Homework 5 Solutions

1

By hand, find eigensystems for each of these matrices. Then specify the algebraic and geometric multiplicity of each eigenvalue.

  1. \(\left[\begin{array}{rr}7 & -10 \\ 5 & -8\end{array}\right]\)

  2. \(\left[\begin{array}{rrr}-1 & 0 & 0 \\ 1 & -1 & 0 \\ 0 & 1 & -1\end{array}\right]\)

  3. \(\left[\begin{array}{lll}2 & 1 & 1 \\ 0 & 3 & 1 \\ 0 & 0 & 2\end{array}\right]\)

  4. \(\left[\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right]\)

  5. \(\left[\begin{array}{rr}0 & -2 \\ 2 & 0\end{array}\right]\)

5.1.1 and 5.1.3

Eigenvalue, algebraic multiplicity, geometric multiplicity, bases:

  1. \(\lambda=-3,1,1,\{(2,1)\}, \lambda=2,1,1,\{(1,1)\}\)

  2. \(\lambda=-1,3,1,\{(0,0,1)\}\),

  3. \(\lambda=2,2,2,\{(1,0,0),(0,-1,1)\}, \lambda=3,1,1,\{(1,1,0)\}\)

  4. \(\lambda=-2,1,1\), \(\{(-1,1)\}, \lambda=2,1,1,\{(1,1)\}(\mathrm{e}) \lambda=-2 \mathrm{i}, 1,1,\{(\mathrm{i},-1)\}, \lambda=2 \mathrm{i}, 1,1,\{(\mathrm{i}, 1)\}\)

2

Compute the eigensystems of these matrices, and identify any defective matrices. (You can do this by hand or on the computer.)

  1. \(\left[\begin{array}{lll}2 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 2\end{array}\right]\)

  2. \(\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 6 & 2\end{array}\right]\)

  3. \(\left[\begin{array}{rr}1+\mathrm{i} & 3 \\ 0 & \mathrm{i}\end{array}\right]\)

  4. \(\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 4 & -2 \\ 0 & 0 & 1\end{array}\right]\)

  5. \(\left[\begin{array}{rrrr}2 & 1 & -1 & -2 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{array}\right]\)

5.1.2 and 5.1.4

Eigenvalue, basis, defective (yes/no)

  1. \(0,\{(-1,0,1)\}, 1,\{(0,1,0)\}\), \(3,\{(1,0,1)\}\), no

  2. \(0,\{(1,0,0)\}, 2,\{(0,1,2)\}, 5,\{(0,-1,3)\}\) no

  3. \(1+\mathrm{i}\), \(\{(1,0)\}\), i, \(\{(-3,1)\}\), no

  4. \(0,\{(2,1,0)\}, 1,\{(-1,2,4)\}, 5,\{(-1,2,0)\}\), no (e) \(-1,\{(1,3,-6,6)\}, 1,\{(-1,1,0,0)\}, 2,\{(1,0,0,0)\}\), yes

3

By hand, find a matrix \(P\) such that \(P^{-1} A P\) is diagonal for each of the following matrices. Then use \(P\) to find a formula for \(A^{k}, k\) a positive integer.

  1. \(\left[\begin{array}{lll}2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{array}\right]\)

  2. \(\left[\begin{array}{lll}1 & 2 & 2 \\ 0 & 0 & 0 \\ 0 & 2 & 2\end{array}\right]\)

  3. \(\left[\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right]\)

  4. \(\left[\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right]\)

  5. \(\left[\begin{array}{rrrr}2 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 1\end{array}\right]\)

5.2.3 and 4

Matrices P:

  1. \(\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]\)
  2. \(\left[\begin{array}{rll}0 & 1 & 2 \\ -1 & 0 & 0 \\ 1 & 0 & 1\end{array}\right]\)
  3. \(\left[\begin{array}{rl}-1 & 2 \\ 1 & 3\end{array}\right]\)
  4. \(\left[\begin{array}{rr}-1 & 1 \\ 1 & 1\end{array}\right]\)
  5. \(\left[\begin{array}{cccc}1 & -1 & 1 & -1 \\ -2 & 1 & 0 & -1 \\ 0 & -1 & 0 & 3 \\ 0 & 2 & 0 & 0\end{array}\right]\)

\(A^k\):

  1. \(\left[\begin{array}{rrr}2^{k} & 0 & -2^{k}+3^{k} \\ 0 & 2 & 0 \\ 0 & 0 & 3^{k}\end{array}\right]\)

  2. \(\left[\begin{array}{ccr}1 & -2+2^{k+1} & -2 * 2^{k+1} \\ 0 & 0 & 0 \\ 0 & 2^{k} & 2^{k}\end{array}\right]\)

  3. \(\frac{1}{5}\left[\begin{array}{rr}3(-1)^{k}+2 \cdot 4^{k} & -2(-1)^{k}+2 \cdot 4^{k} \\ -3(-1)^{k}+3 \cdot 4^{k} & 2(-1)^{k}+3 \cdot 4^{k}\end{array}\right]\)

  4. \(\frac{1}{2}\left[\begin{array}{rr}(-2)^{k}+2^{k} & -(-2)^{k}+2^{k} \\ -(-2)^{k}+2^{k} & (-2)^{k}+2^{k}\end{array}\right]\)

  5. \(\left[\begin{array}{rrrr}2^{k} & 2^{k-1} & 2^{k-1}-3^{k-1}-\frac{1}{2}+2^{k-1}-\frac{1}{2} 3^{k-1} \\ 0 & 0 & -3^{k-1} & \frac{1}{2}-\frac{1}{2} 3^{k-1} \\ 0 & 0 & 3^{k} & -\frac{1}{2}+\frac{1}{2} 3^{k} \\ 0 & 0 & 0 & 1\end{array}\right]\)

4

Compute \(\sin \left(\frac{\pi}{6} A\right)\) and \(\cos \left(\frac{\pi}{6} A\right)\), where \(A=\left[\begin{array}{rr}2 & 4 \\ 0 & -3\end{array}\right]\).

5.2.11

\(\sin \left(\frac{\pi}{6} A\right)=\left[\begin{array}{cc}\frac{1}{2} \sqrt{3} & \frac{4}{5}+\frac{2}{5} \sqrt{3} \\ 0 & -1\end{array}\right], \cos \left(\frac{\pi}{6} A\right)=\left[\begin{array}{cc}\frac{1}{2} & \frac{2}{5} \\ 0 & 0\end{array}\right]\)

5

By hand, find the spectral radius and dominant eigenvalue, if any, for each of these matrices:

  1. \(\left[\begin{array}{rr}-7 & -6 \\ 9 & 8\end{array}\right]\)

  2. \(\frac{1}{3}\left[\begin{array}{ll}1 & 3 \\ 2 & 0\end{array}\right]\)

  3. \(\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]\)

  4. \(\frac{1}{2}\left[\begin{array}{lll}1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1\end{array}\right]\)

  5. \(\left[\begin{array}{rr}1 & 1 \\ -1 & -1\end{array}\right]\)

5.3.2

  1. 2 , dominant eigenvalue 2
  2. 1 , dominant eigenvalue 1
  3. 1 , no dominant eigenvalue (First eigenvalue is \(1\), other eigenvalues are \(-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i\), which have norm 1.)
  4. 1 , dominant eígenvalue 1
  5. 0 , no dominant eigenvalue

6

If the matrices of the previous exercise are transition matrices, for which do all \(\mathbf{x}^{(k)}\) remain bounded as \(k \rightarrow \infty\) ? Are any of these matrices stable?

5.3.6

  1. gives a matrix for which \(\mathbf{x}^{(k)} \rightarrow \mathbf{0}\) as \(k \rightarrow \infty\). (b) and (d) give stable matrices.

7

The three-stage insect model of Example 2.21 yields a transition matrix

\[ A=\left[\begin{array}{ccc} 0.2 & 0 & 0.25 \\ 0.6 & 0.3 & 0 \\ 0 & 0.6 & 0.8 \end{array}\right] \]

Use a technology tool to calculate the eigenvalues of this matrix. Deduce that \(A\) is diagonalizable and determine the approximate growth rate from one state to the next (after much time has passed), given a random initial vector.

Exercise 5.3.11

Eigenvalues are \(\lambda \approx 0.1636 \pm 0.3393 \mathrm{i}, 0.973\) with absolute values 0.3766 and 0.973 . So population will decline at rate of approximately \(2.7 \%\) per time period.

8

The financial model of Example 2.27 gives rise to a discrete dynamical system \(x^{(k+1)}=A x^{(k)}\), where the transition matrix is

\[ A=\left[\begin{array}{rrr} 1 & 0.06 & 0.12 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \]

Use a technology tool to calculate the eigenvalues of this matrix. Deduce that \(A\) is diagonalizable and determine the approximate growth rate from one state to the next, given a random initial vector.

Find a starting vector \(x^{(0)}\) such that growth rate for the first few iterations is quite different from this approximate growth rate, and demonstrate this using a technology tool.

(If you do this using Sympy, and you set up the matrix A using rational numbers, you can get exact results and find a starting vector which will never reach your approximate growth rage. You can see how this result changes if you use floating point numbers at some point in the calculation, e.g. x = N(x).)

from sympy import Matrix, Rational

# Set up the matrix A in symbolic form
A = Matrix([[1, Rational(6, 100), Rational(12, 100)], [1, 0, 0], [0, 1, 0]])

Doing it with numpy:

import numpy as np
A = np.array([[1, 0.06, 0.12], [1, 0, 0], [0, 1, 0]])
# find the eigenvectors and eigenvalues
eigenvalues, eigenvectors = np.linalg.eig(A)
print(eigenvalues)
print(eigenvectors)
[ 1.14411531+0.j         -0.07205766+0.31574076j -0.07205766-0.31574076j]
[[ 0.65266846+0.j          0.08945851+0.04307557j  0.08945851-0.04307557j]
 [ 0.57045689+0.j          0.0682135 -0.29889653j  0.0682135 +0.29889653j]
 [ 0.49860087+0.j         -0.94665171+0.j         -0.94665171-0.j        ]]
x = eigenvectors[2]
# pick a random initial vector
x = np.random.rand(3)
for i in range(10):
    xnew = A @ x
    print(xnew[0] / x[0])
    x = xnew
1.09266475162127
1.145377739168319
1.148268339357237
1.143493401495684
1.1438619487749357
1.1441971003933349
1.1441253124077122
1.1441074974782397
1.1441154930207216
1.1441159133140761

Doing it with sympy:

from sympy import Matrix, symbols, randprime, Rational,N
from sympy.abc import x


A = Matrix([[1, Rational(6, 100), Rational(12, 100)], [1, 0, 0], [0, 1, 0]])


# find the eigenvalues and eigenvectors
eigenvalues = A.eigenvals()
eigenvectors = A.eigenvects()

print("Eigenvalues:")
for value in eigenvalues:
    print(value)

print("\nEigenvectors:")
for vector in eigenvectors:
    print(vector)

# pick a random initial vector
#x = Matrix([randprime(1, 10) for _ in range(3)])

# use the third eigenvector as the initial
x = eigenvectors[0][2][0]

for i in range(100):
    #xnew = N(A * x)
    xnew= A * x
    growth_rate = N(xnew[0] / x[0])
    print(f"Growth rate at iteration {i}: {growth_rate}")
    x = xnew
Eigenvalues:
59/(450*(sqrt(20707)/1500 + 289/2700)**(1/3)) + 1/3 + (sqrt(20707)/1500 + 289/2700)**(1/3)
1/3 + 59/(450*(-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)
1/3 + (-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3) + 59/(450*(-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3))

Eigenvectors:
(1/3 + (-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3) + 59/(450*(-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)), 1, [Matrix([
[(1/3 + (-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3) + 59/(450*(-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)))**2],
[     1/3 + (-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3) + 59/(450*(-1/2 - sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3))],
[                                                                                                                                        1]])])
(1/3 + 59/(450*(-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3), 1, [Matrix([
[(1/3 + 59/(450*(-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3))**2],
[     1/3 + 59/(450*(-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(20707)/1500 + 289/2700)**(1/3)],
[                                                                                                                                        1]])])
(59/(450*(sqrt(20707)/1500 + 289/2700)**(1/3)) + 1/3 + (sqrt(20707)/1500 + 289/2700)**(1/3), 1, [Matrix([
[(59/(450*(sqrt(20707)/1500 + 289/2700)**(1/3)) + 1/3 + (sqrt(20707)/1500 + 289/2700)**(1/3))**2],
[     59/(450*(sqrt(20707)/1500 + 289/2700)**(1/3)) + 1/3 + (sqrt(20707)/1500 + 289/2700)**(1/3)],
[                                                                                              1]])])
Growth rate at iteration 0: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 1: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 2: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 3: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 4: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 5: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 6: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 7: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 8: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 9: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 10: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 11: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 12: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 13: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 14: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 15: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 16: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 17: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 18: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 19: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 20: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 21: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 22: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 23: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 24: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 25: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 26: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 27: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 28: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 29: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 30: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 31: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 32: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 33: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 34: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 35: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 36: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 37: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 38: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 39: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 40: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 41: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 42: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 43: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 44: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 45: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 46: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 47: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 48: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 49: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 50: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 51: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 52: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 53: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 54: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 55: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 56: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 57: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 58: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 59: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 60: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 61: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 62: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 63: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 64: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 65: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 66: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 67: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 68: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 69: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 70: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 71: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 72: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 73: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 74: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 75: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 76: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 77: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 78: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 79: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 80: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 81: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 82: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 83: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 84: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 85: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 86: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 87: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 88: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 89: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 90: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 91: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 92: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 93: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 94: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 95: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 96: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 97: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 98: -0.0720576551574491 - 0.315740759667275*I
Growth rate at iteration 99: -0.0720576551574491 - 0.315740759667275*I

9

On this problem,you may use a computer to find eigensystems etc, but don’t solve it using simulations and trial and error.

A species of bird can be divided into three age groups: age less than 2 years for group 1, age between 2 and 4 years for group 2, and age between 4 and 6 years for the third group. Assume that these birds have at most a 6-year life span. It is estimated that the survival rates for birds in groups 1 and 2 are \(50 \%\) and 75%, respectively. Also, birds in groups 1, 2, and 3 produce 0,1 , and 3 offspring on average in any biennium (period of 2 years).

  1. Model this bird population as a discrete dynamical system and analyze the long-term change in the population – give a percentage by which the population of group 1 changes from one period to the next.

  2. Now suppose that the survival rates for group 1 and 2 are not known (so are no longer assumed to be 50% and 75%), but are assumed to be equal to one another. What value for this survival rate would make the population stable in the long run?

5.3.20

If \(\mathbf{x}^{(k)}=\left(x_{1}^{(k)}, x_{2}^{(k)}, x_{3}^{(k)}\right)\) is the vector of birds in each age group at the \(k\) th biennium, then the model is \(\mathbf{x}^{(k+1)}=\left[\begin{array}{rrr}0 & 1 & 3 \\ 0.5 & 0&0 \\ 0 & 0.75 & 0\end{array}\right] \mathbf{x}^{(k)}\). Eigenvalues of \(A\) are \(\lambda \approx 1.992,-0.600 \pm 0.760 i\). Hence, the population will grow at a rate of \(\% 20\) per year in the long run.

If survival rates are equal, say \(s\), then transition matrix is \(\left[\begin{array}{lll}0 & 1 & 3 \\ s & 0 & 0 \\ 0 & s & 0\end{array}\right]\), which has a dominant eigenvalue of \(\lambda=1\) when \(s\approx 0.434\).

(Problem is simpler when the birds produce 0, 2, and 3 offspring on average; then we find s=6 as the answer and the population growth rate is different. This was the original version in the book but I had a typo when I put the problem in.)