1 S = IU , the identity on U . (Alternatively, one could also explain invertibility in terms of injectivity and surjectivity of S.) (b) [5 points] Let T : V ! V be a linear transformation, and let ↵ = {v1 , v2 } be a basis for the vector space V . Suppose that the matrix of T with respect to ↵ is  3 1 2 1 Explain how you know that T is invertible. Solution: The matrix above has determinant 5 6= 0 so is an invertible matrix. Therefore T is invertible. (c) [10 points] Calculate T 1 (v1 ). Write your answer as a linear combination of the vectors v1 and v2 . Solution: The matrix for T 1 is given by   1 1 1 1 3 1 = . 2 1 2 3 5 Therefore T 1 1 (v1 ) = v1 5 Page 5 of 5 2 v2 . 5 Solutions to the Algebra problems on the Comprehensive Examination of January 27, 2017 1. [25 points] Let G be a group, let H ⊆ G be a subgroup, and let N ⊆ G be a normal subgroup. Define N H = {nh : n ∈ N, h ∈ H}. Prove that N H is a subgroup of G. Solution: (Nonempty): Since N and H are both nonempty, we may choose x ∈ N and h ∈ H. Then xh ∈ N H, so N H = 6 ∅. (Closed under ∗): Given x, y ∈ N H, write x = n1 h1 and y = n2 h2 , for some n1 , n2 ∈ N and h1 , h2 ∈ H. Since N is a normal subgroup of G, we have h1 N = N h1 . Thus, there is some n3 ∈ N such that h1 n2 = n3 h1 . So xy = (n1 h1 )(n2 h2 ) = n1 (h1 n2 )h2 = n1 (n3 h1 )h2 = (n1 n3 )(h1 h2 ) ∈ N H, as desired, since n1 n3 ∈ N and h1 h2 ∈ H. (Closed under inverses): Given x ∈ N H, write x = nh with n ∈ N and h ∈ H. Since N is a normal subgroup of G, we have hN = N h. Thus, there is some n1 ∈ N such that nh = hn1 . So −1 x−1 = (nh)−1 = (hn1 )−1 = n−1 ∈ N H, 1 h −1 as desired, since n−1 ∈ H. 1 ∈ N and h 2. [17 points] Let G be a finite group, and suppose that there is an element a ∈ G with the property that a2 = a−1 but a is not the identity. Prove that the order of G is divisible by 3. Solution: Since a2 = a−1 , we have that a(a2 ) = aa−1 , or equivalently, a3 = e, where e is the identity element in G. Then o(a)|3, and hence o(a) is either 1 or 3. Since a 6= e, we have o(a) 6= 1, so the order of a must be 3. By (a corollary of) Lagrange’s Theorem, o(a) divides the order of G. That is, the order of G is divisible by 3, as desired. 3. Consider the group S8 of permutations of the set {1, 2, 3, 4, 5, 6, 7, 8}. (a) [10 points] Find an element of S8 of order 15. (Don’t forget to explain or prove why it has order 15.) Solution: Let σ = (1, 2, 3)(4, 5, 6, 7, 8) ∈ S8 . Since we have already written σ as a product of disjoint cycles, we have o(σ) = lcm(3, 5) = 15. (b) [15 points] Prove that there are no odd permutations in S8 of order 15. Solution: Suppose σ ∈ S8 satisfies |σ| = 15. Write σ = α1 α2 · · · αk as a product of k disjoint cycles α1 , . . . , αk for some positive integer k. Because these cycles are disjoint, we have 15 = o(σ) = lcm(o(α1 ), o(α2 ), . . . , o(αk )). 1 By definition of the least common multiple, each o(αj ) divides 15. Thus, each o(αj ) ∈ {3, 5, 15}, since these are the only divisors of 15 which are greater than 1. Hence, each αj is a cycle of odd length. We know that any cycle of odd length is an even permutation. Thus, each αj is even. Therefore, σ is also even, because it is the product of even permutations. 4. Let R be a ring. (a) [8 points] Define what it means for a subset I ⊆ R to be an ideal of R. If you use any other technical terms like “closed,” “subring,” “group,” “subgroup,” etc., you must fully define those terms as well. Solution: I ⊆ R is an ideal if: 1. I 6= ∅. 2. For all x, y ∈ I, we have x − y ∈ I 3. For all x ∈ I and r ∈ R, we have rx ∈ I and xr ∈ I. (b) [25 points] Suppose that R is commutative and has a multiplicative identity 1. Let I ⊆ J ⊆ R be ideals, and suppose that the quotient ring R/I is a field. If I ( J, prove that 1 ∈ J. (In fact, it is a Theorem from Math 350 that J = R in this case, but you are only being asked to prove that 1 ∈ J. In particular, however, you may not quote the J = R theorem.) Solution: Because I ( J, there exists r ∈ J r I. By the coset relation, r ∈ / I implies I + r 6= I + 0 = 0R/I . Thus, I + r is nonzero. By the definition of field, every nonzero element of R/I has a multiplicative inverse. Hence, there exists I + s ∈ R/I such that (I + r)(I + s) = I + 1. So I + rs = I + 1, and therefore 1 − rs ∈ I ⊆ J by the coset relation. Note that rs ∈ J, because J is an ideal of R and r ∈ J. Thus, since J is closed under addition, we have 1 = (1 − rs) + rs ∈ J. 2 Analysis Solutions January 2017 1. Suppose we have nonempty subsets A, B ⊆ R that are bounded above. Define A + B = {a + b | a ∈ A, b ∈ B}. (a) [10 points] Prove that A + B is bounded above. Solution: Since A and B are both nonempty and bounded above, we know from the Axiom of Completeness that their supremums exist. Let s = sup(A) and t = sup(B). We aim to show that s + t is an upper bound for A + B. Let x ∈ A + B. There exists some a ∈ A and b ∈ B such that x = a + b. Since s and t are upper bounds for A and B respectively, we know that a ≤ s and b ≤ t. Thus x = a + b ≤ s + t and A + B is bounded above by s + t. (b) [15 points] Prove that sup(A + B) ≤ sup(A) + sup(B). Solution: From part (a), we know that A + B is bounded and clearly nonempty so by the Axiom of Completeness, sup(A + B) exists. Further, from part (a) we know that sup(A) + sup(B) is an upper bound for A + B. Since sup(A + B) is the smallest upper bound for A + B, it must be that sup(A + B) ≤ sup(A) + sup(B). 2. (a) [10 points] State the -N definition of what it means for a sequence (an ) of real numbers to converge to a ∈ R. Solution: A sequence of real numbers is said to converge to a real number a if, for every  > 0, there exists an N ∈ N such that whenever n ≥ N it follows that |a − an | < . Note: One could also give the Topological Version of Convergence. A sequence of real numbers is said to converge to a real number a if, for every  > 0, there exists an N ∈ N such that whenever n ≥ N it follows that an ∈ V (a). (b) [15 points] Assume that sequences (an ) and (bn ) converge to real numbers a and b respectively. In other words, (an ) → a and (bn ) → b. Use the definition given in part (a) to prove that (an + bn ) → a + b. Solution: Let  > 0 be given. By definition of convergence, since (an ) → a we know that there exists an N1 ∈ N such that whenever n ≥ N1 it follows that |a − an | < 2 . Similarly, since (bn ) → b we know that there exists an N2 ∈ N such that whenever n ≥ N2 it follows that |b − bn | < 2 . Letting N = max{N1 , N2 } gives us that, whenever n ≥ N it follows that |(a + b) − (an + bn )| = |(a − an ) + (b − bn )| ≤ |a − an | + |b − bn |   < + = , 2 2 where the last inequality follows form the fact that n ≥ N1 and n ≥ N2 . Thus, by the definition of convergence (an + bn ) → a + b. Page 1 of 3 Analysis Solutions January 2017 (c) [10 points] In the limit notation used in calculus, the result of Problem 2(b) can be stated as the implication (1) if lim an = a and lim bn = b, then lim (an + bn ) = a + b. n→∞ n→∞ n→∞ Suppose that we have k sequences (a1,n ), (a2,n ), . . . (ak,n ). Use (1) and induction on k ≥ 1 to prove that if lim a1,n = a1 , . . . , lim ak,n = ak , then n→∞ n→∞ lim (a1,n + · · · + ak,n ) = a1 + · · · + ak . n→∞ Solution: The base case, when k = 1 is trivially true. Next, assume that the statement is true for k = ` sequences and that we have ` + 1 sequences (a1,n ), (a2,n ), . . . (a`+1,n ) such that lim ai,n = ai for each 1 ≤ i ≤ ` + 1. n→∞ In this case we have that h i lim (a1,n + · · · + a`+1,n ) = lim (a1,n + · · · a`,n ) + a`+1,n n→∞ n→∞ = lim (a1,n + · · · a`,n ) + lim a`+1,n n→∞ n→∞ = (a1 + · · · + a` ) + a`+1 = a1 + · · · + a` + a`+1 , where the second equality follows from 2(b) and the third equality follows from the inductive assumption. 3. (a) [5 points] Suppose that A ⊆ R and for every n ∈ N we have a function fn : A → R. Define what it means for (fn ) to converge pointwise on A to a function f : A → R. Solution: The sequence of functions (fn ) converges pointwise on A to f if for every  > 0 and x ∈ A there exists an N ∈ N such that whenever n ≥ N it follows that |fn (x) − f (x)| < . (b) [10 points] Using the notation of part (a), define what it means for (fn ) to converge uniformly on A to f . Solution: The sequence of functions (fn ) converges uniformly on A to f if for every  > 0 there exists an N ∈ N such that whenever n ≥ N and x ∈ A it follows that |fn (x) − f (x)| < . Note: The difference between pointwise and uniform convergence is that in uniform convergence the choice of N ∈ N can be made independently of the value of x ∈ A. (c) [15 points] Suppose that g : A → R is a bounded function. For each n ∈ N, define fn (x) = g(x) . n There is a function f such that (fn ) converges uniformly to f . Find f and prove that the convergence is uniform. Page 2 of 3 Analysis Solutions January 2017 Solution: Let  > 0 be given. Since g(x) is bounded on A, there exists an M > 0 such that for every x ∈ A we have that |g(x)| ≤ M . By the archimedean property, there exists an N ∈ N such that N1 < M . Let f (x) be the function that is uniformly zero on A. For every n ≥ N and x ∈ A it follows that |fn (x) − f (x)| = |fn (x) − 0| = g(x) M M M |g(x)| ≤ ≤ < 
= . = n n n N M Therefore, fn converges uniformly to f on A. 4. [10 points] Suppose [ that we have open sets Oλ ⊆ R for all λ in some index set Λ. Prove that the union O = Oλ is an open set. λ∈Λ Solution: Let a ∈ O = [ Oλ . By definition there exists a λ0 ∈ Λ such that a ∈ λ∈Λ Oλ0 . Since Oλ0 is open, there exists an  > 0 such thatSthe epsilon neighborhood of a, V (a), is contained in Oλ0 . Therefore, V (a) ⊆ Oλ0 ⊆ λ∈Λ Oλ = O and O is open by definition. Page 3 of 3