finalExamSolutionsPctex

finalExamSolutionsPctex - Math 222 - 200 Solutions Final...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 222 - 200 Solutions Final Exam April 27, 2005 1. (15) What shall I name my pond snake? The top three names are: Eigen from Ashley Hubble Pond, James Pond agent 008 from Amy Hopson Sam, aka ‘Snake at Mike’s’ from Valerie Berlin 2. (25) Let T : V V be a linear transformation from V into V ,where V is assumed to be a finite dimensional vector space over an algebraically closed field F . (a) Suppose M is an invariant subspace under T . Show that M is invariant under T - λI , for any scalar λ I is the identity transformation from V into V .Conv e r se lyshow that if M is invariant under T - λI , for a particular scalar λ ,then M is invariant under T . Since M is an invariant subspace for any ~x M we have λ~x M and T~ x M , and therefore x - M .Thu s , M is invariant under T - λI , for any λ F .I f M is invariant under T - λI , then by the first part of this problem we have M invariant under ( T - λI ) - ( - λ ) I = T ,and M is invariant under T . (b) Let c be an eigenvalue of T . Show that ac is an eigenvalue of the linear transformation aT , and in general if f ( t )= a 0 + a 1 t + ··· + a n t n is any polynomial with coefficients a i F f ( c ) is an eigenvalue of the linear transformation f ( T a 0 I + a 1 T + + a n T n . Let be any eigenvector associated with the eigenvalue c .Tha ti s , x = c~x .Th i s equation implies T n = T n - 1 x = T n - 1 ( cT n - 1 = c ( c n - 1 ) = c n . We also have ( aT )( ~ x a ( )=( ac ) . In summary ac is an eigenvalue of aT and c n is an eigenvalue of T n , and any eigenvector of T associated with the eigenvalue c is an eigenvector of aT and T n associated with the eigenvalues ac and c n respectively. Thus, we have f ( T ~ x a 0 I + a 1 T + + a n T n ~ x ) = a 0 I ( ~ x )+( a 1 T ~ x )+ +( a n T n ~ x ) = a 0 ~ x + a 1 ( + a n ( c n ~ x ) =( a 0 + a 1 c + + a n c n ~ x ) = f ( c ) ~ x . Hence f ( c ) is an eigenvalue of f ( T ) with associated eigenvector .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(c) Let c be an eigenvalue of T ,andle t E c be the generalized eigenspace of T associated with the eigenvalue c .L e t T c denote the linear transformation T restricted to E c .Tha t is, T c : E c E c , where T c ( ~x )= T ( ~ x ). Show that T c has only one eigenvalue, and that it is c . Deduce from this that the characteristic polymomial of T c equals p ( λ )=( λ - c ) m ,where m =dim( E c ). Let q be any eigenvalue of T c with an associated eigenvector .S i n c e ~ x E c there is an integer k such that ( T - cI ) k = ~ 0. But is an eigenvector associated with the eiqenvalue q . So from part b. and this equation we have ~ 0=( T - cI ) k =( q - c ) k ~ x . Since 6 = ~ 0, we must have q = c . Thus, the only eigenvalue of T c is c . Moreover since we’re assuming that the field F is algebraically closed we know that the characteristic polynomial of T c can be factored into linear factors. That is, det ( T c - λI λ - λ 1 ) m 1 ··· ( λ - λ k ) m k , where the λ i ’s are the distinct eigenvalues. Since T c
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

finalExamSolutionsPctex - Math 222 - 200 Solutions Final...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online