Plugin-hw11 - Assignment 11 Math 417 — Winter 2009 Due April 20 7.1#2 We know that A~v = λ~v(Note that 6 = 0 since A is invertible and hence

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assignment 11 Math 417 — Winter 2009 Due April 20 § 7.1#2. We know that A~v = λ~v . (Note that λ 6 = 0 since A is invertible and hence ker( A ) = { ~ } .) Then A- 1 ( λ~v ) = ~v and so A- 1 ~v = λ- 1 A- 1 ( λ~v ) = λ- 1 ~v . Therefore, ~v is an eigenvector of A- 1 with eigenvalue λ- 1 . § 7.1#4. Since A~v = λ~v , we see that (7 A ) ~v = 7( A~v ) = 7( λ~v ) = (7 λ ) ~v . So ~v is an eigenvector of 7 A with eigenvalue 7 λ . § 7.1#8. This amounts to finding all 2 × 2 matrices A such that ~e 1 ∈ ker( A- 5 I 2 ). The 2 × 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 * * § 7.1#38. First compute the product   4 1 1- 5- 3- 1- 1 2     1- 1- 1   =   2- 2- 2   . The resulting vector is exactly twice the original vector, so the eigenvalue is 2. (Note that we didn’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.) § 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  - 3- λ 4- 1- λ- 2 7 3- λ   = (- 1- λ ) det- 3- λ 4- 2 3- λ = (- 1- λ ) ((- 3- λ )(3- λ ) + 8) = (- 1- λ )( λ 2- 1) = (- 1- λ ) 2 (1- λ ) 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. § 7.2#18. The characteristic polynomial is det a- λ b b a- λ = ( a- λ ) 2- b 2 . The roots of this quadratic occur when ( a- λ ) 2 = b 2 , or λ = a ± 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.) § 7.2#24. The characteristic polynomial is det 1 / 2- λ 1 / 4 1 / 2 3 / 4- λ = (1 / 2- λ )(3 / 4- λ )- 1 / 8 = λ 2- (5 / 4) λ + 1 / 4 . The roots of this polynomial are 1 and 1 / 4. These are the two eigenvalues of A . It is a general fact — which is very useful in the theory of Markov chains — that any matrix whose entry in each column (or each row) add up to 1 must have 1 as an eigenvalue. § 7.2#34. Let r and s be the two distinct real eigenvalues of the 4 × 4 matrix A . Then the characteristic polynomial of A has the form ( r- λ )( s- λ ) q ( λ ) where q ( λ ) is a quadratic polynomial whose roots (if any) are equal to r or s . Thus there are four possibilities for q ( λ ): 1. q ( λ ) = ( r- λ ) 2 2. q ( λ ) = ( r- λ )( s- λ ) 3. q ( λ ) = ( s- λ ) 2 4. q ( λ ) has no (real) roots, e.g., q ( λ ) = 1 + λ 2 Corresponding example matrices are the following:...
View Full Document

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 Michigan-Dearborn.

Page1 / 9

Plugin-hw11 - Assignment 11 Math 417 — Winter 2009 Due April 20 7.1#2 We know that A~v = λ~v(Note that 6 = 0 since A is invertible and hence

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