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Unformatted text preview: Assignment 11 Math 417 â€” Winter 2009 Due April 20 Â§ 7.1#2. We know that A~v = Î»~v . (Note that Î» 6 = 0 since A is invertible and hence ker( A ) = { ~ } .) Then A 1 ( Î»~v ) = ~v and so A 1 ~v = Î» 1 A 1 ( Î»~v ) = Î» 1 ~v . Therefore, ~v is an eigenvector of A 1 with eigenvalue Î» 1 . Â§ 7.1#4. Since A~v = Î»~v , we see that (7 A ) ~v = 7( A~v ) = 7( Î»~v ) = (7 Î» ) ~v . So ~v is an eigenvector of 7 A with eigenvalue 7 Î» . Â§ 7.1#8. This amounts to finding all 2 Ã— 2 matrices A such that ~e 1 âˆˆ ker( A 5 I 2 ). The 2 Ã— 2 matrices that have ~e 1 in their kernel are precisely the matrices of the form * * . To obtain a matrix A as desired, we simply add 5 I 2 to any matrix of the form * * . Thus the desired matrices A are simply all matrices of the form 5 * * Â§ 7.1#38. First compute the product ï£® ï£° 4 1 1 5 3 1 1 2 ï£¹ ï£» ï£® ï£° 1 1 1 ï£¹ ï£» = ï£® ï£° 2 2 2 ï£¹ ï£» . The resulting vector is exactly twice the original vector, so the eigenvalue is 2. (Note that we didnâ€™t need to compute all three coordinates of the product to determine the eigenvalue, the first coordinate would be enough since we are told that this is an eigenvector.) Â§ 7.2#10. Here (and in many instances of computing characteristic polynomials) it is easier to use cofactor expansion. We will use the second row since it has only one nonzero entry: det ï£® ï£° 3 Î» 4 1 Î» 2 7 3 Î» ï£¹ ï£» = ( 1 Î» ) det 3 Î» 4 2 3 Î» = ( 1 Î» ) (( 3 Î» )(3 Î» ) + 8) = ( 1 Î» )( Î» 2 1) = ( 1 Î» ) 2 (1 Î» ) 1 (Note how the cofactor expansion automatically gave us one factor of the cubic polynomial.) The eigenvalues of the matrix are 1 and 1, with algebraic multiplicities 1 and 2, respectively. Â§ 7.2#18. The characteristic polynomial is det a Î» b b a Î» = ( a Î» ) 2 b 2 . The roots of this quadratic occur when ( a Î» ) 2 = b 2 , or Î» = a Â± b . Thus a + b and a b are the two eigenvalues of A . (When b = 0, a is the only eigenvalue and it then has algebraic multiplicity 2.) Â§ 7.2#24. The characteristic polynomial is det 1 / 2 Î» 1 / 4 1 / 2 3 / 4 Î» = (1 / 2 Î» )(3 / 4 Î» ) 1 / 8 = Î» 2 (5 / 4) Î» + 1 / 4 . The roots of this polynomial are 1 and 1 / 4. These are the two eigenvalues of A . It is a general fact â€” which is very useful in the theory of Markov chains â€” that any matrix whose entry in each column (or each row) add up to 1 must have 1 as an eigenvalue. Â§ 7.2#34. Let r and s be the two distinct real eigenvalues of the 4 Ã— 4 matrix A . Then the characteristic polynomial of A has the form ( r Î» )( s Î» ) q ( Î» ) where q ( Î» ) is a quadratic polynomial whose roots (if any) are equal to r or s . Thus there are four possibilities for q ( Î» ): 1. q ( Î» ) = ( r Î» ) 2 2. q ( Î» ) = ( r Î» )( s Î» ) 3. q ( Î» ) = ( s Î» ) 2 4. q ( Î» ) has no (real) roots, e.g., q ( Î» ) = 1 + Î» 2 Corresponding example matrices are the following:...
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This note was uploaded on 10/13/2009 for the course MATH 417 taught by Professor Eveet during the Spring '09 term at University of MichiganDearborn.
 Spring '09
 eveet
 Math, Linear Algebra, Algebra

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