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CHAPTER 8 Linear Algebra: Matrix Eigenvalue Problems Prerequisite for this chapter is some familiarity with the notion of a matrix and with the two algebraic operations for matrices. Otherwise the chapter is independent of Chap. 7, so that it can be used for teaching eigenvalue problems and their applications, without first going through the material in Chap. 7. SECTION 8.1. Eigenvalues, Eigenvectors, page 334 Purpose. To familiarize the student with the determination of eigenvalues and eigenvectors of real matrices and to give a first impression of what one can expect (multiple eigenvalues, complex eigenvalues, etc.). Main Content, Important Concepts Eigenvalue, eigenvector Determination of eigenvalues from the characteristic equation Determination of eigenvectors Algebraic and geometric multiplicity, defect Comments on Content To maintain undivided attention on the basic concepts and techniques, all the examples in this section are formal, and typical applications are put into a separate section (Sec. 8.2). The distinction between the algebraic and geometric multiplicity is mentioned in this early section, and the idea of a basis of eigenvectors (“eigenbasis”) could perhaps be mentioned briefly in class, whereas a thorough discussion of this in a later section (Sec. 8.4) will profit from the increased experience with eigenvalue problems, which the student will have gained at that later time. The possibility of normalizing any eigenvector is mentioned in connection with Theorem 2, but this will be of greater interest to us only in connection with orthonormal or unitary systems (Secs. 8.4 and 8.5). In our present work we find eigen values first and are then left with the much simpler task of determining corresponding eigen vectors. Numeric work (Secs. 20.6 2 20.9) may proceed in the opposite order, but to mention this here would perhaps just confuse the student. SOLUTIONS TO PROBLEM SET 8.1, page 338 2. The characteristic equation is D ( l ) 5 ( a 2 )( c 2 ) 5 0. Hence 1 5 a , 2 5 c . Components of eigenvectors can now be determined for 1 5 a from 0 x 1 1 bx 2 5 0, say, x 1 5 1, x 2 5 0 so that an eigenvector is [1 0] T , and for 2 5 c from ( a 2 c ) x 1 1 bx 2 5 0, say, x 1 5 1, x 2 5 ( c 2 a )/ b , so that an eigenvector is [ 1 ] T . c 2 a } b 161 im08.qxd 9/21/05 12:12 PM Page 161

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Here we must assume that b Þ 0. If b 5 0, we have a diagonal matrix with the same eigenvalues as before and eigenvectors [1 0] T and [0 1] T . 4. This zero matrix, like any square zero matrix, has the eigenvalue 0, whose algebraic multiplicity and geometric multiplicity are both equal to 2, and we can choose any basis, for instance [1 0] T and [0 1] T . 6. The characteristic equation is ( a 2 l ) 2 1 b 2 5 0. Solutions are 5 a 6 ib . Eigenvectors are obtained from ( a 2 ) x 1 1 bx 2 57 ibx 1 1 bx 2 5 0. Hence we can take x 1 5 1 and x 2 56 i . Note that b has dropped out, and the eigenvectors are the same as in Example 4 of the text.
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ch08 - im08.qxd 12:12 PM Page 161 CHAPTER 8 Linear Algebra...

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