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# Class Note for MATH 456 at UMass(6)

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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Massachusetts taught by a professor in Fall. Since its upload, it has received 25 views.

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Date Created: 02/06/15
Chapter 1 Overview of Matrix Algebra Usually one of the most important tools in understanding whether two objects are close is that of the norm If the said objects are vectors then we use the vector norm to compute their distance Let us de ne this tool De nition 1 A vector norm denoted by H is ufunction which has vectors as input and produces a number as output In other words E R a R The vector norm has the following properties a Z 0 for all x E R h 0 if and only if x E 0 c for allkER undelR 03 HXYH S HXH HYH f0 all Xiy E R There are several different kinds of vector norms We most often use the following types of norms in our calculations De nition 2 The Euclidean or l2 norm HXH2 and the maximum or loO norm HXHoo Egg W or the l1 norm 7L HXH1 Z W i1 Example Calculate all three the l1 Euclidean and maximum norm for the vector 90 1 732 Solution The l1 norm is HXH1 l1ll 3ll2l 6 the Euclidean norm is llxllz 2902 x x x 17322214 i1 while the maximum norm gives Hxlloo max W maxfl cll 19021719031 max132 3 1991 Therefore using these norms we can now describe distances between vectors as follows MK 7 Yllz As a result we are now also in position to understand the concept of convergence for vectors Conver gence is a concept which we will use later in our numerical calculations and will allow us to know how close we are to the solution of a given problem To do this we essentially need to measure usually via some kind of norm the distance from our current approximate solution to the true solution De nition 3 We say that a vector sequence xn io converges to another vector x in R with respect to given norm if given any 6 gt 0 there exist an integer Ne such that Hxn 7 KM g e for all n 2 Ne The following result can now be shown Theorem 4 The sequence of vectors Kn 0 converges to x in R with respect to the Hoe norm if and only if lim for alli 12 n ngtoo Similarly we can prove the following ordering between different norms Theorem 5 For x E R we have Hxlloo S 11th S x llxlloo There are a couple of well known inequalities which apply to vectors The triangle inequality HX Yll S HXH llYll and the Cauchy Schwarz inequality S llxllzllYllz 11 7 E iyi i1 Example Verify the Cauchy Schwarz inequality for the following vectors x 123 and y 021 Solution According to the Cauchy Schwarz inequality the left hand side of 11 gives7 0 4 3 7 while the right hand side of 117 41 4 940 4 14 70 m 836 Thus the right hand side is bigger as expected We now generalize these results First we de ne the following general p norm7 Hxllp VF l11p 190an Thus the following two inequalities hold7 Holder ineq leyl 1 W HX Yllp llelpllYllq Where i ll llp llyllp l 10 Here xT denotes the transpose of vector x Now that we have put together the basic necessary tools regarding vectors we can start discussing how to handle matrices Matrices are just rows of vectors As such we can essentially use the same tools7 from vectors7 apply them to matrices So the matrix norm is de ned via7 De nition 6 Suppose that A and B denote n gtlt n size matrices and k a constant Then the matrix norm is the real valued function has the following properties 61 HAN Z 0 h 0 if and only if A E 0 c for aZZkER d HABH HAM HBH 8 HABH S HAHHBH Based on the already presented vector norms we can in fact de ne new matrix norms The following result can be shown7 Theorem 7 If is any vector norm then HAM max llA cll HXH1 is the corresponding induced matrix norm Therefore we can now refer to the following types of matrix norms7 llAHz max llAXHz 0r llAlloo max llAXHoo llxllz1 llxlloo1 Note that in fact these norms are not very easy to calculate since you must examine all possible vector of length 1 and nd the maximum possible result We present instead a result which allows us to easily calculate one of these matrix norms7 Theorem 8 Suppose that A is an m gtlt n matrix Then the HAHOO norm is calculated as rn HAHoo 2512 Mel j1 Similarly the HAH1 is found from 7L llAlll lfgfgfn laijl i Let us look at an example using this norm Example Suppose the following matrix is given7 1 2 A Z l s 4 l Calculate Solution The norm is simply found by simply adding all elements in each row and obtaining the maximum result from there7 l1ll2l l3ll4l 7 Thus HAHOO max377 77 while check this llAlll 6 101 Eigenvalues eigenvectors and condition number As you noticed it was not easy to obtain the matrix norm for some cases such as For this kind of norm we can develop another method which will allow us to obtain the value of the matrix norm in an easy way This method relies on eigenvalues and eigenvectors First some de nitions De nition 9 Suppose that A is an n gtlt n square matrix Then the following polynomial p detA 7 AI is the so called eharaeteristie polynomial Note that p is an nth degree polynomial De nition 10 Suppose that the characteristic polynomialp as de ned above The zeros ofp are called eigenvalues for the matrix A Iffor a given eigenvalue A we have that A 7 Ax 0 with 90 3A 0 then x is called an eigenveetor corresponding to the eigenvalue A Let us look at a simple example on how to obtain the eigenvalues and eigenvectors for a given matrix A Example Suppose the following matrix is given 1 1 A 7 l 72 4 l Find the eigenvalues and eigenvectors for A Solution To nd the eigenvalues we solve the following characteristic polynomial p based on De nition 9 17x47x20 The polynomial simpli es to A2 7 5A 6 0 and the solutions are A1 2 and A2 3 To nd the corresponding eigenvectors we must solve the following matrix system for each eigenvalue 153 xlliiHSl where here 221212 represented the unknown eigenvector For A1 2 we must solve the following system 71 1 v1 7 0 i2 2 v2 7 0 21122 which gives that Thus an eigenvector for A1 2 is 11 Similarly the eigenvector for A2 3 is 12 De nition 11 The spectral radius pA of a given matrix A is given by MA max m where A corresponds to the eigenvalues ofA Now we are nally in position to provide an alternate method for evaluating the matrix norm HAllg based on the eigenvalues of A Theorem 12 Suppose that A is an n gtlt n matrix Then PM llAllz pA g for any matrix norm 13 m HA IHZ for A symmetric 14 Let us see this result in more detail through a numerical example Example Given the following matrix A compute the HAllz norm H21 Solution Let us outline this procedure We rst calculate AtA and then nd its eigenvalues then according to the theorem above the square root of the maximum eigenvalue will be equal to Thus we start by rst calculating7 t 7 0 2 0 1 7 4 2 A A T l 1 1 2 1 T 2 2 The eigenvalues for this matrix are A1 3 x and A2 3 7 Thus Hle V3 x5 Another very important quantity for matrices is the condition number The condition number of a matrix describes how good77 that matrix is For instance matrices that are not invertible have condition number 00 which is considered as bad as possible On the other hand the smaller the condition number the better the matrix behaves in our calculations The condition number is de ned through7 De nition 13 Suppose A E Rn and that A is nonsz39ngulan Then the condition number ofA is given by COMM HAHHA IH Note that if we use the Euclidean norm and further assume that A is symmetric then the condition number is given from the eigenvalues of A as follows7 pA 3A 7 male minlM T minlM 00ndAllAllzllA 1Hz The above calculation is true but7 at least here7 we do not show the details Can you show why this is true based on what you learned so far

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