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Unformatted text preview: Eigen Methods Math 246, Fall 2009, Professor David Levermore Eigenpairs. Let A be a real n n matrix. A number (possibly complex) is an eigenvalue of A if there exists a nonzero vector v (possibly complex) such that (1) Av = v . Each such vector is an eigenvector associated with , and ( , v ) is an eigenpair of A . Fact 1: If ( , v ) is an eigenpair of A then so is ( , v ) for every complex negationslash = 0. In other words, if v is an eigenvector associated with an eigenvalue of A then so is v for every complex negationslash = 0. In particular, eigenvectors are not unique. Reason. Because ( , v ) is an eigenpair of A you know that (1) holds. It follows that A ( v ) = Av = v = ( v ) . Because the scalar and vector v are nonzero, the vector v is also nonzero. Therefore ( , v ) is also an eigenpair of A . square Finding Eigenvalues. Recall that the characteristic polynomial of A is defined by (2) p A ( z ) = det( z I A ) . It has the form p A ( z ) = z n + 1 z n 1 + 2 z n 2 + + n 1 z + n , where the coefficients 1 , 2 , , n are real. In other words, it is a real monic polynomial of degree n . One can show that in general 1 = tr( A ) , n = ( 1) n det( A ) . In particular, when n = 2 one has p A ( z ) = z 2 tr( A ) z + det( A ) . Because det( z I A ) = ( 1) n det( A z I ), this definition of p A ( z ) coincides with the books definition when n is even, and is its negative when n is odd. Both conventions are common. We have chosen the convention that makes p A ( z ) monic. What matters most about p A ( z ) is its roots and their multiplicity, which are the same for both conventions. Fact 2: A number is an eigenvalue of A if and only if p A ( ) = 0. In other words, the eigenvalues of A are the roots of p A ( z ). Reason. If is an eigenvalue of A then by (1) there exists a nonzero vector v such that ( I A ) v = v Av = 0 . It follows that p A ( ) = det( I A ) = 0. Conversely, if p A ( ) = det( I A ) = 0 then there exists a nonzero vector v such that ( I A ) v = 0. It follows that v Av = ( I A ) v = 0 , whereby and v satisfy (1), which implies is an eigenvalue of A . square Fact 2 shows that the eigenvalues of a n n matrix A can be found if you can find all the roots of the characteristic polynomial of A . Because the degree of this characteristic polynomial is n , and because every polynomical of degree n has exactly n roots counting multiplicity, the n n matrix A therefore must have at least one eigenvalue and at most n eigenvalues. 1 2 Example. Find the eigenvalues of A = parenleftbigg 3 2 2 3 parenrightbigg ....
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 Spring '10
 LEVERMORE

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