MATH535_HW8 - Serge Ballif MATH 535 Homework 8 The goal of...

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Serge Ballif MATH 535 Homework 8 October 26, 2007 The goal of this homework set is to establish the, perhaps surprising, analogue for finite dimensional vector spaces for the problems of last weeks homework. For this set, we fix a linear transformation L on n given by a square matrix A . This produces what we shall call an action of [ X ] on V : ( P ( x ) , v ) 7→ P ( A ) v := d i =0 a i A i v , when P ( X ) = d i =0 a i X i . Notice that it is easy to verify that 1 v = v , for al v V . P ( X )( v 1 + v 2 ) = P ( X ) v 1 + P ( X ) v 2 , for all P ( X ) [ X ], v 1 , v - 2 V . ( P ( X ) + Q ( X )) v = P ( X ) v + Q ( X ) v , for all P ( X ) , Q ( X ) [ X ], v V . ( P ( X ) Q ( X )) v = P ( X )( Q ( X ) v ), for all P ( X ) , Q ( X ) F [ X ], v V . This is summarized by saying that V is a module over F [ X ]. We say that a non-zero v V has order P ( X ) if P ( X ) is the monic polynomial in [ X ] of least degree such that P ( X ) v = 0. 1. A subspace U of V is said to be cyclic if, for some u U , U = u + Au + · · · + A k u . Show that the subspace U of V is cyclic if and only if it contains an element u whose order has degree equal to the dimension of U . Such an element is called a cyclic vector for U . ( ) Suppose that U is cyclic with minimal polynomial M ( x ). Then for some u U , U = u + Au + · · · + A d u . In particular, 0 = a 0 u + a 1 Au + · · · + a k - 1 A d - 1 u + A d u for suitable choice of a i . Define Q ( x ) = d i =0 a i X i to be the polynomial that gives the above action on u . Then Q ( x ) is a monic polynomial of degree dim U such that Q ( A ) u = 0. We claim that Q ( x ) is the order of u . This is clear because the linear independence of the elements u, Au, . . . , A d u guarantees that there is no polynomial P ( x ) of degree less than d for which P ( x ) u = 0.
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