18.06 Spring 2006 - Exam 3 Review ProblemsSOLUTIONS TO SELECTED PROBLEMS1. Section 6.1, Problem 4Answer: A has λ1= −3 and λ2= 2 (check trace and determinant) withx1= (3, −2) and x2= (1, 1). A2has the same eigenvectors as A, witheigenvalues λ1= 9 and λ2= 4.2. Section 6.1, Problem 5Answer: A and B both have λ1= 1 and λ2= 1. A+B has λ1= 1 and λ2= 3.Eigenvalues of A + B are not equal to eigenvalues of A plus eigenvalues ofB.3. Section 6.1, Problem 25Answer: λ = 0, 0, 6 with eigenvectors x1= (0, −2, 1), x2= (1, −2, 0) andx3= (1, 2, 1).4. Section 6.2, Problem 2Answer: If A = SΛS−1, then A3= SΛ3S−1and A−1= SΛ−1S−1.5. Section 6.2, Problem 18Answer: The rank of A − 3I is one, hence A is not diagonalizable (3 is arepeated eigenvalue but has only one associated eigenvector). Change anyentry except a12= 1 to make A diagonalizable.16. Section 6.3, Problem 6Answer: λ1= 0 and λ2= 2. Now v(t) = 20 + 10e2t→ ∞ as t → ∞.7. Section 6.4, Problem 7Answer:a) A =1 22 1has λ = −1, 3.b) A has a negative pivot because the pivots have the same signs as the λ’s.c) A can’t have two negative eigenvalues because its trace is positive.8. Section 6.4, Problem 18Answer : Suppose A = ATand Ax = λx and Ay = 0y. Then y is in thenullspace and x is in the column space. Since A = AT, the column spaceequals the row space, hence x is in the row space of A. The row space andnullspace are orthogonal subspaces, so y ⊥ x.If the second eigenvalue is a nonzero number β, then shift by β: (A−βI)x =(λ − β)x and (A − βI)y = 0 and again x ⊥ y.9. Section 6.4, Problem 22Answer: If A is skew-symmetric, then AT= −A and ATA = AAT= −A2.Every orthogonal matrix is normal because ATA = AAT= I.A =a 1−1 dis normal if a = d.10. Section 6.5, Problem 1Answer: A4has two positive eigenvalues because a = 1 and det(A4) = 1.xTA1x is zero for x = (1, −1) and xTA1x < 0 for x = (6, −5).211. Section 6.5, Problem 16Answer: xTAx is not positive when x = (0, 1, 0) because of the zero on thediagonal.12. Section 6.6, Problem 2Answer: If C = F−1AF and also C = G−1BG then M = F G−1givesB = M−1AM. If C is similar to A and also to B then A is similar to B.13. Section 6.7, Problem 1Answer: ATA =5 2020 80has σ21= 85, v1=1/√174/√17and v2=4/√17−1/√17.14. Section 6.7, Problem 3Answer: u1=1/√52/√5for the column space; v1=1/√174/√17for the rowspace; u2=2/√5−1/√5for the nullsapce; v2=4/√17−1/√17for the leftnullspace.15. Section 6.7, Problem 10Answer: A = W ΣWT= UΣVT.316. Section 6.7, Problem 12Answer: Since A = AT, we have σ21= λ21and σ22= λ22. So σ1= 3 and σ2= 2(singular values are positive). The unit eigenvectors of ATA = AATare thesame as those for A: u1= v1and u2= −v2(notice the sign change becauseσ2= −λ2).17. Section 6.7, Problem 15Answer:a) If A changes to 4A, multiply Σ by 4.b) AT= V ΣTUT. And if A−1exists, it is square and equal to (VT)−1Σ−1U−1.18. Section 8.3, Problem 1Answer: λ = 1 and .75; steady state eigenvector x = (.6, .4).19. Section 8.3, Problem 9Answer: u1= P u0= (0, 0, 1, 0); u2= P u1= (0, 1, 0, 0); u3= P u2=(1, 0, 0, 0); u4= P u3= u0.The four eigenvalues of P are 1, i, −1,
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