- 10.8.1: . A certain forest is divided into three height classes and has a g...
- 10.8.2: . In Example 1, to what level must the price of trees in the fifth ...
- 10.8.3: In Example 1, what must the ratio of the pricesp2: p3: p4: p5: p6 b...
- 10.8.4: Derive Equation (19) for the nonharvest vector x corresponding to t...
- 10.8.5: For the optimal sustainable harvesting policy described in Theorem ...
- 10.8.6: If all the growth parameters g1, g2,...,gn1 in the growth matrix G ...
- 10.8.T1: A particular forest has growth parameters given by gi = 1 i for i =...
- 10.8.T2: A particular forest has growth parameters given by gi = 1 2i for i ...
Solutions for Chapter 10.8: Forest Management
Full solutions for Elementary Linear Algebra, Binder Ready Version: Applications Version | 11th Edition
peA) = det(A - AI) has peA) = zero matrix.
Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).
Cross product u xv in R3:
Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].
Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then S-I AS = A = eigenvalue matrix.
0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].
Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n - r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.
Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.
Hilbert matrix hilb(n).
Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.
Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.
Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.
Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
Rank r (A)
= number of pivots = dimension of column space = dimension of row space.
Reduced row echelon form R = rref(A).
Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.
Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.
Skew-symmetric matrix K.
The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!
If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.
Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.
Stretch and shift the time axis to create Wjk(t) = woo(2j t - k).