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is a harmonic F* in D, f is holomorphic 0 func U into N{O, 1), find the appropriate function in harmonic back into difficulties than U. This is a strategy rather in the way. As already mentioned, f, and the feasibility the of evaluating F. function of the boundary functions, we finish s . . value with holomorphic func... |
- z the if if and only Thus, as ()goes from 0 to 21r, the image point ioo. The image in the w-plane traverses of the semicircle from 1r /2 to 1 -2i) is the line segment frpm 1 + 3i if the line from Q to P, we note that Re z = 1 from 1 + ioo to 1 line Re z = 1 311" /2 {from the P?int 1 + 2i to the point to 1 -3i. To fin... |
0 x•-1e-"' dx (s > 0), s to be complex 0 {12.2) and here too we Re s > whenever can allow 0. It is easily and regard the function as defined that proved r(s) = (s-1)r(s-1) (Re s> 1), equation {12.3) to define r( s) for Res < 0: backwards and we can use this functional {0, 1), then if Re(s + n) E r(s + n) r (s)= s(s+1) ... |
)1r COS = 2n COS (n � 1r) j =- cosO) Chapter 2 2.1. a) M(a, b)M(c, b) This . is routine c) M(O, 1)M(O, 1) = M(O.O-1.1, d) d) = M(ac-bd, ad+ be). 0.1-1.0) = M( -1, 0). M(a, b)= ( �b ! ) = (: �) + b ( �1 �) =a+ bi. 2.2. X = 1 ± A 2.3. Let z = x + iy. Then iz = ix -y and so Re(iz) = -y = -Im z, Im(iz) = = 1 ± 2i. x = Rez.... |
4 . 1 5. As a multifu -+ nction, zi = ei Log z = { ei(log l zl+i arg z+2mri) {eiloglz le-argz-2mr: n E Z}. For z = -i we have loglzl = 0 and arg z = /2. So ( -i)i = { eC"/2)-2mr : n E Z}. E Z} = : n -7!' 4 . 16 . w E Sin-1 z {=} (eiw-e-iw) = z {=} e2iw-2izeiw -1 = 0 {=} eiw = iz ± Vf=Z2 {=} w E -iLog(iz ± Vf=Z2") . th... |
-1)n � (2n)! oo ( -1)n � (2 + 1)! n 7.11. a) From (7.7) , an= ( 1/27ri) .. (o,r)[/(z)/zn+l] dz; hence J lanl :::; (1/211") . 211"r . M(r)/rn+l = M(r)/rn. ( 13 4) . b) Since 1 /(z)l :::; M for some M, lanl :::; we see that an = 0 for all M/rn for all 2: 1. Thus f is constant. r c) From (13.4) we have laNI :::; KrN-n for... |
2-b2)/b are the roots bz2 + 2az + b = 0. Note that a/3 = 1. Since J/31 = (a+ of the equation ../a2-b2)jb > ajb > 1, it follows that Jaj < 1, and so the only relevant poles are a simple pole at 0. The residue pole at a and a double at a is (a2 -1)2 / [a2(a-/3)]. Now, (a2 -1)/a2 = [a -(1/aW = (a-/3)2, and Near 0 the so i... |
an-1[(7R -3R2)/(R8 + (7R -3R2)/(R8 + 5) , is 0 when throughout stays Hence to 0 as R -+ finite Ll-y3 (arg f) -+ 0. As for 72, by we may use Rouche's Theorem to deduce in the are two roots Ll-y2 ( arg f) = Ll-y2 ( arg z8) = 47r. Hence there by observing are that there r8 + that the roots of f(z) = 0 on the line {rei""/4... |
hematics web site [ 1 2] Heinz- www-history. mc s.s t - and . ac .uk/history/ Hartmut Otto Peitgen, of Science , Advanced Calculus, Fractals, New Frontiers R. Spiegel 13] Murray [ Hill, 1963. Jurgens and Dietmar , Springer- Verlag, Schaum's Outline New York, 1992. Series, McGraw Saupe, Chaos and 256 Complex A n alysis... |
anics, functional analy sis, disordered systems, solid state physics) Consider the linear map Lu(n) = ]C|m-n|=i u(n) + V(n)u(n) on the space of sequences u = (..., u-2, "U-i, uo, ui, u2,...). We assume that V(n) takes random values in {0,1}. The function V is called the potential. The problem is to determine the spectr... |
{1,2,3}, the set C = {{1,2}, {2,3 }} generates the cr-algebra A which consists of all 8 subsets of ft. Definition. If (E, O) is a topological space, where O is the set of open sets in E. then a(0) is called the Borel cr-algebra of the topological space. If A C B, then A is called a subalgebra of B. A set B in B is also... |
e. On ft = {1,2,3,4 } the two cr-algebras A = {0, {1,3 }, {2,4 }, ft } and B = {0, {1,2 }, {3,4 }, ft } are independent. Properties. (1) If a cr-algebra T C A is independent to itself, then P[A fl A] = P[A] = P[A]2 so that for every A e T, P[A] e {0,1}. Such a a-algebra is called P-trivial. (2) Two sets A,B eA are indep... |
implies u(A) = 0. One says that v is absolutely continuous with respect to /i. 2.2 Kolmogorov's 0-1 law, Borel-Cantelli lemma Definition. Given a family {Ai}iei of a-subalgebras of A. For any nonempty set J C I, let Aj := \fjeJAj be the a-algebra generated by \JjeJAj. Define also A® = {0,ft}. The tail a-algebra T of {A}... |
y the meaning of these numbers. The expectation is the "average", "mean" or "expected" value of the variable and the standard deviation measures how much we can expect the variable to deviate from the mean. Example. The m'th power random variable X(x) = xm on ([0,1], B, P) has the expectation E[X}= I': / o Jo xmdx = r ... |
X] = E[(X - E[X})2} for a random variable X e C2. Remark. The Cauchy-Schwarz-inequality can be restated in the form \Cov[X,Y]\<a[X]a[Y] Definition. The regression line of two random variables X, Y is defined as y = ax + b, where Iff2 = {l,...,n}isa finite set, then the random variables X, Y define the vectors X = (X(l),...... |
. a) Given a sequence pn e [0,1] and a sequence Xn of IID random variables taking values in {-1,1} such that P[Xn = 1} = pn and P[Xn = — 1] = 1 — pn. Show that -T(Xk-mk)^0 n f e = i in probability, where mk = 2pk — 1. b) We assume the same set up like in a) but this time, the sequence pn is dependent on a parameter. Gi... |
. Therefore, for n > m also ) => b). We have seen already that Xn -> X in probability if ||Xn-X||i -> 0. We have to show that Xn -> X in C1 implies that Xn is uniformly integrable. Given e > 0. There exists m such that E[|Xn - X|] < e/2 for n > m. By the absolutely continuity property, we can choose S > 0 such that P[A... |
age < ? i n _ 1 n n ^ ^ 2=0 converges almost everywhere to a T-invariant random variable X satisfying E[X] = E[X]. Especially, if T is ergodic, then Sn/n converges to E[X]. Proof. Define X = limsup^^ Sn, X = lim inf n_+oo Sn . We get X = X(T) and X = X(T) because n on — Sn(T) = — . n (i)X = X. Define for a < f3 G R the s... |
hat one can replace integrals on the probability space Cl by integrals on the real line R which is more convenient. Remark. The distribution function Fx determines the law px because the measure u((—oo,a]) = Fx(a) on the 7r-system 1 given by the intervals {(—oo, a]} determines a unique measure on R. Of course, the dist... |
c = S = p For calculating the higher moments one can proceed as in the Poisson case or use the moment generating function. Cantor distribution: because one can realize a random variable with the Cantor distribution asX = X^Li -W3n, where the IID random variables Xn take the values 0 and 2 with probability p = 1/2 each,... |
are independent and satisfy M = sup \\Xi\\s < oo, 6 = liminf - V Var[Xi Then 5* converges in distribution to a random variable with standard normal distribution N(0,1): lim P[S; <x] = --L / e~y2/2 dy, \/.14. The central limit theorem 93 Figure. The probabil ity density function fs* of the random variable X(x) — x on [... |
be used to measure the distance between two distributions. It is not a metric although. The relative entropy is also known under the name Kullback-Leibler divergence or Kullback-Leibler metric, if v = dx [85]. Theorem 2.15.2 (Distributions with maximal entropy). The following dis tributions have maximal entropy. a) If... |
by requiring EMA[Xi] = Ci, then p\ maximizes the entropy H(p) among all measures p satisfying E^X*] = c*. Assume v = p is a probability measure. The measure p\ maximizes F(£)=JJ(£) + AEA[X]. Proof. Take the same proofs as before by replacing AX with A • X = 2.16 Markov operators Definition. Given a not necessarily finite... |
en two continuous distributions F, G with densities h and k. Then the distribution of F • G is given by the convolution h*k(x)= I h(x — y)k(y)dy J r because (F * G)'(x) = ^-JF(x- y)k(y) dy = J h(x - y)k(y) dy . Lemma 2.17.4. If F and G are distribution functions with characteristic functions </> and i/j, then F • G has... |
fc. Both inequalities hold therefore for infinitely many values of fc. For such fc, Snk(u) > Sn^ (u) + cy/2Ank log log Anfc > -2y/2nk-i loglognfc_i + cy/2Ank log log Ank > (-2/ v^ + cy/1 - l/N) ^2nk log log nk > (1 -e)y/2nk log log nk , where we have used assumption (2.10) in the last inequality. □ We know that JV(0,1)... |
C, then E[X\C] = E[X]. \ ] < Proof. (1) For positivity, note that if Y = E[X\B] would be negative on a set of positive measure, then A = Y~l([-l/n, 0]) G B would have positive probability for some n. This would lead to the contradiction 0 < E[1^X] = E[Uy] < -n-1m(A) < 0. (2) Use that P" < P' < P implies P" = Y'P' = Yf... |
is also the name of a part of a horse's harness or a belt on the back of a man's coat. Remark. If A" is a supermartingale, then -X is a submartingale and vice versa. A supermartingale, which is also a submartingale is a martingale. Since we can change X to X - X0 without destroying any of the martingale properties, we... |
dapted to a filtration An and let T be a stopping time with respect to An, define the random variable *tm={V :^)<cc or equivalently Xt — Y^=QXn^{T=n}- The process X% = Xr/\n is called the stopped process. It is equal to Xt for times T <n and equal to Xn if T > n . Proposition 3.2.5. If X is a supermartingale and T is a s... |
on and Xn = E[X|^4n] is a martingale which converges. We will see below with Levys upward theorem (3.4.2 that the limit actually is the random variable X. Example. Let Xk be IID random variables in C1. For 0 < A < 1, the branching random walk Sn = ]C/c=o ^k^k is a martingale which is bounded in C1 because ||S„||l < j-^... |
-algebras, the tail cr-algebra T = f]nBn with Bn = \Jm>n Am is trivial. Proof. Given A G T, define X = 1A G C°°(T) and the a-algebras Cn = cr(A\,... ,An)- By Levy's upward theorem (3.4.2), X = E[X|Coo]= lim E[X\Cn] . n—▶oo But since Cn is independent of An and X is Cn measurable, we have P[A]=E[X] = E[X\Cn]-+X and beca... |
< qs\ip\\Xn\\P . D Corollary 3.7.3. Given a non-negative submartingale X which is bounded in Cp. Then X^ = lim^oo Xn exists in Cp and ||Xoo||P = supn ||Xn||p . Proof. The submartingale X is dominated by the element X* in the Cpinequality. The supermartingale -X is bounded in Cp and so bounded in C1. We know therefore ... |
e the golden ratio (\/5 + l)/2 or the silver ratio y/2 + 1 one has for infinitely many n the relation a ■ log(n) + 0.78 < S„(0) < a • log(n) + 1 with a = 1/(2 log(l + \/2)). It is not known whether Sn{0) grows like log(n) for almost all a. Figure. An almost periodic ran dom walk in one dimensions. In stead of flipping co... |
that the function m(x) is transcendental. 3.11 The free Laplacian on a discrete group Definition. Let G be a countable discrete group and A C G a finite set which generates G. The Cayley graph V of (G, A) is the graph with edges G and sites (ij) satisfying i - j G A or j - i G A. Remark. We write the composition in G ad... |
e, where A = fxx + fyy: A = # * " + * + * ; . The discrete Laplacian in Z3 is defined in the same way as a discretisation of A = fxx + fyy + fzz. The setup is analogue in higher dimensions 1 d (Au)(n) = — ^2(u(n + a) + u(n - a) - 2u(n)) , where ei,..., e<* is the standard basis in Zd. Definition. A bounded region D in Zd... |
th transition probability P, then pn=P[Xn = j}. Example. Sum of independent 5-valued random variables Let S be a count able Abelian group and let tt be a probability distribution on S assigning to each j G S the weight ttj. Define P{j — ttj-i. Now Xn is the sum of n independent random variables with law tt. The sum cha... |
i,t2, • • • G T, this process has independent increments Bti - Bu_x and is a Gaussian process. For each t, we have E[B2] = t and for s <t, the increment Bt - Bs has variance t — s so that E[B,flt] = E[PS2] + E[Ps(Pt - Ba)] = E[P2] = s . This model of Brownian motion has everything except continuity. Theorem 4.1.4 (Kolmo... |
duct. Multiplying out the null vec tors {||v|| = 0 } and doing a completion gives a separable Hilbert space H. Define now as in the construction of Brownian motion the process Xt = X(5t). Because the map X : H -» C2 preserves the inner product, we have E[XuXs] = (5s,5t) = V(s,t). □ Lets look at some examples of Gaussian... |
ch of them logCAr^) > log(2"(l - 5)). We see that £n P[An} converges. By Borel-Cantelli we get for almost every uj an integer ti(oj) such that for n > n(ui) \Bj2-«-Bi2-n\<(l + e)-h(k2-n), where k = j-ieK. Increase possibly n{uS) so that for n > n(w) Y,h(2-m)<e-h(2-(n+1){l-S)). m > n Pick 0 < h < h < 1 such that t = t2 ... |
motion Bt starting at x. Remark. Let K be a compact subset of a Green domain D. The hitting probability p(x)=Px[TK<TSD] is the equilibrium potential of K relative to D. We give a definition of the equilibrium potential later. Physically, the equilibrium potential is obtained by measuring the electrostatic potential, if ... |
P[f)4,fcnC] n—▶oo f c = l = E[/(B0)Unc] = E\f(B0)]-P[AnC] = E[/(J5t)]-P[AHC] Remark. If T < 00 almost everywhere, no conditioning is necessary and Bt-\-T — Bt is again Brownian motion. D Theorem 4.9.2 (Blumental's zero-one law). For every set A G Ao+ we have P[A] = 0 or P[A] = 1. Proof. Take the stopping time T which ... |
of radius r around x0. Since by the previous lemma, Brownian motion hits every ball almost surely, we can assume that x0 = 0 and by scaling that r = 1. Define inductively a sequence of hitting or leaving times Tn, Sn of the annulus {1/2 < \x\ < 2}, where Ti = inf{£ | \Bt\ = 2} and } Tn = inf{*>Sn_i||Bt| = 2}. These are... |
((A*)n_1"o, (A*)n_1"o), which is by induction n(n — l)!(Sn-i,n-i = n!. b) A* fin = \/n + l • fin+i is the definition of fin. AQn = -^=A(A*)nn0 = 4=nfi0 = VnVn-i • V n ! V n ! c) This follows from b) and the definition fin = ^A*fin-i- d) Part a) shows that {fin}n°=o 'lt is an orthonormal set in L2(R). In order to show that... |
ally for continuous martingales. Lets start with a mo tivation. We know by theorem (4.2.5) that almost all paths of Brownian motion are not differentiable. The usual Lebesgue-Stieltjes integral r t /o / *Jo f(Bs)Bs ds can therefore not be defined. We are first going to see, how a stochas tic integral can still be constru... |
ntum mechanics: Proposition 4.16.7. As operators, we have the identity ■Qn ■= ^ ■ V + *T •■- ^ E ( " ) (*?*"-' Definition. Define L = £"=0 ( " ) iA*)jAn~ Proof. Since we know that ftn forms a basis in L2, we have only to verify that : Qn : ttk = 2"n/2LQfc for all fc. From 2"1/2[Q,L] = [A + A*E(^)(A*)M^] = E ( n ) Ji^y-... |
ave Ts »(M) < TAn(M) if n is so large that An contains both s and t. Therefore (M, M) is increasing on \Jn An, which can be chosen to be dense. The continuity Remark. The assumption of boundedness for the martingales is not essen tial. It holds for general martingales and even more generally for so called local marting... |
for r < 0. Remark. The stochastic population model is also important when modeling financial markets. In that area the constant r is called the percentage drift or expected gain and a is called the percentage volatility. The Black-Scholes model makes the assumption that the stock prices evolves according to geometric B... |
were two fixed points x, y of (/>. Then d(x, y) = d(0(x), 0(£)) < A • d(x, y Chapter 5 Selected Topics 5.1 Percolation Definition. Let a be the standard basis in the lattice Zd. Denote with hd the Cayley graph of Zd with the generators A = {eu ..., ed }. This graph hd = (V, E) has the lattice Zd as vertices. The edges or... |
N in L?} GN = {3 no closed path of length > N in L?} . We know that FNnGNC {\C\ = oo}. Since FN and GN are both increas ing, the correlation inequality says Pp[Fjv H GN] > ?p[Fn} • Pp[Gn]- We deduce 0(p) = PP[|C| = oo] = PP[FN H GN] > PP[FN] - PP[GN] . If (1 - p)A(2) < 1, then we know that oo PP[GCN)< £(l-pyW(n-l) n = ... |
terion of Simon-Wolff [96] will be important. In the case of IID potentials with absolutely continuous distribution, a spectral averaging argument will then lead to pure point spectrum also for a = 0 Theorem 5.2.5 (Simon-Wolff criterion). For any interval [a, b] C R, the random operator L has pure point spectrum if for... |
ian. Proof, a) Ix+y < c2Ix + (1 — c)2Iy is proven using the Jensen inequal ity (2.5.1). Take then c = Iy/(IX + W) ) □ Theorem 5.3.7 (Rao-Cramer bound). A random variable X with mean m and variance a2 satisfies: Ix > 1/cr2. Equality holds if and only if X is the Normal distribution. Proof. This is a special case of Rao-C... |
n C(M, N) has a unique solution: because of Lipshitz continuity ||G(/) ~ G(/;)||oo < 2||D(V,V)||oo • 11/ - /'Hoc the standard Piccard existence theorem for differential equations assures local existence of solutions. The Gronwall's lemma assures that ||X(fj)|| can not grow faster than ex Remark. If m is a point measure... |
orsI probability meapure p on Id is unifbrn^;||jbp^ lutely continuous with respect to Lebesgue measure oii\Jd if and only if Proof. Use corollary ^ T h e r e for p > 2. This has an obvious generalization to d dim^M&ns Proposition 5.5.6. Given a bounded positive probability measure p € M(Id) and assume 1 < p < oo. Then ... |
cn with law v and then iterate xo,xx =/(x0,co),a?2 = /(xi,ci)... . Example. If (Sl,A,P,T) is an ergodic dynamical system, and A : ft -▶ 5L(d, R) is measurable map with values in the special linear group SL(d, R) of all d x d matrices with determinant 1. With M = Rd, the random diffeomorphism f(x,v) = A(x)v is called a ... |
rges weakly to a random variable with lattice distribution P[X = fc] = logio(fc + 1) - logio(fc) supported on {fc27r/10 | 0 < fc < 10 }. It is called the distribution of the first significant digit. The interpretation is that if you take a large random number, then the probability that the first digit is 1 is log(2), the ... |
n. It is an example of a compact Abelian topological group. Example. The general linear group G = GZ(n,R) with matrix multiplica tion is a topological group if the topology is the topology inherited as a sub set of the Euclidean space Rn2 of n x n matrices. Also subgroup of Gl(n, R), like the special linear group 5L(n,... |
circle problem asks to estimate the number of 1/n-lattice points gin) = im2 + E{n) enclosed in the unit disk. One believes that an estimate £(n) < Cne holds for every 0 > 1/2. The smallest 6 for which one knows the is 9 = 46/73. • For a smooth curve of length 1 which is not a line, we have a similar result as for the ... |
1 < fc < n. For smaller values of k, the data points appear random. The points are located on the graph of the circle map U t ) n 1 '?> To show the asymptotic independence of Xn with any of its translations, we restrict the random vectors to [1, l/na] with a < 1. Lemma 5.10.3. Let fn be a sequence of smooth maps from ... |
+ y3 + z3 = 1 has infinitely many solutions. It is believed that x3+y3+z3+w3 = n has solutions for all n. For (fc, ra) = (2,3), multiple solutions lead to so called taxi-cab or Hardy-Ramanujan numbers. Remark. For general polynomials, the degree and number of variables alone does not decide about the existence of nontr... |
deduce from the characteristic function whether X is absolutely continuous or not? The first question is completely answered by Wieners theorem given below. The decision about singular or absolute continuity is more subtle. There is a necessary condition for absolute continuity: Theorem 5.12.1 (Riemann Lebesgue-lemma).... |
so sum from -n to n, changing only the constant C. If /x is not uniformly y/h continuous, there exists a sequence of intervals \Ik\ -> 0 with p(It) > ly/h(\Ii\). A property of the Fejer kernel Kn(t) is that for large enough n, there exists S > 0 such that ^Kn(t) > S > 0 if 1 < n\t\ < tt/2. Choose nu so that 1 < nt • \I... |
lson. Radically elementary probability theory. Princeton univer sity text, 1987. [75] T.Lewis N.I. Fisher and B.J. Embleton. Statistical analysis of spher ical data. Cambridge University Press, 1987. [76] B. Oksendal. Stochastic Differential Equations. Universitext. Springer-Verlag, New York, fourth edition edition, 19... |
unt theorem, 222 economics, 14 edge of a graph, 275 edge which is pivotal, 281 Einstein, 194 electron in crystal, 18 elementary function, 40 elementary Markov property, 187 Elkies example, 349 Index ensemble micro-canonical, 102 entropy Boltzmann-Gibbs, 98 circle valued random variable, 320 coarse grained, 99 distribut... |
equation, 271 spectral measure, 178 spectrum, 18 spherical random variable, 327 Spitzer theorem, 244 stake of game, 137 Standard Brownian motion, 192 standard deviation, 40 standard normal distribution, 92, 98 state space, 186 statement almost everywhere, 39 stationary measure, 188 stationary measure discrete Markov pr... |
× 3 matrices. 7. Diagonal matrices: A matrix M can be thought of as an array of numj, known as matrix entries, or matrix components, where i and j bers mi index row and column numbers, respectively. Let M = 1 2 3 4 = mi j . 1, m1 1 and m2 Compute m1 2, m2 2. The matrix entries mi i whose row and column numbers are the... |
ry in the first row does not vanish. Then, using just the three EROs, we could1 then perform the following. 1This is a “brute force” algorithm; there will often be more efficient ways to get to RREF. 42 2.1 Gaussian Elimination 43 Algorithm For Obtaining RREF: • Make the leftmost nonzero entry in the top row 1 by multipli... |
n form is not unique. Once a system is in row echelon form, it can be solved by “back substitution.” Write the following row echelon matrix as a system of equations, then solve the system using back-substitution 50 2.2 Review Problems 51 8. Show that this pair of augmented matrices are row equivalent, assuming ... |
M = P (E−1 1 E−1 2 E−1 3 )(E−1 4 E−1 5 E−1 6 )(E−1 7 )U = P LDU 0 2 −4 2 1 2 1 0 −3 9 0 2 1 −1 0 −1 = 1 0 −.4 Review Problems Webwork: Reading problems Matrix notation LU 3 18 19 61 62 Systems of Linear Equations 1. While performing Gaussian elimination on these augmented matrices write the full... |
· · +ar 1x1 +ar ar Show that your rule for multiplying a matrix by a vector obeys the linearity property. 4. The standard basis vector ei is a column vector with a one in the ith row, and zeroes everywhere else. Using the rule for multiplying a matrix times a vector in problem 3, find a simple rule for multiplying M ei... |
m terminates if all the coefficients in the last row (save perhaps for the last entry which encodes the value of the objective) are positive. To see why we are done, lets write out what our row operations have done in terms of the function f and the constraints (c1, c2). First we have f = 16 − 7y − 6z with both y and z p... |
ectors for any plane in Rn. This now motivates the definition of the dot product. Definition The dot product of u = u1 ... un and v = is v1 ... vn u v := u1v1 + · · · + unvn . 89 90 Vectors in Space, n-Vectors Example 49 of the dot product of two vectors from R100 ... 100 · ... |
sketch the collection of all vectors in two-dimensional Lorentzian space-time with zero length. (c) Find and sketch the collection of all vectors in three-dimensional Lorentzian space-time with zero length. (d) Replace the word “zero” with the word “one” in the previous two parts. The Story of Your Life 5. Create a sys... |
ector and therefore fails (iv). Example 65 The solution set to is + c −1 1 c ∈ R . The vector 0 0 is not in this set. Do notice that if just one of the vector space rules is broken, the example is not a vector space. Most sets of n-vectors are not vector spaces. 106 5.2 Other Fields Example 66 P := since 1 1 ∈ P but −2... |
s is tackled in Chapter 11.) This leads us to write L c1 1 1 + c2 0 0 1 1 = c1 + c2 c1 c2 . This makes sense, but requires a warning: The matrix specifies L so long 0 0 1 1 0 0 as you also provide the information that you are labeling points in the plane V by the two numbers (c1,... |
123 124 Matrices Notice something important: there is no reason to say that β comes before b or vice versa. That is, there is no a priori reason to give these basis elements one order or the other. However, it will be necessary to give the basis elements an order if we want to use them to encode other vectors. We choo... |
ns 1 0 1 1 0 0 0 1 , and . , , 3 , 4 , 2 , 1 , (l) homogeneous first order polynomial function H : R2 → R whose 1 0 graph contains 1 1 0 1 , and . , 3 , 4 , 2 , 132 7.3 Properties of Matrices 133 (m) second order polynomial function J : R2 → R whose graph con- , tains , , , and 0 2 1 1 , 4 . (n) first order polynomial fu... |
mensions. In fact the following matrices built from a 2 × 2 rotation matrix, a 1 × 1 identity matrix and zeroes everywhere else M = cos θ − sin θ 0 sin θ 0 cos θ 0 1 0 and N = 0 1 cos θ 0 0 − sin θ , 0 sin θ cos θ perform rotations by an angle θ in the xy and yz planes, respectively. Because, they rotat... |
pute M −1, we would like M −1, rather than M −1V to appear on the right side of our augmented matrix. This is achieved by solving the collection of systems M X = ek, where ek is the column vector of zeroes with a 1 in the kth entry. I.e., the n × n identity matrix can be viewed as a bunch of column vectors In = (e1 e2 ... |
U X = W011 Back substitution gives z = −11, y = 3, and x = −3. Then X = −3 3 , and we’re done. −11 Using an LU decomposition 161 162 Matrices 7.7.2 Finding an LU Decomposition. In chapter 2, section 2.3.4, Gaussian elimination was used to find LU matrix decompositions. These ideas are presented here again as revi... |
2 Determinants Permutation Example Reading homework: problem 1 We can use permutations to give a definition of the determinant. Definition The determinant of n × n matrix M is det M = σ sgn(σ) m1 σ(1)m2 σ(2) · · · mn σ(n). The sum is over all permutations of n objects; a sum over the all elements of {σ : {1, . . . , n} →... |
RREF(M ) for any λ. Then: det(M N ) = det(E1E2 · · · Ek RREF(M )N ) = det(E1) · · · det(Ek) det(RREF(M )N ) = det(E1) · · · det(Ek) det(Rn(λ) RREF(M )N ) = det(E1) · · · det(Ek)λ det(RREF(M )N ) = λ det(M N ) 181 182 Determinants Figure 8.6: “The determinant of a product is the product of determinants.” Which implies t... |
det − det det 1 2 0 1 3 −1 1 3 −1 2 1 0 T det − det Reading homework: problem 6 Let’s compute the product M adj M . For any matrix N , the i, j entry of M N is given by taking the dot product of the ith row of M and the jth column of N . Notice that the dot product of the ith row of M and the ith co... |
en we can find a solution M −1 x y = z r1 r2 r3 for any vector ∈ R3. x y z Therefore we should choose a so that M is invertible: i.e., 0 = det M = −2a2 + 3 + a = −(2a − 3)(a + 1). Then the span is R3 if and only if a = −1, 3 2 . Linear systems as spanning sets Some other very important ways of bu... |
ithout v4 by making the substitution v4 = v1 + v2. Then: S = span{1 + t, 1 + t2, t + t2, 2 + t + t2, 1 + t + t2} = span{1 + t, 1 + t2, t + t2, 1 + t + t2}. 2 (1 + t2) + 1 Now we notice that 1 + t + t2 = 1 2 (1 + t) + 1 2 (t + t2). So the vector 1 + t + t2 = v5 is also extraneous, since it can be expressed as a linear c... |
s basis is often written {ˆi, ˆj, ˆk} for R3. Note that it is often convenient to order basis elements, so rather than writing a set of vectors, we would write a list. This is called an ordered basis. For example, the canonical ordered basis for Rn is (e1, e2, . . . , en). The possibility to reorder basis vectors is no... |
ace V . We now add some extra information. The string’s behavior in time and space can be modeled by a wave equation ∂2y ∂t2 = ∂2y ∂x2 , which says that the acceleration of a point on the string is equal its concavity at that point. For example, if the string were made of stretched rubber, it would prefer to be in a st... |
e λ called the characteristic polynomial PM (λ) of M . For an n × n matrix, the characteristic polynomial has degree n. Then PM (λ) = λn + c1λn−1 + · · · + cn. Notice that PM (0) = det(−M ) = (−1)n det M . Now recall the following. Theorem 12.2.1. (The Fundamental Theorem of Algebra) Any polynomial can be factored into... |
ce the expression for vj in the basis S is vj itself, then P Q maps each vj to itself. As a result, each vj is an eigenvector for P Q with eigenvalue 1, so P Q is the identity, i.e1 . The matrix P is called a change of basis matrix. There is a quick and dirty trick to obtain it; look at the formula above relating the n... |
th of a vector x = (x1, x2, . . . , xn) ∈ Rn is √ ||x|| = x x = (x1)2 + (x2)2 + · · · (xn)2 . The canonical/standard basis in Rn e1 = 1 0 ... 0 , e2 = 0 1 ... 0 , . . . , en = 0 0 ... 1 , has many useful properties with respect to the dot product and lengths. ... |
span{u, v}, i.e., check that u, v and w are linearly independent. If it does not, we then can define w⊥ := w − u w u u u − v⊥ w v⊥ v⊥ v⊥. We can check that u w⊥ and v⊥ w⊥ are both zero: u w⊥ w v⊥ v⊥ v⊥ v⊥ w v⊥ v⊥ u v⊥ v⊥ w v⊥ v⊥ u v⊥ = 0 = u w − u w − since u is orthogonal to v⊥, and v⊥ w⊥ = v⊥ w v⊥ v⊥ v⊥ v⊥ w v⊥ v⊥ v⊥... |
= v1 0 − 1/2 3/1 . 271 272 Orthonormal Bases and Complements So the set 1 −1 is an orthogonal basis for L⊥. Dividing each basis vector by its length yields , , 1√ 6 1√ 6 − 2√ 6 0 1√ 2 − 1√ 2 0 0 and orthonormal basis ... |
n. Let x† = z1 zn · · · zn ∈ Cn (a 1 × n complex matrix or a row vector). Compute x†x. Using the result of part 1a, what can you say about the number x†x? (E.g., is it real, imaginary, positive, negative, etc.) (d) Suppose M = M T is an n × n symmetric matrix with real entries. Let λ be an eigenvalue of M with eigenv... |
rary functions between sets in that, by looking at just one (very special) vector, we can figure out whether f is one-to-one! 291 292 Kernel, Range, Nullity, Rank 16.2.2 Kernel Let L : V → W be a linear transformation. Suppose L is not injective. Then we can find v1 = v2 such that Lv1 = Lv2. So v1 − v2 = 0, but L(v1 − v2... |
Some were left as review questions or sample final questions. The rest are left as exercises for the reader. Invertibility Conditions 298 16.4 Review Problems 299 16.4 Review Problems Webwork: , 1 Reading Problems Elements of kernel Basis for column space Basis for kernel Basis for kernel and range Orthonomal range basi... |
solutions M T M X = M T V does have solutions. One way to think about this is, since the codomain of M is the direct sum codom M = ranM ⊕ ker M T there is a unique way to write V = Vr+Vk with Vk ∈ ker M T and Vr ∈ ran M , and it is clear that M x = V only has a solution of V ∈ ran M ⇔ Vk = 0. If not, then the closest ... |
= {0, 1} or bits. The rules for addition and multiplication are the usual ones save that 1 + 1 = 0 . 318 C Online Resources Here are some internet places to get linear algebra help: • Strang’s MIT Linear Algebra Course. Videos of lectures and more: http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010... |
u = 0 (at each step in this chain of equalities we have used the fact that V is a vector space and therefore can use its vector space rules). In particular, this means that the zero vector of V is in U and is its zero vector also. (ii) Also, in V , for each u there is an element −u such that u + (−u) = 0. But by additi... |
space, using the rules for vector addition and scalar multiplication inherited from the original vector space. (a) So long as U = U ∪ W = W the answer is no. Take, for example, U to be the x-axis in R2 and W to be the y-axis. Then 1, 0 ∈ U and 0, 1 ∈ W , but 1, 0 + 0, 1 = 1, 1 /∈ U ∪ W . So U ∪ W is not additively clos... |
rix M obey any special properties? Find its eigenvalues. You may call your answers λ and µ for the rest of the problem to save writing. For the rest of this problem we will focus on central conics for which the matrix M is invertible. (c) Your equation in part (a) above should be be quadratic in X. Recall that if m = 0... |
subtracting the first row from all other rows (which leaves the determinant unchanged). 354 355 Now zero out the top row by subtracting x1 times the first column from the second column, x1 times the second column from the third column et cetra. Again these column operations do not change the determinant. Now factor out x... |
Xn+1 = Fn+1 Fn = Fn + Fn−1 Fn = 1 1 1 0 Fn Fn−1 = M Xn . (e) Notice M is symmetric so this is guaranteed to work. det 1 − λ 1 1 −λ = λ(λ − 1 , so the eigenvalues are 1± 2 √ 5 . Hence the eigenvectors are √ respectively (notice that 1+ 1− 2 . 1− 2 ). Thus M = P DP −1 with 2 + 1 = 1+ 2 5 and 1− √ 1+ 2 0 0 √ 1− 2 5 and P... |
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