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Instructor's Solution Manual for Differential Equations and Boundary Value Problems: Computing and Modeling 6th Edition, All Chapters

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Instructor's Solution Manual for Differential Equations and Boundary Value Problems: Computing and Modeling 6th Edition, All ChaptersInstructor's Solution Manual for Differential Equations and Boundary Value Problems: Computing and Modeling 6th Edition, All Chapters

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Institution
Differential Equations, 6th Edition
Course
Differential Equations, 6th Edition











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Institution
Differential Equations, 6th Edition
Course
Differential Equations, 6th Edition

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Uploaded on
February 4, 2025
Number of pages
203
Written in
2024/2025
Type
Exam (elaborations)
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SOLUTION MANUALb




Linear Algebra and Optimization for MachineLearning
b b b b b b




1st Edition by Charu Aggarwal. Chapters 1 – 11
b b b b bb b b b

,Contents


1 Linear Algebra and Optimization: An Introduction
b b b b b 1


2 Linear Transformations and Linear Systems
b b b b 17


3 Diagonalizable Matrices and Eigenvectors b b b 35


4 Optimization Basics: A Machine Learning View
b b b b b 47


5 Optimization Challenges and Advanced Solutions
b b b b 57


6 Lagrangian Relaxation and Duality
b b b 63


7 Singular Value Decomposition
b b 71


8 Matrix Factorization
b 81


9 The Linear Algebra of Similarity
b b b b 89


10 The Linear Algebra of Graphs
b b b b 95


11 Optimization in Computational Graphs
b b b 101

,Chapter 1 b




Linear Algebra and Optimization: An Introduction
b b b b b




1. For any two vectors x and y, which are each of length a, show that (i) x
b b b b b b b b b b b b b b b b




− y is orthogonal to x+y, and (ii) the dot product of x−3y and x+3y is
b b b b b b b b b b b b b b b b b b b b b b




negative.
b




(i) The first is simply
b · −x · x y y using the distributive property of matrix b b b b b b b b b b b b b b b




multiplication. The dot product of a vector with itself is its squared length.
b b b b b b b b b b b b b




Since both vectors are of the same length, it follows that the result is 0. (ii) In
b b b b b b b b b b b b b b b b b




the second case, one can use a similar argument to show that the result is a2 −
b b b b b b b b b b b b b b b b b




9a2, which is negative.
b b b b




2. Consider a situation in which you have three matrices A, B, and C, of sizes b b b b b b b b b b b b b b




10×2, 2×10,and 10×10, respectively.
b b b b b b b b b b b




(a) Suppose you had to compute the matrix product ABC. From an efficiency b b b b b b b b b b b




per- spective, would it computationally make more sense to compute (AB)C or
b b b b b b b b b b b b




would it make more sense to compute A(BC)?
b b b b b b b b




(b) If you had to compute the matrix product CAB, would it make more sense to
b b b b b b b b b b b b b b




compute (CA)B or C(AB)?
b b b b




The main point is to keep the size of the intermediate matrix as small as
b b b b b b b b b b b b b b




possible in order to reduce both computational and space requirements.
b b b b b b b b b b




In the case of ABC, it makes sense to compute BC first. In the case of CAB it
b b b b b b b b b b b b b b b b b b




makes sense to compute CA first. This type of associativity property is
b b b b b b b b b b b b




used frequently in machine learning in order to reduce computational
b b b b b b b b b b




requirements.
b




3. Show that if a matrix A satisfies A = b b b b b b b b AT, then all the diagonal elements
b b b b b b




of the matrix are 0.
b b b b b




Notethat A + AT = 0.However,this matrix also contains twice the diagonal
b b b b b b b b b b b b b b




elements of A on its diagonal. Therefore, the diagonal elements of A must
b b b b b b b b b b b b b




be 0.
b b




4. Show that if we have a matrix satisfying A= AT,then for any column vector x,
b b b b b b b b b b b b b b b b




we have xT Ax = 0.
b b b b b b




Note that the transpose of the scalar xT Ax remains unchanged. Therefore,
b b b b b b b b b b b




1

, b we have
b




xTAx = (xTAx)T = xTATx = −xTAx. Therefore, we have 2xTAx = 0.
b b b b b b b b b b b b b b b b b b




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