Math 110. Final Exam December 12, 2002

Transcription

1 C Math 110 PROFESSOR KENNETH A. RIBET Final Exam December 12, :30 3:30 PM The scalar field F will be the field of real numbers unless otherwise specified. Please put away all books, calculators, electronic games, cell phones, pagers,.mp3 players, PDAs, and other electronic devices. You may refer to a single 2-sided sheet of notes. Please write your name on each sheet of paper that you turn in. Don t trust staples to keep your papers together. Explain your answers as is customary and appropriate. Your paper is your ambassador when it is graded. Disclaimer: These solutions were written by Ken Ribet. As usual, sorry if they re a bit terse and apologies also if I messed something up. If you see an error, send me and I ll post an updated document. 1. Let T : V V be a linear transformation. Suppose that all non-zero elements of V are eigenvectors for T. Show that T is a scalar multiple of the identity map, i.e., that there is a λ R such that T (v) = λv for all v V. We can and do assume that V is non-zero. Choose v non-zero in V and let λ be the eigenvalue for v. We must show that T w = λw for all w W. This is clear if w is a multiple of v. If not, w and v are linearly independent, so that w + v is non-zero, in particular. In this case, let µ be the eigenvalue of w and let a be the eigenvalue of w + v. Then a(w + v) = T (w + v) = T w + T v = µw + λv, and so (a µ)w = (λ a)v. Because v and w are linearly independent, we get a = µ and a = λ. Hence µ = λ. 2. Let V be a 7-dimensional vector space over R. Consider linear transformations T : V V that satisfy T 2 = 0. What are the possible values of nullity(t )? (Be sure to justify your answer: don t just reply with a list of numbers.) 1

2 2 FINAL EXAMINATION NAME: I believe that 4, 5, 6, and 7 are the possible values. The condition T 2 = 0 may be rephrased as the inclusion R(T ) N(T ). Since the dimensions of R(T ) and N(T ) need to sum to 7, we must have nullity(t ) 4. Conversely, for desired value of nullity(t ) between 4 and 7, we can fabricate a T that gives this value. For example, to have nullity(t ) = 5, we define T on the standard basis vectors e 1,..., e 7 by having T (e 1 ) = T (e 5 ) = 0, T (e 6 ) = e 1 and T (e 7 ) = e 2. This works: the rank is at least 2 because the range contains the space spanned by e 1 and e 2 while the nullity is at least 5 because the first 5 basis vectors are sent to 0 by T. 3. Let A be a real n n matrix and let A t be its transpose. Prove that L A and L A t A have the same null space. In other words, for x in R n, regarded as a column vector, show that A t Ax = 0 if and only if Ax = 0. Prove also that the linear transformations L A t and L A t A have the same range. (It may help to introduce an inner product on R n.) Use the standard inner product on R n. If Ax is non-zero, then the inner product Ax, Ax is positive. Re-write as x, A t Ax to see that A t Ax is non-zero. The statement about the null space is what we have just proved. the one about ranges follows by dimension considerations, since one range is a priori contained in the other. 4. Let T : V V be a linear transformation. Suppose that v 1, v 2,..., v k V are eigenvectors of T that correspond to distinct eigenvalues. Assume that W is a T -invariant subspace of V that contains the vector v 1 + v v k. Show that W contains each of v 1, v 2,..., v k. This is a recycled quiz problem. See Tom s web page for a solution. Let A Mn n(r) be a real square matrix of size n. Let v 1,..., v n be the rows of A. Assume that v i, v j = δ ij for 1 i, j n; here,, is the standard inner product on R n and δ ij is the Kronecker delta. Show that the corresponding statement holds for the columns of A, i.e., that the inner product of the ith and the jth columns of A is δ ij. I discussed this in class a week ago. After you write down what s involved, you see that you need to know only that BA = I if AB = I when A and B are square matrices. 5. Let A = (a ij ) be an n n real matrix with the following properties: (1) the diagonal entries are positive; (2) the non-diagonal entries are negative; (3) the sum of the entries in each row is positive. Suppose, for each i = 1,..., n, that we have a i1 x 1 + a i2 x a in x n = 0, where the x i are real numbers. Show that all x i are 0. [Assume the contrary and let i be such that x i is at least as big in absolute value as the other x k. Consider the ith equation.] Prove that det A is non-zero.

3 FINAL EXAMINATION NAME: 3 We assume the contrary and let i be such that x i is at least as big in absolute value as the other x k. After changing the sign of the x k, we can and do assume that x i is positive. Then a ii x i = ( a ij )x j ( a ij )x i = x i ( a ij ). j i j i j i Dividing by x i, we get a ii a ij, which is contrary to the hypothesis that the sum of j i the entries in the ith row is positive. This business with the x k shows that the null space of L A is {0}. We conclude that A is invertible, so that its determinant is non-zero. 6. Decide whether or not each of the following real matrices is diagonalizable over the field of real numbers: , , Choose one of the above three matrices call it A. Exhibit a matrix Q of the same size as A such that Q 1 AQ is diagonal. The first matrix is a quintessential example of a non-diagonalizable matrix, as discussed in class. Its only eigenvalue is 10. If it were diagonalizable, it would be 10 times the identity matrix. The third matrix is diagonalizable because it is symmetric; all I want here is that you quote the theorem to the effect that symmetric matrices can be diagonalized over R. The middle matrix is diagonalizable because R 3 has a basis of eigenvectors for it: there s some eigenvector with eigenvalue 3, and we see that e 1 and e 2 are eigenvectors with eigenvalue 1. For this matrix, we can find Q by finding an eigenvector with eigenvalue 3. Such a vector has to have a non-zero third coefficient, since otherwise it would have eigenvalue 1. We can scale it so that it s of the form (a, b, 1). Apply the matrix to this vector and see what condition on a and b we have if we want the vector to be turned into 3 times itself. The answer is that we need a = b = 1. And, indeed, you can check instantly that (1, 1, 1) is an eigenvector with eigenvalue 3. We take Q here to be the matrix whose colums are the three eigenvectors, namely Q =

4 Math 110 Professor K. A. Ribet Final Exam May 18, 2005 This exam was a 180-minute exam. It began at 5:00PM. There were 7 problems, for which the point counts were 8, 9, 8, 7, 8, 7, and 7. The maximum possible score was 54. Please put away all books, calculators, electronic games, cell phones, pagers,.mp3 players, PDAs, and other electronic devices. You may refer to a single 2-sided sheet of notes. Explain your answers in full English sentences as is customary and appropriate. Your paper is your ambassador when it is graded. At the conclusion of the exam, please hand in your paper to your GSI. 1. Let T be a linear operator on a vector space V. Suppose that v 1,..., v k are vectors in V such that T (v i ) = λ i v i for each i, where the numbers λ 1,..., λ k are distinct elements of F. If W is a T -invariant subspace of V that contains v v k, show that W contains v i for each i = 1,..., k. See my solutions for homework set # Assume that T : V W is a linear transformation between finite-dimensional vector spaces over F. Show that T is 1-1 if and only if there is a linear transformation U : W V such that UT is the identity map on V. One direction is obvious; if UT = 1 V and T (v) = 0, then v = U(T (v)) = 0, so that T must be injective. The harder direction is to construct U when T is given as 1-1. Choose a basis v 1,..., v n of V and let w i = T (v i ) for each i. Because T is injective, the w i are linearly independent. Complete w 1,..., w n to a basis w 1,..., w n ; w n+1,..., w m of W. We can define U : W V by declaring the images U(w i ) of the basis vectors w i ; if we want w i to go to x i V, then we define U( a i w i ) = a i x i. We take x i = v i for i = 1,..., n and take (for instance) x i = 0 for i > n. It is clear that (UT )(v i ) = v i for each basis vector v i of V. It follows from this that UT is the identity map on V. 3. Let T be a self-adjoint linear operator on a finite-dimensional inner product space V (over R or C). Show that every eigenvalue of T is a positive real number if and only if T (x), x is a positive real number for all non-zero x V.

5 Because T is self-adjoint, there is an orthonormal basis β of V in which T is diagonal. Let λ 1,..., λ n be the diagonal entries of the diagonal matrix [T ] β. The λ i are real numbers even if V is a complex vector space. The issue is whether or not these real numbers are all positive. If x has coordinates a 1,..., a n in the basis β, then T (x), x = a i 2 λ i. In particular, we can take x to be i the ith element of β, so that its ith coordinate is 1 and its other coordinates are 0. Then T (x), x = λ i. Thus if T (x), x is always positive, λ i is positive for each i. Conversely, if the λ i are positive, the sum a i 2 λ i is non-negative for i all n-tuples (a 1,..., a n ) and is positive whenever (a 1,..., a n ) is non-zero, i.e., whenever the vector x corresponding to (a 1,..., a n ) is a non-zero element of V. 4. Let V be an inner product space over F and let X and Y be subspaces of V such that x, y = 0 for all x X, y Y. Suppose further that V = X + Y. Prove that Y coincides with X = { v V x, v = 0 for all x X }. Under the assumptions of the problem, everything in Y is perpendicular to everything in X, so we have Y X. If v is perpendicular to all vectors in X, we must show that v lies in Y. Because V = X +Y, we may write v = x+y with x X, y Y. We have 0 = v, x = x + y, x = x, x + y, x = x, x + 0 = x, x. Because x, x = 0, x = 0 by the axioms of an inner product. Hence v = y does indeed lie in Y. (This problem was inspired by a comment of Chu- Wee Lim, who pointed out to me that the definition on page 398 of the book, and the comments following the definition, are extremely bizarre.) 5. Let V be the space of polynomials in t with real coefficients. Use the Gram Schmidt process to find non-zero polynomials p 0 (t), p 1 (t), p 2 (t), p 3 (t) in V such that 1 1 p i (t)p j (t) dt = 0 for 0 i < j 3. (It may help to note that 0 when i is an odd positive integer.) There is an implicit inner product here: f, g = t i dt = f(t)g(t) dt. The vectors 1, t, t 2, and t 3 are linearly independent elements of V and we can apply G S to this sequence of vectors to generate an orthogonal set of vectors; this is what the problem asks for. The computations, which I won t reproduce, are easy because of the remark about the integrals of odd powers of t. The answer that I got is that the p i are in order: 1, t, t 2 1 3, t3 3 t. I did these computations in class; 5 the p i are called Legendre polynomials. 110 first midterm page 2

6 6. Let A be an n n matrix over F and let A t be the transpose of A. Using the equality row rank = column rank, show for each λ F that the vector spaces { x F n Ax = λx } and { x F n A t x = λx } have the same dimension. Combining the row rank = column rank theorem with the formula relating rank and nullity, we see that the linear transformations L A and L A t have equal nullities. These nullities are the dimensions of the null spaces of the two transformations; meanwhile, the null spaces are exactly the two vector spaces of the problem in the case λ = 0. To treat the case of arbitrary λ, we have only to replace A by A λi n. 7. Let A be an element of the vector space M n n (F ), which has dimension n 2 over F. Show that the span of the set of matrices { I n, A, A 2, A 3,... } has dimension n over F. Direct application of Cayley Hamilton: the set { I n, A, A 2, A 3,..., A n 1 } has the same span as the full set of all powers of A. 110 first midterm page 3

7 Math 110 PROFESSOR KENNETH A. RIBET Final Examination December 20, :40 3:30 PM, 101 Barker Hall Please put away all books, calculators, and other portable electronic devices anything with an ON/OFF switch. You may refer to a single 2-sided sheet of notes. When you answer questions, write your arguments in complete sentences that explain what you are doing: your paper becomes your only representative after the exam is over. All vector spaces are finite-dimensional over the field of real numbers or the field of complex numbers (except for the space P(R) of all real polynomials, which occurs in the first problem). Problem Possible points 1 9 points 2 6 points 3 7 points 4 7 points 5 7 points 6 7 points 7 7 points Total: 50 points 1. Exhibit examples of: (a.) A linear operator D : P(R) P(R) and a linear operator I : P(R) P(R) such that DI is the identity but ID is not the identity. (b.) A (non-zero) generalized eigenvector that is not an eigenvector. (c.) A normal operator on a positive-dimensional real vector space whose characteristic polynomial has no real roots. 1

8 In the first part, D was intended to evoke differentiation and I was intended to suggest integration (definition integration with constant term 0, for example). It s hard to give answers to these questions because different people will have different examples. 2. On the vector space P 2 (R) of real polynomials of degree 2 consider the inner product given by p, q = 1 1 p(x)q(x) dx. Apply the Gram Schmidt procedure to the basis (1, x, x 2 ) to produce an orthonormal basis of P 2 (R). When you apply the Gram Schmidt progress to the sequence of polynomials 1, x, x 2, x 3,..., the resulting sequence of orthonormal polynomials is the sequence of Legendre polynomials. According to the Wikipedia article on Orthogonal Polynomials, the first three of them are 1, x and (3x 2 1)/2. I haven t done this calculation lately, but I ll be wading through 38 such calculations in the very near future. 3. Suppose that P is a linear operator on V satisfying P 2 = P and let v be an element of V. Show that there are unique x null P and y range P such that v = x + y. First, let x = v P v and y = P v. Then clearly v = x+y, and y = P v is in the range of P. Since P x = P v P 2 v = P v P v = 0, x is in the null space of P. Secondly, suppose that v = x + y with x null P and y range P. Then P v = P x + P y = 0 + P y = 0. Further, P y = y because y is in the range of P. Indeed, if y = P w, then P y = P (P w) = P 2 w = P w = y. Hence P v = y, which implies that x = v y = v P v. In other words, the x and y that we are dealing with are the ones that we knew about already. Conclusion: x and y are unique. Note: this was Exercise 21 of Chapter Let T be a linear operator on an inner product space for which trace(t T ) = 0. Prove that T = 0. Choose an orthonormal basis for the space, and let A = [a ij ] be the matrix of T in this basis. The matrix of T is the conjugate-transpose A of A. The matrix of T T is then A A; for each i, i = 1,..., n, the (i, i)th entry of this matrix is a ji aji = a ji 2. The trace of A A is thus a ji 2. Each j j i,j term a ji 2 is a non-negative real number. If the sum is 0, then each term is 0. This means that all a ij are 0, i.e., that A = 0. If A = 0, then of course T = 0. Note: See problem 18 of Chapter 10, where a somewhat more sophisticated proof of the indicated assertion was contemplated by the author of our book. 2

9 5. If X and Y are subspaces of V with dim X dim Y, show that there is a linear operator T : V V such that T (X) = Y. Let d = dim Y and let n = dim V. Choose a basis (v 1,..., v e ) of X; note that e d by hypothesis. Complete this basis to a basis (v 1,..., v n ) of V. Choose a basis (y 1,..., y d ) of Y. We define T : V V so that T (v i ) = y i n for i = 1,..., d and T (v i ) = 0 for i > d. Namely, if v = a i v i, we set T v = d a i y i. The range of T is then the span of the y i, which is Y. Already, i=1 however, the span of (v 1,..., v d ) is mapped onto Y by T. Thus X is mapped onto Y by T. 6. Let N be a linear operator on the inner product space V. Suppose that N is both normal and nilpotent. Prove that N = 0. [The case F = C will probably be easier for you. Do it first do ensure partial credit.] In the complex case, we can invoke the spectral theorem and find an orthonormal basis in which N has a diagonal matrix representation. Since some power of N is 0, the diagonal entries are all 0. Hence N = 0, as required. In the real case, the required assertion follows from the statement of Exercise 24 of Chapter 8, or even better from Exercise 7 of Chapter 7. (You have Axler s solution to that exercise.) Alternatively, we can argue (cheat) as follows: choose an orthonormal basis for the space, so that N becomes a nilpotent matrix of real numbers that commuters with its transpose. We can think of this as a nilpotent matrix of complex numbers that commutes with its adjoint. Such a matrix corresponds to a complex normal nilpotent operator, which we already know to be 0. Hence it s Suppose n is a positive integer and T : C n C n is defined by T (x 1,..., x n ) = (x x n,..., x x n ); in other words, T is the linear operator whose matrix (with respect to the standard basis) consists of all 1 s. Find all eigenvalues and eigenvectors of T. This was Exercise 7 of Chapter 5. You should have a solution available to you. i=1 3

10 C Math 110 PROFESSOR KENNETH A. RIBET Final Examination May 10, :30AM 2:30 PM, 10 Evans Hall Please put away all books, calculators, and other portable electronic devices anything with an ON/OFF switch. You may refer to a single 2-sided sheet of notes. For numerical questions, show your work but do not worry about simplifying answers. For proofs, write your arguments in complete sentences that explain what you are doing. Remember that your paper becomes your only representative after the exam is over. Please turn in your exam paper to your GSI when your work is complete. The point values of the problems were 12, 7, 7, 8, 8, 8 for a total of 50 points. 1. Label each of the following statements as TRUE or FALSE. Along with your answer, provide a clear justification (e.g., a proof or counterexample). a. Each system of n linear equations in n unknowns has at least one solution. Obviously false: for example we could have the system x + y = 1, x + y = 0, which is clearly inconsistent (no solutions). b. If A is an n n complex matrix such that A = A, every eigenvalue of A has real part 0. Because A = A, A commutes with its adjoint and thus is diagonalizable in an orthonormal basis. In this basis, we compute the adjoint by conjugating the elements on the diagonal. These elements must be the negatives of their conjugates, which implies that they are indeed purely imaginary. So the answer is true! 1

11 c. If W and W are 5-dimensional subspaces of a 9-dimensional vector space V, there is at least one non-zero vector of V that lies in both W and W. Yes, this is true. A fancy way to see this is to consider the linear transformation W W V taking a pair (w, w ) to w w. Because W W has dimension 5+5 = 10 and V has dimension 9 < 10, there must be a non-zero element (w, w ) in the null space of this map. We then have w w = 0, i.e., w = w. Because w is in W and w in W and because these elements are equal, w lines in the intersection W W. So the correct answer is true. d. If T is a linear transformation on V = C 25, there is a T -invariant subspace of V that has dimension 17. Again, this is true: Schur s theorem tells you that there is an orthonormal basis of V in which T is upper-triangular. The span of the first 17 elements of this basis will be a T -invariant subspace of V of dimension Let T be a linear transformation on an inner-product space. Show that T T and T have the same null space. This problem was done in class toward the end of the semester. If T v = 0, then of course T T v = 0 as well. The problem is to prove that if T T v = 0, then already T v = 0. However, if T T v = 0, then T T v, v = 0. Using the definition of adjoint, we convert the inner product to T v, T v. Since this quantity is 0, the vector T v must be zero (in view of the definition of an inner-product space). 3. Let T : V W be a linear transformation between finite-dimensional vector spaces over a field F. Show that there is a subspace X of V such that the restriction of T to X is 1-1 and has the same range as T. If we admit the rank nullity theorem, then we can do this problem in the following fairly brain-dead way. Choose a basis v 1,..., v m for the null space of T and extent this basis to a basis v 1,..., v n of V. Let X be the span of the last n m vectors in this basis. Clearly, the range of the restriction of T to X is the same as the range of T ; indeed, the range of T is the set of all vectors T (a 1 v 1 + +a n v n ), which is the set of all vectors T (a m+1 v m+1 + +a n v n ), i.e., the range of the restriction. Since the range has dimension n m, which is the dimension of X, the restriction has to be 1-1. A more enlightened way to do this is to redo the proof of the rank nullity theorem. 4. Suppose that T is a self-adjoint operator on a finite-dimensional real vector space V and that S : V V is a linear transformation with the following property: every eigenvector of T is also an eigenvector of S. Show that there is a basis of V in which both T and S are diagonal. Conclude that S and T commute. Choose an orthonormal basis of V in which T is diagonal. The elements of this basis must be eigenvectors of T and thus will be eigenvectors of S. Accordingly, S (as well as T ) is diagonal in this basis. Since diagonal matrices commute with each other, S and T commute.

12 5. Use mathematical induction and the definition of the determinant to show for all n n complex matrices A that the determinant of the complex conjugate of A is the complex conjugate of det A. (The complex conjugate of a matrix A is the matrix whose entries are the complex conjugates of the entries of A.) Let B be the complex conjugate of A. Work by induction as instructed. For the 1 1 case, A and B have one entry each, and these entries are conjugates of each other. The determinant of a 1 1 matrix is just the single element in the matrix, so we re good. For the induction step, we assume n > 1 and that the result is known for the (n 1) (n 1) case. We have, by definition, n det B = ( 1) 1+j b 1j det B 1j. j=1 In the sum, b 1j is the complex conjugate of a 1j by the definition of B. Also, det B 1j is the complex conjugate of det Ã1j by the inductive assumption. (It s best to remark first that B 1j is the complex conjugate of the matrix Ã1j.) It follows that det B is the complex conjugate n of det A = ( 1) 1+j a 1j det Ã1j, as required. j=1 6. Let W be a subspace of V, where V is a finite-dimensional vector space. Assume that W is a proper subspace of V (i.e., that it is not all of V ). Show that there is a non-zero element of V that is 0 on each element of W. (A harder version of this problem was on the second midterm.) Take a basis v 1,..., v m of W and extend it to a basis v 1,..., v n of V. Let f 1,..., f n be the basis of V that s dual to v 1,..., v n. If f = f n, f is non-zero but it s zero on v 1,..., v m and therefore on W. 3

13 MATH 110, mock final test. Name Student ID # All the necessary work to justify an answer and all the necessary steps of a proof must be shown clearly to obtain full credit. Partial credit may be given but only for significant progress towards a solution. Show all relevant work in logical sequence and indicate all answers clearly. Cross out all work you do not wish considered. Books and notes are allowed. Calculators, computers, cell phones, pagers and other electronic devices are not allowed during the test. 1. Using reduced row echelon form, determine whether the linear system has a solution. x 1 + 2x 2 x 3 = 1 2x 1 + x 2 + 2x 3 = 3 x 1 4x 2 + 7x 3 = 4 2. Let A and B be linear maps on a vector space V such that AB = 0. Prove that rank(a) + rank(b) dim V. 3. Let A be an n n matrix such that A(i, j) = 0 for more than n 2 n pairs of values of i and j. Prove that det A = Squares are labelled 1 through 4 consecutively from left to right. A player begins by placing a marker in square 2. A die is rolled and the marker is moved one square to the left if 1 or 2 is rolled or one square to the right if 3, 4, 5 or 6 is rolled. This process continues until the marker ends in square 1 (winning the game) or in square 4 (losing the game). What is the probability of winning? 5. Find the general solution to the following system of differential equations: x 1 = 8x x 2 x 2 = 5x 1 7x 2 6. Let V be the vector space of polynomials in x and y of (total) degree at most 2, and let 1

14 T : V V be a linear map defined by (T f)(x, y) = x f(x, y) + f(x, y). y Find the Jordan canonical form of T, the corresponding Jordan basis, and the minimal polynomial of T. 7. Let A : V V be a linear map on an n-dimensional vector space V. Prove that the set of all linear maps B on V satisfying the condition AB = 0 is a subspace of the space of all linear maps on V. Can every subspace of the space of all linear transformations on V be obtained in that manner, by the choice of a suitable A? What is the dimension of that subspace when A is a Jordan block of order n with eigenvalue 0? 8. Using the Gram-Schmidt procedure, find an orthonormal basis for the real vector space span{sin t, cos t, 1} equipped with the inner produt f, g = π 0 f(t)g(t)dt. 9. Let T be an inner product space, and let y, z V. Define T : V V by T x := x, y z. Prove that T is a linear map and find an explicit expression for T. 10. Let V be the inner product space of complex-valued continuous functions on [0, 1] with the inner product f, g = 1 0 f(t)g(t)dt. Let h V, and define T : V V by T f := hf. Prove that T is a unitary operator if and only if h(t) = 1 for all t [0, 1]. 2

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