Documents about Linear Combinations
Lesson - 19
Western Michigan, ECE 380
Excerpt: ... ECE 3800 Lesson Nineteen Density Functions of Linear Combinations of Random Variables Vocabulary: Convolution Linear combinations Topics: Finding the density function for sums of random variables Uniform distributions Exponential distributions Applications Examples: 1. a. b. 2. Find Find , , Practice Example: See the Convolution Example in the lecture notes. Bring a copy of this to the next lecture! ...
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Lesson - 20
Western Michigan, ECE 380
Excerpt: ... ECE 3800 Lesson Twenty - Circuit Applications Vocabulary: Convolution Linear combinations Topics: Finding the density function for sums of random variables Uniform distributions Standard operational amplifier configurations summers inverters Example: See the Convolution Example posted in the Lecture Notes. This example should be completed for independent study. Practice Example: See the Convolution Example Solution in the lecture notes. ...
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10l
University of Hawaii - Hilo, MATH 311
Excerpt: ... Math 311 Lecture 10 Subspaces A subset W of a vector space V is also a vector space under the operations of V iff 0LW and W is closed under addition and scalar multiplication, i.e., if u,vLW and cLR implies u+vLW and cuLW. PROOF. As noted above, it suffices to check just the conditions: u, vLW and c a scalar u+vLW, cuLW. A sum of linear combinations is also a linear combination. A scalar multiple of a linear combination is also a linear combination. DEFINITION. W is a subspace of a vector space V iff WV, CGiven v1 = [1, 1, 0], v2 = [0, 1, 1], v3 = [1, 0, -1], write the following as linear combinations of v1, v2, v3 if and for all u, vLW and all scalars c: possible, otherwise write . 0LW, u+vLW, and cuLW. D v = [1, 3, 2] E v = [1, 1, 1]. The last two conditions alone suffice since cuLW for all c 0uLW 0LW. Solution. It is easier to solve the equivalent column For any vector space V, {0} is a subspace, called the zero subspace. Since VV, V is also a subspace of itself. CWrite "subspace" if the given se ...
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l3
UCSD, MATH 20F
Excerpt: ... Math 20F Linear Algebra Lecture 3: Vector Equations, Linear Combinations and Span. In linear algebra we think of vectors in Rn as column vectors or n 1 matrices u1 u2 u = . , . . un v1 v2 v= . . . vn Addition and scalar multiplication are defined by u1 + v1 u2 + v2 , u+v = . . . un + vn u1 u2 u = . , . . un R. Linear Combination: Given vectors v1 , . . . , vk , and scalars 1 , . . . , k , the vector w = 1 v 1 + + k v k is called a linear combination of the vectors v1 , . . . , vk , (with weights 1 , . . . k ). Question: If you give me a vector w, and vectors v1 , . . . , vk , how can I figure out whether w is a linear combination of v1 , . . . vk ? Geometric Interpretation: In R2 and R3 We think of vectors as arrows with a length and a direction. The parallelogram law says that the sum u + v is given by placing the start of v 1 2 1+2 3 where u ends. Check this by drawing u = , v= , and u+v = = . 3 1 3+1 4 If > 0 then scalar multiplication u is the vector in the ...
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PCA_CCA_PLS
Texas Tech, ISQS 6348
Excerpt: ... Comparison of Principal Components, Canonical Correlation, and Partial Least Squares for the Job Salience/Job Satisfaction data analysis. In all analyses, we find linear combinations Lx and Ly where LX = a1X1+amXm, And LY = b1Y1+bpYp. All Xs and Ys are standardized by default. The goals are 1) Find linear combinations that represent the original variables as well as possible 2) Find linear combinations that correlate highly. Principal components is optimal for goal 1) since the linear combinations capture a maximum amount of variance from the original variables. However, these linear combinations are not chosen to relate the two sets of variables, so they might not work well for goal 2). Canonical correlation is optimal for goal 2) since the linear combinations are chosen to optimize the correlation between the two sets of variables. However, the linear combinations may not be optimal for goal 1), since there is no objective to explain or capture variance. Partial least squares attempts to achieve ...
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Math221Lecture003BSlides
UMBC, MATH 221
Excerpt: ... Vectors Linear Combinations Span Lecture 3: Vectors Vectors Vectors in Rn Operations on Vectors in Rn Geometric and Physical Interpretations of Vectors Algebraic Properties of Vectors in Rn Linear Combinations A Linear Combination of a Set of Vectors Example 7 Linear Combinations and Linear Systems Linear Combinations and Linear Systems Span The Span of a Set of Vectors Text Examples Section 1.3 Clint Lee Math 221 Lecture 3: Vectors 1/18 Vectors Vectors in Rn Linear Combinations Span Vectors in Rn (Column) Vectors A (column) vector is a matrix with only one column. For example, Clint Lee Math 221 Lecture 3: Vectors 2/18 Vectors Vectors in Rn Linear Combinations Span Vectors in Rn (Column) Vectors A (column) vector is a matrix with only one column. For example, 1 u = 2 , 3 Clint Lee Math 221 Lecture 3: Vectors 2/18 Vectors Vectors in Rn Linear Combinations Span Vectors in Rn (Column) Vectors A (column) vector is a matrix with only one column. For example, 1 u ...
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ReviewexamI
Kentucky, MA 322
Excerpt: ... Exam I Review 1. Read/Study the sections and the corresponding notes. 2. Study the quiz questions. If you did not answer correctly, go back and nd the answer(s) in the book or the notes. 3. Go back and look over HW problems. 4. Be careful of your not ...
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exam2guide
Kentucky, AS 603
Excerpt: ... Study guide for exam 2 Exam 2 is closed book and computer o. You are allowed to bring in two 8.5in 11in pieces of paper. You may have both sides of the paper lled with notes and formulas. Sections of the book that roughly coincide with material presented in class are Appendix C, 1.2, 1.3, 2.2, 2.3, 2.4, 2.6, 2.8, 3.2 (except 3.2.1), 3.6, 4.1, and 4.2. Of course material in appendices A and B may be needed indirectly. It is essential that you know the answer to this exercise: Y N(, V ); What is the distribution of BY + b? Material not in book but covered in class includes 1. Deriving formulas for estimating linear combinations from normal equations. 2. Estimating linear combinations based on reduction of the model to echelon from. 3. Deriving formulas for comparing models from normal equations. 4. Comparing models based on reduction of the model to echelon from. 1 ...
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problems_II(contd)
Vanderbilt, PHYS 330
Excerpt: ... PROBLEMS II (cont.) 6. Suppose a boson has an angular momentum j=2 and therefore mj = 0, 1, 2; there is also a fermion that has the angular momentum j=5/2, and hence mj = 1/2, 3/2, 5/2. For N=2 in both cases, evaluate the total J and find the allowed values of J for which a physical realization is possible. Note that MJ = mj1 + mj2 (consider only the non-negative values of MJ). Compare the results. 7. Determine the total values and also the allowed values of J for the jjcoupled configurations A. (5/2 3/2) B. (3/2)2 8. Knowing that [Jx , Jy] = i J z , [J y ,Jz] = i J x , and [Jz ,J x] = i J y show that if J = Jx i Jy J2 = 1/2 ( J+ J- + J- J+) + Jz2 9. In the case of f2 determine the functions of good symmetry |1I04>, |3H04>, |1G04>, |3H14> as linear combinations of determinantal states (use the results from the lecture) 10. Discuss how you could determine eigenstates |3HJM> as linear combinations of the states |3HMSML>. Hint: use the fact that J = L + S ...
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week9
N.C. State, ST 511
Excerpt: ... Statistics 511 Relevant Sections: 9.2-9.3 Recall the ANOVA F Test Hypotheses: Suppose we reject H0. Which means differ? (9.2) Preliminaries: Linear Contrasts Linear combination of means: Ex. 1 Compare 2 to 3 . Ex. 1 Compare 2 to the average o ...
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hw1
Berkeley, HISTORY 190
Excerpt: ... Physics H190 Spring 2003 Homework 1 Due Wednesday, January 29, 2003 Reading Assignment: Read pp. 112 of the book, and study lecture notes for Jan. 22. 1. Prove the following theorem in quantum mechanics. If A and B are two observables such that [A, ...
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studyguide-final
N.C. State, ST 512
Excerpt: ... Study/reading list for ST512 Exam 3 Osborne All of the lecture notes are fair game for exam 3, but most of the questions will come from material in Chapters 14-16, An outline of the topics covered in this latter section appears below, with some indications of where certain topics are covered in Rao's text. 1. Mixed models: experiments with two or more factors, some of which may be random (14) nesting of factors (13.7,14.6) variance components (14) correlation structures constructing the right F -ratio (Table 14.1) expected mean squares (14.7) standard errors of contrasts involving random effects Satterthwaite approximation for linear combinations of mean squares 2. Designs with blocking factors (15) the RCBD ANOVA the ways in which inference agrees or disagrees according to fixed or random effects modelling of blocking factors latin squares (15.4) 3. Split-plot experiments (16.1-16.3) Definitions Whole plot factors Whole plot units subplot factors subplot units (All o ...
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Sept9notes
UNL, MATH 189
Excerpt: ... 8 Joy of Numbers Tuesday, September 9 5. Linear Combinations We recalled the Euclidean Algorithm from last class: Let a and b be two integers (assume b = 0). Dividing b into a we get a = bq1 + r1 with 0 r1 < b. If r1 is not zero, we can divide r1 into b: b = r1 q2 + r2 with 0 r2 < r1 . If r2 = 0, we repeat the process: r1 = r2 q3 + r3 with 0 r3 < r2 . Eventually, we get down to a remainder of zero: rn1 = rn qn+1 + 0. The rst homework question was, why do we eventually get to a remainder of zero? In other words, why must the Euclidean Algorithm terminate? Megan explained that since the remainders are getting smaller and are always nonnegative, eventually we must reach a remainder of zero. In other words, since we have a sequence of nonnegative integers b > r1 > r2 > r3 > , we must have r1 b 1, r2 b 2, r3 b 3 so that the process must terminate in at most b steps. The next homework problem was to explain why the last nonzero remainder in the Euclidean Algo ...
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211lec16
ECCD, MATH 211
Excerpt: ... e type of structure as the vector space Rn , it is just smaller. By using the second and third properties, it is easy to see that the following property also holds. Proposition: If x1 , . . . , xk are in N and a1 , . . . , ak are real numbers, then the vector a1 x1 + + ak xk is also in N . The type of sum in the last proposition is called a linear combination of the vectors x1 , . . . , xk . Examples: Find the nullspaces of -1 -2 3 1 2 1 2 4 1 and 1 1 0 0 -1 1 1 1 1 3 1 1 0 4 2 0 In all of these examples, the nullspace is given by linear combinations of just a few vectors. To formalize this we use the definition. Suppose that we are given vectors x1 , . . . , xk . The span of these vectors is the set of all of their linear combinations . Sometimes this is denoted span(x1 , . . . , xk ). It is not difficult to see that the span of a set of vectors is also a subspace of Rn . It would be nice to be able to pick a smallest subset of the nullspace which spans it. This will enable us to avoid repet ...
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hw9
BYU, STAT 512
Excerpt: ... Statistics 512 Winter 2003: Grimshaw Homework 9: Due Tuesday 15 April 2003 Chapter 17 # 4 Note: Add a short description of cluster analysis and explain why cluster analysis is not appropriate for this problem. #7 # 9 Note: Entertain rotating the axes to obtain more interpretable linear combinations for this problem. # 13 1 ...
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h-ReviewTest1-M340LFall2007
University of Texas, M 340
Excerpt: ... w") and scalar multiplication of vectors. Linear combinations of vectors. Parametric representation of a straight line in the plane. Sketching straight lines in the plane. Linear geometry in 3-space: vectors, their geometric representations (arrows), vector addition, scalar multiplication. Linear combinations of vectors. Review arithmetic properties (distribution laws) of the real numbers. Algebraic properties of vector addition and scalar multiplication of vectors in Rn (i.e. analogues of the distribution laws of numbers hold for vector addition and scalar multiplication in Rn ). The span of a set of vectors in R2 , R3 , and Rn . The same span can have many different spanning sets. Zero vector 0n = origin = additive identity of Rn . 3. Matrix-vector multiplication. Vector form of a system of linear equations. Vector notation (x1 a1 + +xn an = b) and matrix notation (Ax = b) for a linear system. Definition of matrix/vector multiplication (linear combination of columns). A linear system A x ...
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lect10
Washington, B 533
Excerpt: ... 10. ESTIMABLE FUNCTIONS AND GAUSS-MARKOV THEOREM 1 10.1. Best Linear Unbiased Estimates Definition: The Best Linear Unbiased Estimate (BLUE) of a parameter based on data Y is 1. a linear function of Y. That is, the estimator can be written as b Y, 2. unbiased (E[b Y] = ), and 3. has the smallest variance among all unbiased linear estimators. ^ Theorem 10.1.1: For any linear combination c , c Y is the ^ BLUE of c , where Y is the least-squares orthogonal projection of Y onto R(X). Proof: See lecture notes # 8 ^ Corollary 10.1.2: If rank(Xnp) = p, then, for any a, a is the BLUE of a . Note: The Gauss-Markov theorem generalizes this result to the less than full rank case, for certain linear combinations a (the estimable functions). 2 10. ESTIMABLE FUNCTIONS AND GAUSS-MARKOV THEOREM Proof of Corollary 10.1.2: X (X X)-1 X a = = = = X X X (X X)-1X X = a (X X)-1X c So a = c where c = a (X X)-1X . ^ Now, a = a (X X)-1X Y and ^ ^ c Y = a (X X)-1X Y = a (X X)-1X X(X X)-1 X Y = a (X X)-1X Y ^ ^ Therefo ...
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SAO
Laurentian, CHEM 3820
Excerpt: ... Symmetry-Adapted Linear Combinations of Atomic Orbitals 1 2 3 4 5 6 7 ...
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L18_LinearCombinations
Laurentian, MATH 1057
Excerpt: ... Linear Combinations of Vectors Lecture 18, MATH 1057E Julien Dompierre Dpartement de mathmatiques et d'informatique e e Universit Laurentienne e 27 fvrier 2007, Sudbury e Julien Dompierre 1 Linear Combination (p. 208) Definition Let v1 , v2 , . . . , vm be vectors in a vector space V . We sat that v, a vector in V , is a linear combination of v1 , v2 , . . . , vm if there exist scalars c1 , c2 , . . . , cm such that v can be written as v = c 1 v1 + c 2 v2 + + c m vm . Julien Dompierre 2 Spanning Set (p. 212) Definition The vectors v1 , v2 , . . . , vm are said to span a vector space if every vector in the space can be expressed as a linear combination of these vectors. A spanning set of vectors in a sense defines the vector space, since every vector in the space can be obtained from this set. Julien Dompierre 3 Generating a Vector Space (p. 214) Theorem Let v1 , v2 , . . . , vm be vectors in a vector space V . Let U be the set consisting of all linear combinations of v1 , v2 , . . . ...
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ma265 lecture notes
Purdue, MA 265
Excerpt: ... AMPLE 3.2.1 Let A Mmn (F ). Then the set of vectors X F n satisfying AX = 0 is a subspace of F n called the null space of A and is denoted here by N (A). (It is sometimes called the solution space of A.) 55 56 CHAPTER 3. SUBSPACES Proof. (1) A0 = 0, so 0 N (A); (2) If X, Y N (A), then AX = 0 and AY = 0, so A(X + Y ) = AX + AY = 0 + 0 = 0 and so X + Y N (A); (3) If X N (A) and t F , then A(tX) = t(AX) = t0 = 0, so tX N (A). 1 0 , then N (A) = {0}, the set consisting of 0 1 1 2 just the zero vector. If A = , then N (A) is the set of all scalar 2 4 multiples of [-2, 1]t . For example, if A = EXAMPLE 3.2.2 Let X1 , . . . , Xm F n . Then the set consisting of all linear combinations x1 X1 + + xm Xm , where x1 , . . . , xm F , is a subspace of F n . This subspace is called the subspace spanned or generated by X1 , . . . , Xm and is denoted here by X1 , . . . , Xm . We also call X1 , . . . , Xm a spanning family for S = X1 , . . . , Xm . Proof. (1) 0 = 0X1 + + 0Xm , so 0 X1 , . . . ...
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notes_Lecture_12&13
Washington, CHEM 455
Excerpt: ... at linearity of A was used "several times" in this simple demonsteation. Note also that if any one of the funcions fi in the sum didn't have eigenvalue "a", then the a would not factor outsidet he sum, and the result would be false. Said specifically: " linear combinations of non-degenerate eigenfuctions" of an operator are NOT eigenfunctions of that opertor. As the above result has nothing to do with "what linear combination" has been taken, it applies equally to all linear combinations . Can we then generate an infintie number of eigenfuctions of operator A? Of course Note that linearity of A was used "several times" in this simple demonsteation. Note also that if any one of the funcions fi in the sum didn't have eigenvalue "a", then the a would not factor outsidet he sum, and the result would be false. Said specifically: " linear combinations of non-degenerate eigenfuctions" of an operator are NOT eigenfunctions of that opertor. Chem455A_Lecture12,13.nb 5 As the above result has nothing to do with "what l ...
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lecture1
University of Illinois, Urbana Champaign, JVANHA 225
Excerpt: ... ors from top to bottom would be confusing with vectors from R(mn) . Instead, they are organized from left to right. Addition and scalar multiplication of vectors are organized similar to addition and scalar multiplication of vectors. Example: 0 1 3 1 2 4 1 1 1 2 3 4 + -2 3 1 = 0 6 5 0 -2 10 1 -2 9 1 0 -1 1.5 2 0 4 -6 -10 8 0 112 = 3 0 6 -9 -15 0 12 -18 2 2.1 Matrix-Vector Product Linear Combinations Vectors of the same length can be added, and multiplied with a scalar. Together these are called linear combinations . For example: 1 0 2 2 2 - 4 -1 = 8 3 2 -2 1 0 is a linear combination of 2 and -1 . 3 2 Definition. A linear combination of two vectors u, v in Rm is any vector that can be written as au + bv for some choice of weights a, b in R. Linear combinations of linear combinations of u, v are still linear combinations of u, v. 2 Proof. Take two linear combinations a1 u + b1 v and a2 u + b2 v. Both of these are vectors in Rm , so we can take a linear combination with weights ...
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