- 8.5.6E: In Exercises 5–8, find the minimal representation of the polytope d...
- 8.5.7E: In Exercises 5–8, find the minimal representation of the polytope d...
- 8.5.8E: In Exercises 5–8, find the minimal representation of the polytope d...
- 8.5.10E: Find an example of a closed convex set S in R2 such that its profil...
- 8.5.9E: Is the origin an extreme point of conv S? Is the origin a vertex of...
- 8.5.3E: Repeat Exercise 1 where m is the minimum value of f on S instead of...
- 8.5.4E: Repeat Exercise 2 where m is the minimum value of f on S instead of...
- 8.5.5E: In Exercises 5–8, find the minimal representation of the polytope d...
- 8.5.11E: Find an example of a bounded convex set S in R2 such that its profi...
- 8.5.12E: a. Determine the number of k-faces of the 5-dimensional simplex S5 ...
- 8.5.16E: In Exercises 16 and 17, mark each statement True or False. Justify ...
- 8.5.17E: In Exercises 16 and 17, mark each statement True or False. Justify ...
- 8.5.18E: Let v be an element of the convex set S. Prove that v is an extreme...
- 8.5.20E: Find an example to show that the convexity of S is necessary in Exe...
- 8.5.21E: If A and B are convex sets, prove that A + B is convex.
- 8.5.22E: A polyhedron (3-polytope) is called regular if all its facets are c...
Solutions for Chapter 8.5: Linear Algebra and Its Applications 5th Edition
Full solutions for Linear Algebra and Its Applications | 5th Edition
Adjacency matrix of a graph.
Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected). Adjacency matrix of a graph. Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected).
Augmented matrix [A b].
Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.
Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A
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].
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.
Incidence matrix of a directed graph.
The m by n edge-node incidence matrix has a row for each edge (node i to node j), with entries -1 and 1 in columns i and j .
Krylov subspace Kj(A, b).
The subspace spanned by b, Ab, ... , Aj-Ib. Numerical methods approximate A -I b by x j with residual b - Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.
Length II x II.
Square root of x T x (Pythagoras in n dimensions).
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.
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.
Every v in V is orthogonal to every w in W.
Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.
Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
Reflection matrix (Householder) Q = I -2uuT.
Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.
Row space C (AT) = all combinations of rows of A.
Column vectors by convention.
Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!
Tridiagonal matrix T: tij = 0 if Ii - j I > 1.
T- 1 has rank 1 above and below diagonal.
Stretch and shift the time axis to create Wjk(t) = woo(2j t - k).