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# Eigen - Eigen Methods Math 246 Fall 2009 Professor David...

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Eigen Methods Math 246, Fall 2009, Professor David Levermore Eigenpairs. Let A be a real n × n matrix. A number λ (possibly complex) is an eigenvalue of A if there exists a nonzero vector v (possibly complex) such that (1) Av = λ v . Each such vector is an eigenvector associated with λ , and ( λ, v ) is an eigenpair of A . Fact 1: If ( λ, v ) is an eigenpair of A then so is ( λ, α v ) for every complex α negationslash = 0. In other words, if v is an eigenvector associated with an eigenvalue λ of A then so is α v for every complex α negationslash = 0. In particular, eigenvectors are not unique. Reason. Because ( λ, v ) is an eigenpair of A you know that (1) holds. It follows that A ( α v ) = α Av = αλ v = λ ( α v ) . Because the scalar α and vector v are nonzero, the vector α v is also nonzero. Therefore ( λ, α 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 + π 1 z n - 1 + π 2 z n - 2 + · · · + π n - 1 z + π n , where the coefficients π 1 , π 2 , · · · , π n are real. In other words, it is a real monic polynomial of degree n . One can show that in general π 1 = tr( A ) , π 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 book’s 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 λ is an eigenvalue of A if and only if p A ( λ ) = 0. In other words, the eigenvalues of A are the roots of p A ( z ). Reason. If λ is an eigenvalue of A then by (1) there exists a nonzero vector v such that ( λ I A ) v = λ v Av = 0 . It follows that p A ( λ ) = det( λ I A ) = 0. Conversely, if p A ( λ ) = det( λ I A ) = 0 then there exists a nonzero vector v such that ( λ I A ) v = 0. It follows that λ v Av = ( λ I A ) v = 0 , whereby λ and v satisfy (1), which implies λ 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

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2 Example. Find the eigenvalues of A = parenleftbigg 3 2 2 3 parenrightbigg . Solution. The characteristic polynomial of A is p A ( z ) = z 2 6 z + 5 = ( z 1)( z 5) . By Fact 2 the eigenvalues of A are 1 and 5. Example. Find the eigenvalues of A = parenleftbigg 3 2 2 3 parenrightbigg . Solution. The characteristic polynomial of A is p A ( z ) = z 2 6 z + 13 = ( z 3) 2 + 4 = ( z 3) 2 + 2 2 .
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Eigen - Eigen Methods Math 246 Fall 2009 Professor David...

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