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chap04 - symmetric matrices

# chap04 - symmetric matrices - Symmetric matrices and the...

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Symmetric matrices and the second derivative test 1 Chapter 4 Symmetric matrices and the second derivative test In this chapter we are going to finish our description of the nature of nondegenerate critical points. But first we need to discuss some fascinating and important features of square matrices. A. Eigenvalues and eigenvectors Suppose that A = ( a ij ) is a fixed n × n matrix. We are going to discuss linear equations of the form Ax = λx, where x R n and λ R . (We sometimes will allow x C n and λ C .) Of course, x = 0 is always a solution of this equation, but not an interesting one. We say x is a nontrivial solution if it satisfies the equation and x 6 = 0. DEFINITION. If Ax = λx and x 6 = 0, we say that λ is an eigenvalue of A and that the vector x is an eigenvector of A corresponding to λ . EXAMPLE. Let A = 0 3 1 2 . Then we notice that A 1 1 = 3 3 = 3 1 1 , so 1 1 is an eigenvector corresponding to the eigenvalue 3. Also, A 3 - 1 = - 3 1 = - 3 - 1 , so 3 - 1 is an eigenvector corresponding to the eigenvalue - 1. EXAMPLE. Let A = 2 1 0 0 . Then A 1 0 = 2 1 0 , so 2 is an eigenvalue, and A 1 - 2 = 0 0 , so 0 is also an eigenvalue. REMARK. The German word for eigenvalue is eigenwert . A literal translation into English would be “characteristic value,” and this phrase appears in a few texts. The English word “eigenvalue” is clearly a sort of half translation, half transliteration, but this hybrid has stuck.

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2 Chapter 4 PROBLEM 4–1. Show that A is invertible ⇐⇒ 0 is not an eigenvalue of A . The equation Ax = λx can be rewritten as Ax = λIx , and then as ( A - λI ) x = 0. In order that this equation have a nonzero x as a solution, Problem 3–52 shows that it is necessary and sufficient that det( A - λI ) = 0 . (Otherwise Cramer’s rule yields x = 0.) This equation is quite interesting. The quantity det a 11 - λ a 12 . . . a 1 n a 21 a 22 - λ . . . a 2 n . . . a n 1 a n 2 . . . a nn - λ can in principle be written out in detail, and it is then seen that it is a polynomial in λ of degree n . This polynomial is called the characteristic polynomial of A ; perhaps it would be more consistent to call it the eigen polynomial, but no one seems to do this. The only term in the expansion of the determinant which contains n factors involving λ is the product ( a 11 - λ )( a 22 - λ ) . . . ( a nn - λ ) . Thus the coefficient of λ n in the characteristic polynomial is ( - 1) n . In fact, that product is also the only term which contains as many as n - 1 factors involving λ , so the coefficient of λ n - 1 is ( - 1) n - 1 ( a 11 + a 22 + · · · + a nn ). This introduces us to an important number associated with the matrix A , called the trace of A : trace A = a 11 + a 22 + · · · + a nn . Notice also that the polynomial det( A - λI ) evaluated at λ = 0 is just det A , so this is the constant term of the characteristic polynomial. In summary, det( A - λI ) = ( - 1) n λ n + ( - 1) n - 1 (trace A ) λ n - 1 + · · · + det A.
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chap04 - symmetric matrices - Symmetric matrices and the...

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