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### math-2011-note10

Course: ECON 205, Spring 2011
School: Korea University
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Word Count: 2247

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6. Example Suppose that we have a production function Q = kxa yb . Then, Q = akxa1 yb , x 2Q = abkxa1 yb1 , x y Q = bkxa yb1 y 2Q = abkxa1 yb1 y x 14 Some Linear Algebra 14.1 Deniteness of Quadratic Forms (16.2) Denition 27. Let A be an m m symmetric matrix, then A is (a) positive denite if xT A x &gt; 0 for all x = 0 in Rm , (b) positive semidenite if xT A x 0 for all x in Rm , (c) negative denite if...

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6. Example Suppose that we have a production function Q = kxa yb . Then, Q = akxa1 yb , x 2Q = abkxa1 yb1 , x y Q = bkxa yb1 y 2Q = abkxa1 yb1 y x 14 Some Linear Algebra 14.1 Deniteness of Quadratic Forms (16.2) Denition 27. Let A be an m m symmetric matrix, then A is (a) positive denite if xT A x > 0 for all x = 0 in Rm , (b) positive semidenite if xT A x 0 for all x in Rm , (c) negative denite if xT A x < 0 for all x = 0 in Rm , (d) negative semidenite if xT A x 0 for all x in Rm, (e) indenite if xT A x > 0 for some x in Rm and < 0 for some other x in Rm . Q(x, y) = 3x2 6y2 0 and strictly negative except at 0. Hence, it is negative denite, and (x, y) = 0 is its global maximizer. A =? Q3 (x, y) = 3x2 6y2 is indenite. A =? Q4 (x, y) = x2 + 2xy + y2 is positive semidenite. A =? Q5 (x, y) = (x y)2 is negative semidenite. A =? 14.2 Symmetric Matrices (23.7) Denition 28 (p.579, 582). Let A be a m m matrix, v = 0 be a m-vector and r be a number. If Av = rv, then v is called an eigenvector (or characteristic vector) of A, and r is called an eigenvalue (or characteristic value) of A. Example 7. Consider A = Av = 2 2 24 42 = 2v, and Aw = ,v = 6 6 1 1 ,w = 1 1 . Then, = 6w. Hence, the eigenvalues of A are -2 and 6. Theorem 44 (Thm 23.16, p.621). Let A be a symmetric m m matrix. Then there is a nonsingular matrix P whose columns w1 , , wm are eigenvectors of A such that (i) w1 , , wm are mutually orthogonal to each other 40 (ii) P1 = PT , and (iii) P1 AP = PT AP = r1 0 0 0 r2 0 . . .. . .. .. .. . 0 0 rm If we write the diagonal matrix as D, then we have A = PDPT . Hence, xT Ax = xT PDPT x = yT Dy, where y = PT x. A is positive(negative) denite D is positive(negative) denite. Denition 29 (p.165). Let A be an m m matrix. A is said to be nonsingular if there exists an m m matrix A1 , called the inverse of A, such that A1 A = AA1 = I Theorem 45 (Thm 8.5, p.165). If A has an inverse, it is unique. Theorem 46 (Thm 23.16, p.621). Let A be a symmetric m m matrix. Then there exists a nonsingular matrix P such that (i) P1 = PT , and (ii) P1 AP = PT AP = r1 0 0 0 r2 0 . . .. . .. .. .. . 0 0 rm If we write the diagonal matrix as D, then we have A = PDPT . Hence, xT Ax = xT PDPT x = yT Dy, where y = PT x. A is positive(negative) denite D is positive(negative) denite. 15 Unconstrained Optimization 2 (ch. 17) 15.1 Second Order Conditions Sufcient Conditions Theorem 47 (Thm 17.2, p.399). Let F : U R1 be a C2 function dened on an open set U Rm . Suppose that DF (x ) = 0. (Such x is called a critical point of F.) (1) If D2 F (x ) is negative denite, then x is a strict local max of F; (2) If D2 F (x ) is positive denite, then x is a strict local min of F; 41 (3) If D2 F (x ) is indenite, then x is neither a local max nor a local min of F. Denition 30 (p.399). A critical point x for which D2 F (x ) is indenite is called a saddle point of F. Q3 (x, y) = 3x2 6y2 is indenite. (0, 0) is a saddle point. 15.2 Proof of the Sufciency Proof. When DF (x ) = 0, we have F (x + h) = F (x ) + where R2 (h) h 2 1 T2 h D F (x )h + 2R2 (h) 2 0 as h 0. There exists a nonsingular matrix P such that P1 = PT and D2 F (x ) = PDPT , where D is a diagonal matrix with all negative (positive) diagonal elements. Since hT h hT I h R2 (h)I R2 (h) = R2 (h) = R2 (h) = hT h, 2 2 h h h2 we have hT D2 F (x )h + 2R2 (h) = hT D2 F (x )h + 2R2 (h) 2R2 (h)I T = hT PDPT h + hT PPT PP h h2 = hT P D + = yT D + 2R2 (h)PT IP h 2R2 (h)I h 2 2 PT h y < (>)0 (See the Laws of Matrix Algebra in the note 0923p) 16 More Linear Algebra (ch.11 and Rudin ch.9) 16.1 Determinant of a Matrix (Rudin, ch.9) Denition 31 (p.240). Vectors v1 ,v2 , , vk Rm are linearly dependent if there exists c1 , c2 , , ck R, not all zero, such that c1 v1 + c2 v2 + + ck vk = 0. They are linearly independent if c1 v1 + c2 v2 + + ck vk = 0, c1 , c2 , , ck R implies that c1 = c2 = = ck = 0. 42 Denition 32 (Rudin p.232). For an ordered m-tuple of integers ( j1 , j2 , , jm), dene s( j1 , j2 , , jm) = sgn( j p jq ), p >q where sgn x = 1 if x > 0, = 1 if x < 0, = 0 if x = 0. Example 8. s(1, 3, 2) = sgn[(2 1)(2 3)(3 1)] = 1 s(1, 3, 2, 4) = sgn[(4 1)(4 3)(4 2)]s(1, 3, 2) = 1 s( j1, , jh, jh+1 , , jm) = s( j1, , jh+1, jh , , jm) s( j1, , jh, , jk , , jm) = s( j1 , , jk, , jh , , jm) Denition 33 (Rudin p.232). For an m m matrix A = (ai j ), dene det A = s( j1, j2 , , jm)a1 j1 a2 j2 am jm , where the sum is taken over all ordered m-tuple of integers ( j1 , j2 , , jm) with 1 j r m. Example 9. det a11 a12 a21 a22 = s(1, 1)a11a21 + s(1, 2)a11a22 + s(2, 1)a12a21 + s(2, 2)a12a22 = a11 a22 a12 a21 j h = j k s( j 1 , j 2 , , j n ) = 0. Hence, the sum contains n! (the number of ways to order n different integers) non-zero terms. The determinant is the sum of all the multiplications of one element in each row and each column multiplied by the signs dened by s( j1, j2 , , jn). Example 10. a11 a12 a13 det a21 a22 a23 a31 a32 a33 = s(1, 2, 3)a11a22 a33 + s(1, 3, 2)a11a23 a32 + s(2, 1, 3)a12a21 a33 + s(2, 3, 1)a12a23 a31 + s(3, 1, 2)a13a21 a32 + s(3, 2, 1)a13a22 a31 = a11 a22 a33 a11 a23 a32 a12 a21 a33 + a13 a22 a31 + a13 a21 a32 a13 a22 a31 Lemma 3. If A is a matrix obtained from A by switching two columns, then det A = det A. 43 Proof. Suppose that A is obtained by switching h-th and k-th columns. Then det A = s( j1, , h, , k, , jn )a1 j1 aih h aik k an jn = s( j1, , h, , k, , jn )a1 j1 aih k aik h an jn = s( j1, , k, , h, , jn )a1 j1 aih k aik h an jn = s( j1 , , k, , h, , jn )a1 j1 aih k aik h an jn = det A Lemma 4. If two columns of A are identical, then det A = 0. Proof. det A = det A = det A. Hence, det A = 0. Theorem 48. det (a1 , , am) is linear a function of each column vector a j . That is det (a1 , , ka j + hb j , , am ) = k det (a1 , , a j , , am ) + h det (a1 , , b j , , am) Proof. det (a1 , , ka j + hb j , , an ) = s( j1 , , j, , jn )a1 j1 ai1 ji1 (kai j + hbi j )ai+1 ji+1 an jn = s( j1 , , j, , jn )a1 j1 ai1 ji1 (kai j )ai+1 ji+1 an jn + s( j1 , , j, , jn )a1 j1 ai1 ji1 (hbi j )ai+1 ji+1 an jn = k det (a1 , , a j , , an ) + h det (a1 , , b j , , an ) Corollary 3. If A is obtained from A by adding a column a j of matrix A multiplied by some number r to another column ah , then det A = det A. That is det (a1 , , ah + ra j , , am ) = det (a1 , , ah , , am) Proposition 1. Let A = (a1 , a2 , , am). If a1 , a2 , , am are linearly dependent, then det A = 0. Proof. If a1 , a2 , , am are linearly dependent, then there exist c1 , , cm , with ch = 0 for some h, such that j c j a j = 0. Then, one can write as cj aj j=h ch ah = c Hence, det A = det (a1 , , ah , , am) = det a1 , , j=h chj a j , , am = 0 Theorem 49 (Rudin Thm 9.35, p.233). If A and B are m m matrices, then det AB = det A det B. We do not prove this theorem. The proof is in Rudin. 44 Theorem 50 (Rudin Thm 9.36, p.233). An m m matrix A is invertible if and only if det A = 0. Proof. A is invertible there is A1 such that A1 A = I det A1 det A = det I = 1. Hence det A = 0. This proves the only if part. The if part will be proved later. 16.2 Basis Denition 34 (Rudin pp.206-207). Given vectors v1 , , vk Rm and scalars c1 , , ck R, the vector c1 v1 + + ck vk is called a linear combination of v1 , , vk . A nonempty set X Rm is a linear space if v1 , , vk X implies that any linear combination of v1 , , vk belongs to X If S Rm and if E is the set of all linear combinations of elements of S, we say that S spans E, or that E is the span of S. If a linear space X contains an independent set of r vectors but contains no independent set of r + 1 vectors, we say that X has dimension r, and write dim X = r. The set consisting of 0 is a linear space; its dimension is 0. An independent subset of a linear space X which spans X is called a basis of X. Proposition 2. If B = {x1 , , xr } is a basis of X, then every x X has a unique representation of the form x = ci xi Proof. Since B spans X , any x X can be written as a linear combination of xi s, x = ci xi . (2) x = ci xi . (3) Suppose it is also written as Subtracting equation (3) from equation (2), one obtains 0 = (ci ci )xi. Since xi s are linearly independent, this implies ci = ci for all i. Theorem 51. If a linear space X is spanned by a set of r vectors, then dim X r. Proof. According to the denition of the dimension it is enough to show that any set of r + 1 vectors in X is linearly independent. Suppose that {y1 , , yr+1 } is linearly independent. Let {x1 , , xr } span X . Then y1 is written as a linear combination of xi s, y1 = ci xi . Some ci is not 0, since y1 = 0. Without loss of generality lets assume that c1 = 0. Then x1 can be written as a linear combination of y1 , x2 , , xr . Hence, S1 {y1 , x2 , , xr } spans X . Then y2 is written as a linear combination of vectors in S1 , y2 = b1 y1 + r=2 ci xi . Some ci is not 0, because ci = 0, i = 2, , r i implies y2 = b1 y1 which means {y1 , , yr+1} is linearly dependent. Without loss of generality we can assume that c2 = 0. Then, we can replace x2 by y2 in S1 and span X by S2 {y1 , y2 , x3 , , xr }. We can continue this process and show that Sr {y1 , , yr } spans X . But then yr+1 can be written as a linear combination of y1 , , yr , a contradiction. 45 Corollary 4. dim Rm = m. Proof. Consider e1 , , em . They are linearly independent. Hence, dim Rm m. They also span Rm . Hence, dim Rm m. Theorem 52. Suppose X is a linear space, and dim X = m. a) A set E of m vectors in X spans X if and only if E is independent. b) X has a basis and every basis consists of m vectors. Proof. a) only if part. Suppose that E spans X and that E is dependent. Then one of the vectors in E is written as a linear combination of the others, hence one can delete that vector and the remaining vectors can still span X , which implies dim X < m, a contraction. if part. Consider x X . The set E {x} is linearly dependent according to the denition of the dimension. That is there are scalars c0 , c1 , , cm not all equal to 0 such that c0 x + m 1 ci xi = 0, where xi E . We must have c0 = 0, otherwise E is i= ci dependent. Hence x = c0 xi . Since x is chosen arbitrarily, E spans X . b) According to the denition of dimension there is an independent set E of m vectors in X . By a) E spans X , hence it is a basis of X . A basis cannot contain more than m vectors because it would not be independent. A basis cannot contain less than m vectors, because dim X would be less than m according to Theorem 51 Given a vector a Rm , the function f (x) = a x is a linear function from Rm to R. Given a k m matrix A = (a1 , a2 , , am ), F (x) Ax is a linear function from Rm into Rk , A(rx + sy) = rAx + sAy. Denition 35 (p.297). A function F : X Y is said to be one-to-one (or injective) if x, y X and x = y implies F (x) = F (y). We say F is onto (or surjective) if F (X ) {y| y = F (x) for some x X } = Y . Question: When is an m m matrix A = (a1 , a2 , , am) onto as a linear function? Answer: When the column vectors a1 , a2 , , am are linearly independent. Reason: The set {y Rm | y = Ax = x1 a1 + + xm am, x Rm } spans Rm if and only if a1 , a2 , , am are linearly independent. Question: When is an m m matrix A = (a1 , a2 , , am ) one-to-one? Answer: When the column vectors a1 , a2 , , am are linearly independent. Proof? A is one-to-one x = y implies Ax = Ay Ax = Ay implies x = y Ac = 0 implies c = 0 a1 , a2 , , am are linearly independent 46
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Korea University - ECON - 205
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Korea University - ECON - 205
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University of Regina - STAT - 351
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University of Regina - STAT - 351
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St. Johns - PHI - 2240C
St. Johns - PHI - 2240C
HW#7AContraception941509B936HW#7A:CONTRACEPTION(02/10/12for02/14/12)CHCE,2ndEd.,Ch.7,Chemical,Barrier,andSurgicalContraception,pp.87101Q1:Conjugallovehasatwofoldmeaning.Themaritalactisboth_unitive_and_procreative_bynature.Contraception_inhibits_thepro
St. Johns - PHI - 2240C
HW#9 The Principle of the Double EffectHW #9: THE PRINCIPLE OF THE DOUBLE EFFECT (02/17/12 for 02/21/12)CHCE, 2nd Ed., Ch. 3, Selected Moral Principles, D: Double Effect, pp. 23-26Q1: The Principle of Double Effect (PDE) justifies some good actions tha
St. Johns - PHI - 2240C
HW#8WarrenOnRedefiningforAbortionUNIT#8:WARRENONREDEFININGFORABORTION(02/14/12for02/17/12)MaryAnneWarren,OntheMoralandLegalStatusofAbortion,TheMonist,vol.57,no.1,pp.4361.Q1:SinceProf.MaryAnneWarrenwantstodisenfranchisefetal_people_,shefaultsthemfornot
St. Johns - PHI - 2240C
HW#5A The Cardinal VirtuesHW#5A THE CARDINAL VIRTUES (02/03/12 for 02/07/12)Q1: A part of human motions responds to threatening or unpleasant things. It is called the_irascible_ appetite. Another part deals with the enjoyment and sorrow concerning plea
St. Johns - PHI - 2240C
HW#4A How Actions Become Good or BadHW #4A HOW ACTIONS BECOME GOOD OR BAD? (01/31/12 for 02/03/12)Q1: The morality of a human action is a human actions quality of being morally good or bad.The object or nature of the act is the _primary_ source of the
St. Johns - PHI - 2240C
HW#6AWhenDoesHumanLifeBegin?HW#6AWHENDOESHUMANLIFEBEGIN?(02/07/12for02/10/12)NationalCatholicBioethicsQuarterly,Vol.9,No.1(Spring2009),pp.131149Q1:Thetwocriteriausedtodistinguishonecelltypefromanotherareunique_cellcomposition_and_behavior_.Q2:Followi
St. Johns - PHI - 2240C
HW#3A Psychology of the Human ActHW#3A PSYCHOLOGY OF THE VOLUNTARY ACT (01/27/12 for 01/31/12)Q1: In Step 2, there is a simple, vague wishing for, or desire of, a _good_ thing that is not yetan end. On the other hand, there is in step 4 a focused inten
St. Johns - PHI - 2240C
HW#2 IsHappines the Ultimate End of Man?HW#2 IS HAPPINESS THE ULTIMATE END OF MAN? (01/24/12 for 01/27/12)Q1: An end is that for the sake of which an agent acts, i.e., it is a _final_ cause. It issomething that is sought for _itself_?Q2: An appetites
St. Johns - PHI - 2240C
HW1B The Human Persson in Medical EthicsHW 1B THE HUMAN PERSON IN MEDICAL ETHICS (01/20/11 for 01/25/11)Read CHCE, 2nd Ed., Ch. 1: The Human Person (Moraczewski) pp. 3-8Fill in the missing word(s).Q1: In Catholic health care, the individual person is
St. Johns - PHI - 2240C
HW1A Hippocratic OathHW1A HIPPOCRATIC OATH (01/20/11 for 01/25/11)Read the following oath and answer the questions below.I swear by Apollo the Physician and Asclepius and Hygieia and Panacea and all the gods, andgoddesses, making them my witnesses, th
St. Johns - PHIL - 100C
Aristotle 56-59Sunday, September 18, 20113:33 PMPlato was an idealist, who took mathematics as his model and believed in a transcendentreality, the Forms, governed by the form of the Good.Aristotle was a realist, who took biology as his working model
St. Johns - PHIL - 100C
Aristotle 59-63Wednesday, September 21, 201110:16 PMAristotle believed that humanity had a function juts as doctors had the function to cure the sick,or the function of a ruler to govern society well, or the function of a knife to cut well, it was the
St. Johns - PHIL - 100C
Aristotle 63-67Saturday, September 24, 20116:42 PMThe objectivist view : everything has a function and the function of humans is to live a life ofhappiness through reason.This view lost favor when the evolutionary account of human nature became popul
St. Johns - PHIL - 100C
Aristotle 67-70Sunday, September 25, 20118:58 PMVirtues - characteristics that enable individuals to live well in communities.In order to achieve a state of human flourishing, proper social institutions are necessary.Therefore, the moral person needs
St. Johns - PHIL - 100C
Augustine 72Sunday, October 09, 20113:37 PMAugustine believed that not reason but faith is the way to truth. &quot;unless you believe you shall notunderstand&quot;Faith Seeking understanding (fides querens intellectum)Augustine rejects Socrates' theory that i
St. Johns - PHIL - 100C
Bible P. 5-24Sunday, September 04, 20119:09 AMThe Hebrew BibleThe bible depicts God as the maker of heaven and earth, a very powerful, justbenevolent being who sees into the hearts of humankind. He is holy, inscrutable,mysterious, tremendously awe-i
St. Johns - PHIL - 100C
Darwin 204-222Saturday, November 26, 20118:23 PM4 theses of naturalist evolutionary theory:The ancient earth thesis - The Universe is about 15 billion years old and the Earth isabout 4.5 billion years old, not 6,000 as the biblical genealogical recor
St. Johns - PHIL - 100C
Dualism/Materialism 225-246Wednesday, November 30, 20119:50 PMDualistic Interaction - body and the mind are different substances that casually interact on eachother.Ex. Steeping on a nail causes brain state which causes metal state (feeling pain) whi