eigenv5b

# eigenv5b - Matrices A(t depending on a Parameter t Math 412...

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Matrices A ( t ) depending on a Parameter t Math. 412, Fall 1995 Jerry L. Kazdan If a square matrix A ( t ) depends smoothly on a parameter t , are its eigenvalues and eigen- vectors also smooth functions of t ? The answer is “yes” most of the time, but not always. This story, while old, is interesting and elementary — and deserves to be better known. One can also ask the same question for objects such as the Schr¨odinger operator whose potential depends on a parameter, where much of current understanding arose. Warm-up Exercise Given a polynomial p ( x , t ) = x n + a n - 1 ( t ) x n - 1 + ··· + a 1 ( t ) x + a 0 ( t ) whose coef±cients depend smoothly on a parameter t . Assume at t = 0 the number x = c is a simple root of this polynomial, p ( c , 0 ) = 0. Show that for all t suf±ciently near 0 there is a unique root x ( t ) with x ( 0 ) = c that depends smoothly on t . Moreover, if p ( x , t ) is a real analytic function of t , that is, it has a convergent power series expansion in t near t = 0, then so does x ( t ) . S OLUTION : Given that p ( c , 0 ) = 0 we want to solve p ( x , t ) = 0 for x ( t ) with x ( 0 ) = c . The assertions are immediate from the implicit function theorem. Since x ( 0 ) = c is a simple zero of p ( x , 0 ) = 0, then p ( x , 0 ) = ( x - c ) g ( x ) , where g ( c ) 6 = 0. Thus the derivative p x ( c , 0 ) 6 = 0. The example p ( x , t ) : = x 3 - t = 0, so x ( t ) = t 1 / 3 , shows x ( t ) may not be a smooth function at a multiple root. In this case the best one can get is a Puiseux expansion in fractional powers of t (see [Kn, ¶15]). The Generic Case: a simple eigenvalue In the following, let λ be an eigenvalue and X a corresponding eigenvector of a matrix A . We say λ is a simple eigenvalue if λ is a simple root of the characteristic polynomial. We will use the equivalent version: if ( A - λ ) 2 V = 0 , then V = cX for some constant c . The point is to eliminate matrices such as the zero 2 × 2 matrix A = 0, where λ = 0 is a double eigenvalue and any vector V 6 = 0 is an eigenvector, as well as the more complicated matrix A = ( 0 1 0 0 ) which has λ = 0 as an eigenvalue with geometric multiplicity one but algebraic multiplicity two. Theorem

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## This note was uploaded on 02/07/2011 for the course MATH 412 taught by Professor Davidson during the Spring '11 term at UC Riverside.

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eigenv5b - Matrices A(t depending on a Parameter t Math 412...

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