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Unformatted text preview: that the matrix A can be “diagonalized” to the matrix D= 2i 0 0 −2i . We also state without proof the following useful result which will guarantee many examples where the characteristic polynomial has only real roots. PROPOSITION 7D. Suppose that A is an n × n matrix, with entries in R. Suppose further that A is symmetric. Then the characteristic polynomial det(A − λI) has only real roots. We conclude this section by discussing an application of diagonalization. We illustrate this by an example. Example 7.3.2. Consider the matrix 17 A = 45 −30 −10 −28 20 −5 −15 , 12 as in Example 7.2.3. Suppose that we wish to calculate A98 . Note that P −1 AP = D, where 1 P = 3 −2 It follows that A = P DP −1 , so that A98 = (P DP −1 ) . . . (P DP −1 ) = P D98 P −1
98 1 2 0 3 3 0 and −3 D= 0 0 0 2 0 0 0. 2 398 =P 0 0 0 298 0 0 0 P −1 . 298 This is much simpler than calculating A98 directly. 7.4. An Application to Genetics In this section, we discuss very brieﬂy the problem of autosomal inheritance. Here we consider a set of two genes designated by G and g. Each member of the population inherits one from each parent, resulting in possible genotypes GG, Gg and gg. Furthermore, the gene G dominates the gene g, so that in the case of human eye colours, for example, people with genotype GG or Gg have brown eyes while people with genotype gg have blue eyes. It is also believed that each member of the population has equal probability of inheriting one or the other gene from each parent. The table below gives these peobabilities in detail. Here the genotypes of the parents are listed on top, and the genotypes of the oﬀspring are listed on the left. GG − GG GG Gg gg 1 0 0 GG − Gg
1 2 1 2 GG − gg 0 1 0 Gg − Gg
1 4 1 2 1 4 Gg − gg 0
1 2 1 2 gg − gg 0 0 1
page 10 of 12 0 Chapter 7 : Eigenvalues and Eigenvectors Linear Algebra c W W L Chen, 1982, 2005 Example 7.4.1. Suppose that a plant breeder has a large population consisting of all three genotypes. At regular intervals, each plant he owns is fertilized with a plant known to have genotype GG, and is then disposed of and replaced by one of its oﬀsprings. We would like to study the distribution of the three genotypes after n rounds of fertilization and replacements, where n is an arbitrary positive integer. Suppose that GG(n), Gg(n) and gg(n) denote the proportion of each genotype after n rounds of fertilization and replacements, and that GG(0), Gg(0) and gg(0) denote the initial proportions. Then clearly we have GG(n) + Gg(n) + gg(n) = 1 for every n = 0, 1, 2, . . . . On the other hand, the left hand half of the table above shows that for every n = 1, 2, 3, . . . , we have GG(n) = GG(n − 1) + 1 Gg(n − 1), 2 Gg(n) = 1 Gg(n − 1) + gg(n − 1), 2 and gg(n) = 0, so that GG(n) 1 Gg(n) = 0 gg(n) 0 0 GG(n − 1) 1 Gg(n − 1) . 0 gg(n − 1) 1/2 1/2 0 It follows that GG(n) GG(0) Gg(n) = An Gg(0) gg(n) gg(0) where the matrix 1 A = 0 0 1/2 1/2 0 0 1 0 for every n = 1, 2, 3, . . . , has eigenvalues λ1 = 1, λ2 = 0, λ3 = 1/2, with respective eigenvectors 1 v1 = 0 , 0 We therefore write 1 1 1 P = 0 −2 −1 0 1 0 1 v2 = −2 , 1 0 0 , 1/2 1 v3 = −1 . 0 1 1 . −2 and 1 D = 0 0 0 0 0 with P −1 1 = 0 0 1 0 −1 Then P −1 AP = D, so that A = P DP −1 , and so 1 1 1 1 1 0 0 An = P Dn P −1 = 0 −2 −1 0 0 0 0 0 0 1 0 0 0 1/2n n n−1 1 1 − 1/2 1 − 1/2 = 0 1/2n 1/2n−1 . 0 0 0
Chapter 7 : Eigenvalues and Eigenvectors 1 0 −1 1 1 −2 page 11 of 12 Linear Algebra c W W L Chen, 1982, 2005 It follows that GG(n) 1 1 − 1/2n 1 − 1/2n−1 GG(0) Gg(n) = 0 1/2n 1/2n−1 Gg(0) gg(n) 0 0 0 gg(0) GG(0) + Gg(0) + gg(0) − Gg(0)/2n − gg(0)/2n−1 = Gg(0)/2n + gg(0)/2n−1 0 1 − Gg(0)/2n − gg(0)/2n−1 1 = Gg(0)/2n + gg(0)/2n−1 → 0 as n → ∞. 0 0 This means that nearly the whole crop will have genotype GG. Problems for Chapter 7 1. For each of the following 2 × 2 matrices, ﬁnd all eigenvalues and describe the eigenspace of the matrix; if possible, diagonalize the matrix: 3 4 2 −1 a) b) −2 −3 1 0 2. For each of the following 3 × 3 matrices, ﬁnd all eigenvalues and describe the eigenspace of the matrix; if possible, diagonalize the matrix: −2 9 −6 2 −1 −1 a) 1 −2 0 b) 0 3 2 3 −9 5 −1 1 2 1 1 0 c) 0 1 1 0 0 1 −10 6 3 0 −6 −16 3. Consider the matrices A = −26 16 8 and B = 0 17 45 . 16 −10 −5 0 −6 −16 a) Show that A and B have the same eigenvalues. b) Reduce A and B to the same disgonal matrix. c) Explain why there is an invertible matrix R such that R−1 AR = B. 4. Find A8 and B 8 , where A and B are the two matrices in Problem 3. 5. Suppose that θ ∈ R is not an integer multiple of π. Show that the matrix not have an eigenvector in R2 . 6. Consider the matrix A = cos θ sin θ sin θ , where θ ∈ R. − cos θ cos θ sin θ − sin θ cos θ does a) Show that A has an eigenvector in R2 with eigenvalue 1. b) Show that any vector v ∈ R2 perpendicular to the eigenvector in part (a) must satisfy Av = −v. 7. Let a ∈ R be nonzero. Show that the matrix 1 0 a 1 cannot be diagonalized. Chapter 7 : Eigenvalues and Eigenvectors page 12 of 12...
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