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(i) Show that where à = WA, b {wi}/²1. m Σw? ((xWLS + xWLSai) – bi)² = ||ÃxWLS — Ã||², i=1 = Wb and W Rmxm is the diagonal matrix with diagonal entries

(i) Show that where à = WA, b {wi}/²1. m Σw? ((xWLS + xWLSai) – bi)² = ||ÃxWLS — Ã||², i=1 = Wb and W Rmxm is the diagonal matrix with diagonal entries

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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(ii) Show that XWLS solves ATW2AXWLS ATW²b. (iii) Suppose rank(A) = 2. Since W is invertible, we then also have rank(WA) = 2 Verify that XWLS – XLS = (ATW²A)-¹A¹(W² — I)(b – Ax¹S). = You do not need to derive the expression on the right hand side from scratch, it is sufficient to show that it simplifies to the quantity on the left hand side. (iv) Based on the expression in part (iii), briefly discuss your expectations on ||XWLS - XLS || in the following situations: (a) All of the weights {w} are chosen equal to or close to 1. (b) There is a straight line that exactly passes through all of the points {ai, bi}1.
Transcribed Image Text:(ii) Show that XWLS solves ATW2AXWLS ATW²b. (iii) Suppose rank(A) = 2. Since W is invertible, we then also have rank(WA) = 2 Verify that XWLS – XLS = (ATW²A)-¹A¹(W² — I)(b – Ax¹S). = You do not need to derive the expression on the right hand side from scratch, it is sufficient to show that it simplifies to the quantity on the left hand side. (iv) Based on the expression in part (iii), briefly discuss your expectations on ||XWLS - XLS || in the following situations: (a) All of the weights {w} are chosen equal to or close to 1. (b) There is a straight line that exactly passes through all of the points {ai, bi}1.
Exercise 1. Suppose we are given a set of points {ai, bi}. Then, as in the lecture notes, we can find the straight line b = x₁ + x2a that best fits these observations by choosing xLS € R² to minimise m Σ ((xLS + x²Sai) − b;)² = ||AxLS — b||², - i=1 where b Rm is the vector containing {b}₁, and the matrix A € Rmx2 is the Vandermonde matrix with entries aij = a-¹. In practice, some of the data points {ai, bi} might be more reliable or more important than others. In this case, we can choose a set of positive weights {w}₁CR and instead choose XWLSER2 to minimise m Σw ((xWLS + xy i=1 xWLSai) – bi) ². A large weight we will force the minimiser xWLS to make ((xWLS + xWLSai) – bi)² small, whereas a small weight will allow ((xWLS + xWLSai) – bi) ² to be quite large. (i) Show that m Σw ((xWLS + xWLS ai) – bi)² = ||ÃxWLS – B||², i=1 where à = WA, b = Wb and We Rmxm is the diagonal matrix with diagonal entries {wi}=1. 1
Transcribed Image Text:Exercise 1. Suppose we are given a set of points {ai, bi}. Then, as in the lecture notes, we can find the straight line b = x₁ + x2a that best fits these observations by choosing xLS € R² to minimise m Σ ((xLS + x²Sai) − b;)² = ||AxLS — b||², - i=1 where b Rm is the vector containing {b}₁, and the matrix A € Rmx2 is the Vandermonde matrix with entries aij = a-¹. In practice, some of the data points {ai, bi} might be more reliable or more important than others. In this case, we can choose a set of positive weights {w}₁CR and instead choose XWLSER2 to minimise m Σw ((xWLS + xy i=1 xWLSai) – bi) ². A large weight we will force the minimiser xWLS to make ((xWLS + xWLSai) – bi)² small, whereas a small weight will allow ((xWLS + xWLSai) – bi) ² to be quite large. (i) Show that m Σw ((xWLS + xWLS ai) – bi)² = ||ÃxWLS – B||², i=1 where à = WA, b = Wb and We Rmxm is the diagonal matrix with diagonal entries {wi}=1. 1
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