# plugin-hw11 - Assignment 11 Math 417 Winter 2009 Due April...

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Assignment 11 Math 417 — Winter 2009 Due April 20 § 7.1#2. We know that Av = λv . (Note that λ = 0 since A is invertible and hence ker( A ) = { 0 } .) Then A - 1 ( λv ) = v and so A - 1 v = λ - 1 A - 1 ( λv ) = λ - 1 v . Therefore, v is an eigenvector of A - 1 with eigenvalue λ - 1 . § 7.1#4. Since Av = λv , we see that (7 A ) v = 7( Av ) = 7( λv ) = (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 0 * 0 * . To obtain a matrix A as desired, we simply add 5 I 2 to any matrix of the form 0 * 0 * . Thus the desired matrices A are simply all matrices of the form 5 * 0 * § 7.1#38. First compute the product 4 1 1 - 5 0 - 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 0 4 0 - 1 - λ 0 - 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

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(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: 1. A = r 0 0 0 0 s 0 0 0 0 s 0 0 0 0 s , det( A - λI 4 ) = ( r - λ )( s - λ ) 3 .
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