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# cse11 - CHAPTER 1 APPLIED LINEAR ALGEBRA 1.1 FOUR SPECIAL...

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CHAPTER 1 APPLIED LINEAR ALGEBRA 1.1 FOUR SPECIAL MATRICES An m by n matrix has m rows and n columns and mn entries. We operate on those rows and columns to solve linear systems Ax = b and eigenvalue problems Ax = λx . From inputs A and b (and from software like MATLAB) we get outputs x and λ .A fast stable algorithm is extremely important, and this book includes fast algorithms. One purpose of matrices is to store information, but another viewpoint is more important for applied mathematics. Often we see the matrix as an “operator.” A acts on vectors x to produce Ax . The components of x have a meaning— displacements or pressures or voltages or prices or concentrations. The operator A also has a meaning—in this chapter A takes di±erences. Then Ax represents pressure di±erences or voltage drops or price di±erentials. Before we turn the problem over to the machine—and also after, when we interpret A \ b or eig(A) —it is the meaning we want, as well as the numbers. This book begins with four special families of matrices —simple and useful, absolutely basic. We look ²rst at the properties of these particular matrices K n ,C n , T n ,and B n . (Some properties are obvious, others are hidden.) It is terri²c to practice linear algebra by working with genuinely important matrices. Here are K 2 ,K 3 4 in the ²rst family, with 1 and 2 and 1 down the diagonals: K 2 = · 2 1 12 ¸ K 3 = 2 10 1 0 K 4 = 2 1 00 1 0 0 1 What is signi²cant about K 2 and K 3 and K 4 , and eventually the n n matrix K n ? I will give six answers in the same order that my class gave them—starting with four properties of the K ’s that you can see immediately. 1

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2 Chapter 1 Applied Linear Algebra 1. These matrices are symmetric . The entry in row i ,co lumn j also appears in row j i .T h u s K ij = K ji , on opposite sides of the main diagonal. Symmetry can be expressed by transposing the whole matrix at once: K = K T . 2. The matrices K n are sparse . Most of their entries are zero when n gets large. K 1000 has a million entries, but only 1000 + 999 + 999 are nonzero. 3. The nonzeros lie in a “band” around the main diagonal, so each K n is banded . The band has only three diagonals, so these matrices are tridiagonal . Because K is a tridiagonal matrix, Ku = f can be quickly solved. If the unknown vector u has a thousand components, we can Fnd them in a few thousand steps (which take a small fraction of a second). ±or a full matrix of order n = 1000, solving = f would take hundreds of millions of steps. Of course we have to ask if the linear equations have a solution in the Frst place. That question is coming soon. 4. The matrices have constant diagonals . Right away that property wakes up ±ourier. It signiFes that something is not changing when we move in space or time. The problem is shift-invariant or time-invariant. Coeﬃcients are constant.
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## This note was uploaded on 02/22/2012 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.

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cse11 - CHAPTER 1 APPLIED LINEAR ALGEBRA 1.1 FOUR SPECIAL...

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