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a2 C be bases for f1; f2; f3g D f d3, d2 C a vector space V , and suppose f1 D 2d1 2d3. 3d1 C f2 D a. Find the change-of-coordinates matrix from F to D. b. Find x D f1 be bases for b1; b2g In Exercises 7–10, let B R2. In each exercise, find the change-of-coordinates matrix from B to C and the change-of-coordinates matri... |
4/; cos.2=4/; cos.3=4/; : : : g yk f g D f : : : ; 1; :7; 0; k D 0 :7; 1; :7; 0; :7; 1; :7; 0; : : : g Table 1 shows a calculation of the output sequence for .p2=4/.p2=2/ :25. The output is ·k f , where :35.:7/ is an abbreviation g yk f , shifted by one term. g D SECOND REVISED PAGES k2–2–2–4 4.8 Applications to Differ... |
ssume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems. 7. 1k; 2k; . 8. 2k; 4k; . 9. 1k; 3k cos k yk yk 2/k yk I I C 3 5/k yk 3 2 ; 3k sin k 2 I 1/k; 5k yk C C C yk 4yk C 22yk ... |
is given by x0 in (1) above. What is the distribution of the population in 2015? In 2016? FIGURE 1 Annual percentage migration between city and suburbs. SOLUTION In Example 3 of Section 1.10, we saw that after one year, the population vector 600;000 400;000 changed to :95 :05 :03 :97 600;000 400;000 582;000 418;000 D ... |
ply only P by 10. Instead, multiply the augmented matrix for equation .P 0 by 10. I /x D SECOND REVISED PAGES 262 CHAPTER 4 Vector Spaces You may have noticed that if xk 0; 1; : : : ; then P xk for k P .P x0/ D P 2x0; D x2 D D 1 C P x1 D P kx0 and, in general, D xk for k To compute a specific vector such as x3, fewer ar... |
all linear combinations of v1; : : : ; vp is a vector D f . v1; : : : ; vpg space. SECOND REVISED PAGES WEB b. c. d. e. f. If 1g 1g v1; : : : ; vp f v1; : : : ; vp f spans V , then S spans V . is linearly independent, then so is S. If If S is linearly independent, then S is a basis for V . If Span S If dim V D dependen... |
private communication from Professor Lamberson, 1993. 267 SECOND REVISED PAGES 268 CHAPTER 5 Eigenvalues and Eigenvectors dynamical system (or a discrete linear dynamical system) because it describes the changes in a system as time passes. The 18% juvenile survival rate in the Lamberson stage matrix is the entry affec... |
description of a sequence If A is an n in Rn. whose formula for each xk does not A solution of (8) is an explicit description of depend directly on A or on the preceding terms in the sequence other than the initial term x0. 0; 1; 2; : : :/ xk f xk f The simplest way to build a solution of (8) is to take an eigenvector... |
rtible Matrix Theorem (continued) Let A be an n s. The number 0 is not an eigenvalue of A. t. The determinant of A is not zero. n matrix. Then A is invertible if and only if: 3 matrix, When A is a 3 turns out to be the volume of the parallelepiped determined by the columns a1, a2, and a3 of A, as in Figure 1. (See Sect... |
alues are considered.) /.2 .n .1 I / / / D 20. Use a property of determinants to show that A and AT have the same characteristic polynomial. In Exercises 21 and 22, A and B are n statement True or False. Justify each answer. n matrices. Mark each 21. a. The determinant of A is the product of the diagonal entries in A. ... |
ng eigenvectors of A. By Theorem 5, A is not diagonalizable. The following theorem provides a sufficient condition for a matrix to be diagonalizable. T H E O R E M 6 An n n matrix with n distinct eigenvalues is diagonalizable. SECOND REVISED PAGES 5.3 Diagonalization 287 PROOF Let v1; : : : ; vn be eigenvectors correspo... |
s just a change-of-coordinates matrix (see T .x/ D Section 4.7). FIGURE 3 Linear Transformations from V into V In the common case where W is the same as V and the basis C is the same as B, the matrix M in (4) is called the matrix for T relative to B, or simply the B-matrix for T, and is denoted by T B The B-matrix for ... |
with a basis B ; W D f , and I c1; : : : ; cng be the same space as V with a basis C be the identity transformation I W . Find the matrix for I relative to B and C. What was this matrix called in Section 4.7? D f ! V W 29. Let V be a vector space with a basis B the B-matrix for the identity transformation I D f . Find ... |
5 Eigenvalues and Eigenvectors The phenomenon displayed in Example 7 persists in higher dimensions. For instance, if A is a 3 3 matrix with a complex eigenvalue, then there is a plane in R3 on which A acts as a rotation (possibly combined with scaling). Every vector in that plane is rotated into another point on the sa... |
e .8 and .64, with eigenvectors v1 1 0 D and v2 0 1 D . If x0 c1v1 C D c2v2, then xk D c1.:8/k 1 0 C c2.:64/k 0 1 Of course, xk tends to 0 because .:8/k and .:64/k both approach 0 as k . But the way ! 1 xk goes toward 0 is interesting. Figure 1 shows the first few terms of several trajectories that begin at points on th... |
(6) and (7), following Example 1, apply here. Also, it can be shown that the constant c1 in the initial decomposition of x0 is positive when the entries in x0 are nonnegative. Thus the owl population will grow slowly, with a long-term growth rate of 1.01. The eigenvector v1 describes the eventual distribution of the o... |
. [If v identically zero and hence satisfies x Ax.] Observe that 0 D Ax, a solution might be a (4) 0, the function x.t/ is D x0.t/ Ax.t/ D By calculus, since v is a constant vector vet Avet Multiplying both sides of (4) by A D Since et is never zero, x Av, that is, if and only 0.t/ will equal Ax.t/ if and only if v if i... |
aused by the sine and cosine functions that arise from a complex eigenvalue. The trajectories spiral inward because the factor e 2 is the real part of the eigenvalue in Example 3. When A has a complex eigenvalue with positive real part, the trajectories spiral outward. If the real part of the eigenvalue is zero, the tr... |
Axk k 0 1 5 2 1 :4 1 :225 1 :2035 1 :2005 1 :20007 8 1:8 7:125 1:450 7:0175 1:4070 7:0025 1:4010 7:00036 1:40014 5 8 7.125 7.0175 7.0025 7.00036 The evidence from Table 2 strongly suggests that k g f approaches 7. If so, then .1; :2/ is an eigenvector and 7 is the dominant eigenvalue. This is easily verified by computi... |
h eigenvector of an invertible matrix A is also an 1. eigenvector of A g. Eigenvalues must be nonzero scalars. h. Eigenvectors must be nonzero vectors. i. Two eigenvectors corresponding to the same eigenvalue are always linearly dependent. If A is an n n diagonalizable matrix, then each vector in Rn can be written as a... |
five sections of this chapter. In order to find an approximate solution to an inconsistent system of equations that has no actual solution, a well-defined notion of nearness is needed. Section 6.1 introduces the concepts of distance and orthogonality in a vector space. Sections 6.2 and 6.3 show how orthogonality can be us... |
A. That is, an arbitrary element x in Nul A is shown to be in (Row A) ? , and then an arbitrary element x in (Row A) (Row A) is established by showing that Nul A is a subset of (Row A) is shown to be in Nul A Let A be an m n matrix. The orthogonal complement of the row space of A is the null space of A, and the orthogo... |
necessary to solve a system of linear equations in order to find the weights, as in Chapter 1. We turn next to a construction that will become a key step in many calculations involving orthogonality, and it will lead to a geometric interpretation of Theorem 5. SECOND REVISED PAGES 342 CHAPTER 6 Orthogonality and Least S... |
rough u and the origin. In Exercises 17–22, determine which sets of vectors are orthonormal. If a set is only orthogonal, normalize the vectors to produce an orthonormal set. 17. 2 4 1=3 1=3 1=3 3 5 , 2 4 1=2 0 1=2 3 5 19. :6 :8 , :8 :6 18. 20=3 1=3 2=3 3 , 5 2 4 1=3 2=3 0 3 5 21. 22. 2 4 2 4 1=p10 3=p20 3=p20 3 5 , 2 ... |
rojections of y onto one-dimensional subprojection O spaces that are orthogonal to each other. The vector y in Figure 3 corresponds to the O y that is in W . vector y in Figure 4 of Section 6.2, because now it is O and y O y O 1 2 FIGURE 3 The orthogonal projection of y is the sum of its projections onto one-dimensiona... |
Since dim W v1; v2 Then f is a basis for W . v1; v2 g f g 2, the set D The next example fully illustrates the Gram–Schmidt process. Study it carefully. 2 2 3 EXAMPLE 2 Let x1 0 1 6 1 6 4 1 clearly linearly independent and thus is a basis for a subspace W of R4. Construct an orthogonal basis for W . x1; x2; x3 f , and ... |
Suppose A QR.] QR, where R is an invertible matrix. Show that A and Q have the same column space. [Hint: Given y in Col A, show that y Qx for some x. Also, given y in Col Q, show that y Ax for some x.] D D D D D QR as in Theorem 12, describe how to find an m (square) matrix Q1 and an invertible n n D orthogonal m upper... |
(2, 0, 11)b084(0, 2, 1)(4, 0, 1)Ax ˆx1x2x3Col A 6.5 Least-Squares Problems 367 matrix A with linearly independent columns, let A Given an m QR be a QR factorization of A as in Theorem 12. Then, for each b in Rm, the equation Ax b has a unique least-squares solution, given by D D R 1QT b x O D (6) PROOF Let x O D R 1QT... |
problem, Ax y1 y2 ::: yn 0 1 1 1 ::: 1) D D D D 2 3 The square of the distance between the vectors X and y is precisely the sum of the squares of the residuals. The that minimizes this sum also minimizes the distance between X and y. Computing the least-squares solution of X y is equivalent to finding the that determine... |
; .xn; yn/ must pass through .x; y/. That is, show that x and y satisfy the linear equation y D O0 C O1x. [Hint: Derive this equation from the vector equation y . Denote the first C column of X by 1. Use the fact that the residual vector is orthogonal to the column space of X and hence is orthogonal to 1.] O D X Given d... |
approximation in V to p by polynomials in P2 is D p O proj P2 p p; p0 i h p0; p0 i h 8 31 14 p2 5 p0 C This polynomial is the closest to p of all polynomials in P2, when the distance between polynomials is measured only at p; p1 i h p1; p1 i h 14 .t 2 31 8 5 p; p2 i h p2; p2 h i 2/: 1, 0, 1, and 2. See Figure 1. p0 p1... |
h 0; v 0v; v v; v 0 h , by Axiom 3, so ; v h C w w; u i C h w; u i D 6.8 APPLICATIONS OF INNER PRODUCT SPACES The examples in this section suggest how the inner product spaces defined in Section 6.7 arise in practical problems. The first example is connected with the massive leastsquares problem of updating the North Ame... |
. Thus i D (7) The coefficient of the (constant) function 1 in the orthogonal projection is 2 Z 1 2 f; 1 i h 1; 1 i h where a0 is defined by (7) for k as a0=2. D 0 f .t/ 1 dt D 1 2 1 Z 0 2 f .t/ cos.0 t/ dt a0 2 D 0. This explains why the constant term in (4) is written D SECOND REVISED PAGES 390 CHAPTER 6 Orthogonality... |
becomes b V , which may also be written as b. Find the least-squares solution of this system of V n equations in the one unknown , and write this solution using the original symbols. The resulting estimate for is called a Rayleigh quotient. See Exercises 11 and 12 in Section 5.8. D D 13. Use the steps below to prove th... |
gonalizable as in (1), then AT .PDPT /T P T T DT P T PDPT D D D A D Thus A is symmetric! Theorem 2 below shows that, conversely, every symmetric matrix is orthogonally diagonalizable. The proof is much harder and is omitted; the main idea for a proof will be given after Theorem 3. T H E O R E M 2 An n matrix. n matrix ... |
encountered in Chapter 6 when computing xTx. Such sums and more general expressions, called quadratic forms, occur frequently in applications of linear algebra to engineering (in design criteria and optimization) and signal processing (as output noise power). They also arise, for example, in physics (as potential and k... |
n is one approach.) Such a Cholesky factorization is possible if and only if A is positive definite. See Supplementary Exercise 7 at the end of Chapter 7. PRACTICE PROBLEM Describe a positive semidefinite matrix A in terms of its eigenvalues. 7.2 EXERCISES 1. Compute the quadratic form xTAx, when A and a. x x1 x2 D b. x ... |
the (eigenvector) columns of P so that P 3 matrix with eigenvalues a u1 u2 u3 and b c Given any unit vector y in R3 with coordinates y1, y2, y3, observe that ay2 by2 cy2 1 D 2 3 ay2 1 ay2 2 ay2 3 and obtain these inequalities: D yTDy by2 2 C ay2 2 C y2 2 C a a, by definition of M . However, yTDy a. By (3), the x that co... |
The new quadratic form is yTDy 4y2 . 2. The maximum of Q.x/, for a unit vector x, is 4 and the maximum is attained at . This vector the unit eigenvector 1=p2 1=p2 maximizes the quadratic form yTDy instead of Q.x/.] . [A common incorrect answer is 1 0 D 1 C 2y2 2 The maximum value of Q.x/ subject to xT x 1 is 4. D 7.4 T... |
osition 421 The square roots of the eigenvalues are the singular values: 1 6p10; 2 3p10; 3 0 D The nonzero singular values are the diagonal entries of D. The matrix is the same size as A, with D in its upper left corner and with 0’s elsewhere. D D " 6p10 0 D D # ; 0 3p10 D 0 D " 6p10 0 D 0 3p10 # 0 0 Step 3. Construct ... |
with r positive entries n orthogonal matrix. 17. Show that if A is square, then singular values of A. det A j j is the product of the 18. Suppose A is square and invertible. Find a singular value decomposition of A 1. 15. Suppose the factorization below is an SVD of a matrix A, with the entries in U and V rounded to t... |
the second principal component is the eigenvector corresponding to the second largest eigenvalue, and so on. The first principal component u1 determines the new variable y1 in the following is the first row of P T , the equation way. Let c1; : : : ; cp be the entries in u1. Since uT 1 Y P T X shows that D D y1 X uT 1 c1... |
e written in a form similar to (1). [Hint: Use partitioned matrix multiplication to write S as 1=.N 1/ times the sum of N matrices of size p M in place of write Xk Xk.] O p. For 1 k N , SOLUTIONS TO PRACTICE PROBLEMS 1. First arrange the data in mean-deviation form. The sample mean vector is easily . Subtract M from th... |
instance, Sections 8.2 and 8.6 include applications to computer graphics, and Section 8.5 outlines a proof (in Exercise 22) that there are only five regular polyhedra in R3. 437 SECOND REVISED PAGES 438 CHAPTER 8 The Geometry of Vector Spaces FIGURE 1 The five Platonic solids. Most applications in earlier chapters involv... |
u1 Let y s2 in S (by Theorem 2). But then .1 This shows that W is a subspace of Rn. Thus S is a flat, because S C Rn and some subspace W . To show that S is affine, it suffices to show that for any pair s1 and s2 of points in S , the line through s1 and s2 lies in S. By definition of W , there exist u1 and u2 u2 in W such ... |
e) coefficient must be nonzero. , the first equation in (1) is just c1 g Exercise 13 asks you to show that an indexed set D is affinely dependent if and only if v1 v2. The following theorem handles the general case and shows how the concept of affine dependence is analogous to that of linear dependence. Parts (c) and (d) g... |
on of the color data. The RGB values for p are :25v3. Use the barycentric coordinates of p to make a linear :25v1 :5v2 C D C 2 :25 4 1 0 1 3 5 C :50 2 4 1 :4 1 3 5 C :25 2 4 :6 0 1 3 2 5 D 4 :9 :2 1 3 5 red green blue One of the last steps in preparing a graphics scene for display on a computer screen is to remove “hid... |
From Example 1, the problem is to determine if the points are affinely dependent. Use the method of Example 2 and subtract one point from the other two. If one of these two new points is a multiple of the other, the original three points lie on a line. 2. The proof of Theorem 5 essentially points out that an affine depe... |
echnique of Section 8.2 to obtain an affine dependence relation 5v1 4v2 3v3 4v4 0 (3) C Next, choose the points v2 and v4 in (3), whose coefficients are positive. For each point, compute the ratio of the coefficients in equations (2) and (3). The ratio for v2 is 1 . The ratio for v4 is smaller, so subtract 1 1 48 48 times... |
e 3x v. EXAMPLE 3 Let n f : 12, where f .x; y/ D description of the parallel hyperplane (line) H1 SOLUTION First, find a point p in H1. To do this, find a point in H and add v to it. For instance, 0 is in H1. Now, compute 3 9. See Figure 2, which also shows the sub- , so H 12 g D 12. Find an implicit 1 6 D f : is in H , ... |
hat some other hyperplane not parallel to H would strictly separate them. \ ¿. D PRACTICE PROBLEM Let , n1 2 D 4 3 5 , and n2 2 D 4 2 1 3 3 5 ; let H1 be the hyper- 1 1 2 plane (plane) in R3 passing through the point p 1 H2 be the hyperplane passing through the point p an explicit description of H1 in H1 and having nor... |
can lie on H , so H P \ (b) that H p g exist points x and y in P such that p p D f (c) Let p be a vertex of P . Then there exists a hyperplane H ) P \ f : d such d . If p were not an extreme point, then there would cy with 0 < c < 1. That is, and p is a vertex of P . and f .P / c/x D f .1 D g cy p .1 D c/x and D y C 1 ... |
angle S 2, and the other three come from stretching the edges of S 2 out to the new point v4. Notice too that the vertices of S 2 get stretched out into edges in S 3. The other edges in S 3 come from the edges in S 2. This suggests how to “visualize” the fourdimensional S 4. The construction of S 4, called a pentatope,... |
11 < 12, so pbc is in P . The intersection of (a) and (c) is at 9:6 > 9. So pac is not in P . D Finally, the five vertices (extreme points) of the polytope are .0; 0/, .6; 0/, .5; 2/ .3; 3/, and .0; 4/. These points form the minimal representation of P . This is displayed graphically in Figure 13. D .4:8; 2:4/. Testing... |
ting and ending points of the curves and the tangent vectors to the curves at those points.1 The Bézier curve in equation (4) can also be “factored” in another way, to be used in the discussion of Bézier surfaces. For convenience later, the parameter t is replaced 1 The term basis matrix comes from the rows of the matr... |
control points. Adaptive recursion is possible in this setting, too, but there are some subtleties involved.2 2 See Foley, van Dam, Feiner, and Hughes, Computer Graphics—Principles and Practice, 2nd Ed. (Boston: Addison-Wesley, 1996), pp. 527–528. SECOND REVISED PAGES q0 = p0q3 = r0p3 = r3p1p2r1r2q1q2(p1 + p2)12 492 C... |
esults of Exercises 13 and 14, write an algorithm that computes the control points for both y.t/ and z.t/ in an efficient manner. The only operations needed are sums and division by 2. 16. Explain why a cubic Bézier curve is completely determined 0.1/. 0.0/, x.1/, and x by x.0/, SOLUTIONS TO PRACTICE PROBLEMS 1. From eq... |
nd c1v1 C C affinely independent set: A set affine set (or affine subset): A set S of points such that if p and S for each real number t . Rm of the form affine transformation: A mapping T q are in S, then .1 t/p Rn t q C 2 T .x/ Ax b, with A an m W n matrix and b in Rm. ! algebraic multiplicity: The multiplicity of an eig... |
he form xk C tion is a sequence of vectors, x0; x1; : : : : 1 D D dilation: A mapping x dimension: 7! rx for some scalar r, with 1 < r. of a flat S : The dimension of the corresponding parallel subspace. of a set S: The dimension of the smallest flat containing S. of a subspace S: The number of vectors in a basis for S, ... |
ost nonzero entry in a row of a matrix. x from b to A x, least-squares error: The distance O O k b. D D O0 C O1x that minimizes the C A x is a least-squares solution of Ax O least-squares line: The line y least-squares error in the equation Glossary A11 x such that O least-squares solution (of Ax b): A vector b k A x O... |
osition in a matrix A that corresponds to a leading entry in an echelon form of A. plane through u, v, and the origin: A set whose parametric t v (s, t in R/, with u and v linearly su equation is x independent. D C polar decomposition (of A): A factorization A PQ, where n positive semidefinite matrix with the same rank ... |
s spiral about 0. stage-matrix model: A difference equation xk Axk where xk lists the number of females in a population at time k, with the females classified by various stages of development (such as juvenile, subadult, and adult). 1 D C standard basis: The basis E of the columns of the n 1; t; : : : ; t n for Pn. f g ... |
of entries (2) in the 3 (3) in the vector. Usually the intermediate steps are not displayed. 7. a u 2v, b 2u 2v, c D 2u 3:5v, d 3u 4v D 9. x1 x2 4 1 6 3 3 5 C 2 4 x3 11. Yes, b is a linear combination of a1, a2, and a3. 13. No, b is not a linear combination of the columns of A. 15. Noninteger weights are acceptable, o... |
.y/ d C mx b C D b. When f .x/ b, with b nonzero, b D m.0/ c. ⁄ D C 0. f .0/ In calculus, f is called a “linear function” because the graph of f is a line. v1; v2; v3g f a certain equation and work with it. is linearly dependent, you can write 31. Hint: Since 33. One possibility is to show that T does not map the zero... |
The abbreviation IMT (here and in the Study Guide) denotes the Invertible Matrix Theorem (Theorem 8). 1. Invertible, by the IMT. Neither column of the matrix is a multiple of the other column, so they are linearly independent. Also, the matrix is invertible by Theorem 4 in Section 2.2 because the determinant is nonzero... |
p 1 linear transformation A, and then translate by p. sin2 ’ cos ’ D 0 1 A 0T I 0T cos ’ p 1 D D 13. A 0T . First apply the 2 6; 3/ 15. .12; 0 1 0 1=2 0 p3=2 0 19. The triangle with vertices at .7; 2; 0/, .7:5; 5; 0/, .5; 5; 0/ 0 p3=2 1= 17. 3 0 21. [M] 2 4 2:2586 1:3495 :0910 1:0395 2:3441 :3046 :3473 :0696 1:2777 Se... |
9. a. 41. 12 det A 375 e/d b. .a .ad C 43. Hint: Compute det A by a cofactor expansion down D f c/ C .ed D D bc/ D C ad ed C det B bc det C C .b f /c column 3. 45. [M] See the Study Guide after you have made a conjecture about ATA and AAT . Section 3.3, page 186 1. 5=6 1=6 3. 4=5 3=10 5. 2 4 1=4 11=4 3=8 3 5 7. s ⁄ 9. ... |
entation as a linear combination of elements of S . 21. 9 4 2 1 23. Hint: Suppose that uB denote the entries in uB by c1; : : : ; cn. Use the definition of uB. D wB for some u and w in V , and 25. One possible approach: First, show that if u1; : : : ; up are linearly dependent, then u1B; : : : ; upB are linearly depende... |
2816 :3355 :1819 :2009 :2816 :3355 :1819 :2009 :2816 :3355 :1819 :2009 :2816 :3355 :1819 :2009 2816 :3355 :1819 :2009 3 7 7 5 Note that, due to round-off, the column sums are not 1. b. To four decimal places, Q80 2 D 4 :7354 :0881 :1764 :7348 :0887 :1766 :7351 :0884 :1765 3 5 ; Q116 Q117 D 2 D 4 :7353 :0882 :1765 :7353... |
2 C 1 c2 C 1 C i e. i/t 2 (real solution): repulsion: the line through .0; 0/ and . greatest attraction: the line through .0; 0/ and .1; 4/ 11. Attractor; eigenvalues: .9, .8; greatest attraction: line 1; 1/; direction of through .0; 0/ and .5; 4/ 13. Repellor; eigenvalues: 1.2, 1.1; greatest repulsion: line through .0... |
date works. u u y u u. Replace u by cu with c y O D 31. Suppose 0; then ⁄ y .cu/ .cu/ .cu/ .cu/ D u/ u .c/u c.y c2u y D O SECOND REVISED PAGES 33. Let L T .x/ D D Span proj u f x. By definition, , where u is nonzero, and let g T .x/ D .x D u/.u u/ 1u L u u u x u For x and y in Rn and any scalars c and d , properties of ... |
ion of A, or appeal to Theorem 2. 31. The Diagonalization Theorem in Section 5.3 says that the columns of P are (linearly independent) eigenvectors corresponding to the eigenvalues of A listed on the diagonal of D. So P has exactly k columns of eigenvectors corresponding to . These k columns form a basis for the eigens... |
ints of A. That is, aff A B. 23. Since A aff A aff A .A [ aff .A [ aff B 25. To show that D [ B/, it follows from Exercise 22 that B/. Similarly, aff B B/. aff .A aff .A [ B/, so E and D The complete proof is presented in the Study Guide. E \ F . [ F , show that D 0 v2 Section 8.2, page 454 1. Affinely dependent and 2v1... |
arbitrarily. 3 p p 5 2 9. Write a vector of the polynomial weights for x.t/, expand the polynomial weights, and factor the vector as MB u.t/: 2 6 6 6 6 4 4t 1 4t C 12t 2 6t 2 4t 3 12t 3 t 4 C 4t 4 6t 2 C 12t 3 6t 4 4t 3 C 4t 12 6 0 0 6 12 6 0 0 4 12 12 4 0 4 12 12 MB D , 1 11. See the Study Guide. 13. a. Hint: Use the ... |
equation, 314–315 Eigenspace, 270–271, 399 CONFIRMING PAGES Eigenvalue, 269 characteristic equation of a square matrix, 276 characteristic polynomial, 279 determinants, 276–278 finding, 278 Evolution, dynamical system, 303 Existence linear transformation, 73 matrix equation solutions, 37–38 matrix transformation, 65 sys... |
ranspose, 101–102 Matrix equation, 2 b, 35–36 D Ax computation of Ax, 38, 40 existence of solutions, 37–38 properties of Ax, 39–40 Matrix factorization, 94, 125–126 LU factorization algorithm, 127–129 overview, 126–127 permuted LU factorization, 129 Matrix of a linear transformation, 71–73 Matrix multiplication, 96–99 ... |
actorization, 132 Spectral Theorem, 399 Spiral point, dynamical system, 319 Spline, 492 B-spline, 486–487, 492–493 natural cubic, 483 Index I7 Standard matrix, 290 Standard matrix of a linear transformation, 72 Standard position, 406 State-space design, 303 Steady-state response, 303 Steady-state vector, 259–262 Stiffn... |
x, 133 Vandermonde matrix, 162, 188, 329 111, 127, 235, 240 361, 101 129 129, 174 Vector pipeline architecture, 122 23, 162 Physical Sciences Cantilevered beam, 254 Center of gravity, 34 Chemical reactions, 52, 55 Crystal lattice, 220, 226 Decomposing a force, 344 Digitally recorded sound, 247 Gaussian elimination, 12 ... |
t change. For the total sum to be 0, X Y + Z X = −λX, Y = −µY, Z = (λ + µ)Z. The signs before λ and µ are there just to make our life easier later on. We can solve these to obtain X = a sin λx + b cos λx, √ √ √ √ λy + d cos Y = c sin Z = g exp(λ + µz) + h exp(−λ + µz). λx, We now impose the homogeneous boundary conditi... |
pole, and the second term is known as the dipole, in analogy to charges in electromagnetism. The monopole term is what we get due to a single charge, and the second term is the result of the interaction of two charges. 29 4 Partial differential equations IB Methods 4.6 Laplace’s equation in cylindrical coordinates We le... |
1 − φ2 and E(t) = 1 2 Ω Φ2 dV. Then we know that E(t) ≥ 0. Since φ1, φ2 both obey the heat equation, so does Φ. Also, on the boundary and at t = 0, we know that Φ = 0. We have dE dt = Ω = κ Φ dΦ dt dV Φ∇2Φ dV Ω = κ ∂Ω = −κ Φ∇Φ · dS − κ (∇Φ)2 dV Ω ≤ 0. Ω (∇Φ)2 dV So we know that E decreases with time but is always non-n... |
hen since ψ obeys the wave equation with fixed boundary conditions, we know Eψ is constant. Initially, at t = 0, we know that ψ = ∂ψ ∂t = 0. So Eψ(0) = 0. At time t, we have Eψ = 2 1 2 Ω ∂ψ ∂t + c2(∇ψ) · (∇ψ) dV = 0. Hence we must have ∂ψ is always 0. ∂t = 0. So ψ is constant. Since it is 0 at the beginning, it 42 4 Par... |
these two regions. 48 5 Distributions IB Methods Suppose {y1(x), y2(x)} are a basis of solutions to LG = 0 everywhere on [a, b], with boundary conditions y1(a) = 0, y2(b) = 0. Then we must have G(x, ξ) = A(ξ)y1(x) B(ξ)y2(x So we have a whole family of solutions. To fix the coefficients, we must decide how to join these s... |
. We first took the limit as L → ∞ to turn the integral from L −∞. Then we take the limit as ∆k → 0. However, we can’t really do this, since ∆k is defined to be 2π/L, and both limits have to be taken at the same time. So we should really just take this as a heuristic argument for why this works, instead of a formal proof... |
rform the integral. In real life, f might be some radio signal, and we are just given some data of what value f takes at different points.. There is no hope that we can do the integral exactly. Hence, we first give ourselves a simplifying assumption. We suppose that f is mostly concentrated in some finite region [−R, S]. ... |
to our family of curves, and we can label the members of our family by the value of t at which they intersect B. B(t) C4 C3 C2 C1 We can thus label our family of curves by (x(s, t), y(s, t)). If the Jacobian J = ∂x ∂s ∂y ∂t − ∂x ∂t ∂y ∂s = 0, then we can invert this to find (s, t) as a function of (x, y), i.e. at any p... |
tions IB Methods This is not too helpful in general. However, in the case where we have two dimensions, we can find the characteristic curves explicitly. Suppose our curve is given by f (x, y) = 0. We can write y = y(x). Then since f is constant along each characteristic, by the chain rule, we know Hence, we can compute... |
+ c(t − τ )) does not contribute. On the other hand, δ(|x − y| − c(t − τ )) is non-zero only if t > τ . So Θ(t − τ ) is always positive in this region. So we can write our Green’s function as G(x, t; y, τ ) = δ(|x − y| − c(t − τ )). 1 4πc 1 |x − y| As always, given our Green’s function, the general solution to the forc... |
(x), ∂t(x, 0) = 0, ∂xφ(0, t) = 0. On R1,1 d’Alembert’s solution gives φ(x, t) = 1 2 [b(x − ct) + b(x + ct)] This is not what we want, since eventually we will have a wave moving past the x = 0 line. x = 0 To compensate for this, we introduce a mirror wave moving in the opposite direction, such that when as they pass th... |
Recall that if Y ⊆ X, there is an inclusion function ι : Y → X that sends y → y. We can use this to obtain the following defining property of a subspace. Proposition. If Y has the subspace topology, f : Z → Y is continuous iff ι ◦ f : Z → X is continuous. Proof. (⇒) If U ⊆ X is open, then ι−1(U ) = Y ∩ U is open in Y . ... |
e ous. Then let g(x) = f (x) − 1 intermediate value property. 3.2 Path connectivity The other notion of connectivity is path connectivity. A space is path connected if we can join any two points with a path. First, we need a definition of a path. Definition (Path). Let X be a topological space, and x0, x1 ∈ X. Then a pat... |
l spaces, these notions can be different. The first is just known as “compactness” and the second is known as “sequential compactness”. The actual definition of compactness is rather weird and unintuitive, since it is an idea we haven’t seen before. To quote Qiaochu Yuan’s math.stackexchange answer (http://math.stackexcha... |
s a finite subcover {Vx1, · · · , Vxm }. Then V = m i=1 Vxi is a finite subset of V, which covers all of X × Y . In the general, case, suppose V is an open cover of X × Y . For each (x, y) ∈ X × Y , ∃Uxy ∈ V with (x, y) ∈ Uxy. Since Uxy is open, ∃Vxy ⊆ X, Wxy ⊆ Y open with Vxy × Wxy ⊆ Uxy and x ∈ Vxy, y ∈ Wxy. Then Q = {... |
∈ D. Since z ∈ R, we know Z + Zz = {cz + d : c, d ∈ Z} is a discrete subgroup of C. So we know {|cz + d| : c, d ∈ Z} is a discrete subset of R, and is in particular bounded away from 0. Thus, we know Im(z) |cz + d|2 : γ = is a discrete subset of R>0 and is bounded above. Thus there is some γ ∈ ¯Γ∗ with Im γ(z) maximal.... |
convenient to introduce a normalized Eisenstein series Definition (Normalized Eisenstein series). We define Ek(z) = (2ζ(k))−1Gk(z) = 1 − 2k Bk n≥1 σk−1(n)qn = 1 2 (m,n)=1 m,n∈Z 1 (mz + n)k . The last line follows by taking out any common factor of m, n in the series defining Gk. Thus, to figure out the (normalized) Eisenst... |
tructure theorem, but doesn’t really explain why the congruence is true. The function τ (n) was studied extensively by Ramanujan. He proved the 691 congruence (and many others), and (experimentally) observed that if (m, n) = 1, then Also, he observed that for any prime p, we have τ (mn) = τ (m)τ (n). |τ (p)| < 2p11/2, ... |
, the fact that we can row and column reduce. Corollary. The set Γ r1 0 Γ 0 r2 : r1, r2 ∈ Q>0, ∈ Z r1 r2 is a basis for H(G, Γ) over Z. So we have found a basis. The next goal is to find a generating set. To do so, we define the following matrices: For 1 ≤ n2 | n1, we define T (n1, n2) = Γ n1 0 Γ 0 n2 For n ≥ 1, we write ... |
If f ∈ Sk then, L(f, s) is entire, i.e. has an analytic continuation to all of C. Define Λ(f, s) = (2π)−sΓ(s)L(f, s) = M (f (iy), s). Then we have Λ(f, s) = (−1)k/2Λ(f, k − s). The proof follows from the following more general fact: Theorem. Suppose we have a function 0 = f (z) = anqn, n≥1 58 7 L-functions of eigenforms... |
mple. Take Γ0(p) = a b d c ∈ SL2(Z) : b ≡ 0 (mod p) Recall there is a canonical map SL2(Z) → SL2(Fp) that is surjective. Then Γ0(p) is defined to be the inverse image of H = a 0 c b ≤ SL2(Fp). So we know (Γ(1) : Γ0(p)) = (SL2(Fq) : H) = |SL2(Fq)| |H| = p + 1, where the last equality follows from counting. In fact, we ha... |
k 2 −1 δγj, g f | k j 2 −1 k = p = p f | k δγj, g | k γj k j 2 −1(p + 1)f | k δ, g, using the fact that g | k γj = g. 8.3 Examples of modular forms We now look at some examples of (non-trivial) modular forms for different subgroups. And the end we will completely characterize M2(Γ0(4)). This seems like a rather peculia... |
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