20105ee205A_1_2010ee205A_1_HW7_sol

# 20105ee205A_1_2010ee205A_1_HW7_sol - Chapter 9 Eigenvalues...

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Chapter 9 Eigenvalues and Eigenvectors 1. Let A C n × n have distinct eigenvalues λ 1 ,...,λ n with corresponding right eigenvectors x 1 ,...,x n and left eigenvectors y 1 ,...,y n , respec- tively. Let v C n be an arbitrary vector. Show that v can be expressed (uniquely) as a linear combination v = n ± i =1 y H i v y H i x i x i of the right eigenvectors. Find the appropriate expression for v as a linear combination of the left eigenvectors as well. Answer 9.1 Since the eigenvalues of A are distinct, the set of right eigenvectors { x i : i n } spans C n ; thus, any vector in C n can be expressed uniquely as a linear combination of the right eigenvectors. Hence we can write v = n j =1 α j x j . Since y H i x j =0 for i ± = j and y H i x i ± , we have y H i v = n j =1 α j y H i x j = α i y H i x i for each i . This implies that α i = y H i v y H i x i for each i . Therefore, v = n i =1 y H i v y H i x i x i . Similarly, we can write v = n i =1 β i y i , which leads to v = n i =1 x H i v x H i y i y i . 2. Suppose A C n × n is skew-Hermitian, i.e., A H = - A . Prove that all eigenvalues of a skew-Hermitian matrix must be pure imaginary. 33

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34 CHAPTER 9. EIGENVALUES AND EIGENVECTORS Answer 9.2 Let ( λ,x ) be an arbitrary eigenpair for A , i.e., Ax = λx . Then x H Ax = λx H x . Taking Hermitian transposes of this equation yields x H A H x = λx H x . Now using the fact that A is skew-Hermitian we see that λx H x = - λx H x . Since x is an eigenvector, x H x ± =0 , from which we conclude λ = - λ , i.e., λ is pure imaginary. 3. Suppose A C n × n is Hermitian. Let λ be an eigenvalue of A with corresponding right eigenvector x . Show that x is also a left eigenvector for λ . Prove the same result if A is skew-Hermitian. Answer 9.3 A = A H , Ax = λx = x H A = x H A H = λx H = λx H . A = - A H , Ax = λx = x H A = - x H A H = - λx H = λx H . 4. Suppose a matrix A IR 5 × 5 has eigenvalues { 2 , 2 , 2 , 2 , 3 } . Determine all possible Jordan canonical forms for A . Answer 9.4 The characteristic equation for A must be π ( λ ) = ( λ - 2) 4 ( λ - 3) , thus candidates for the minimal polynomial α ( λ ) are ( λ - 2)( λ - 3) , ( λ - 2) 2 ( λ - 3) , ( λ - 2) 3 ( λ - 3) , and ( λ - 2) 4 ( λ - 3) . DeFne the following Jordan blocks for notational ease: J 4 = 2 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 , J 3 = 210 021 002 , J 2 = ± 21 02 ² , J 1 =2 , and J 0 =3 . There exist Fve possible JC±s for A , not counting reorderings of their diagonal blocks: diag( J 4 ,J 0 ) , diag( J 3 1 0 ) , diag( J 2 2 0 ) , diag( J 2 1 1 0 ) , diag( J 1 1 1 1 0 ) . 5. Determine the eigenvalues, right eigenvectors and right principal vec- tors if necessary, and (real) Jordan canonical forms of the following matrices: (a) ± 2 - 1 10 ² , (b) ± 12 ² , (c) ± 45 - ² . (a) Call the matrix A . Then π ( λ )= λ 2 - 2 λ + 1 = ( λ - 1) 2 , so A has two eigenvalues at λ = 1. 2 - rank( A - I ) = 1, so A has one eigenvector and one principal vector associated with the eigenvalues at 1.
35 ( A - I ) x = 0 = x (1) 1 = (1 , 1) T is a right eigenvector of A asso- ciated with 1.

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20105ee205A_1_2010ee205A_1_HW7_sol - Chapter 9 Eigenvalues...

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