Comprehensive Exam Solutions: Multivariable Calculus & Linear Algebra

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