- 6.11.1: Label the following statements as true or false. Assume that the un...
- 6.11.2: Prove that rotations, reflections, anel composites of rotations and...
- 6.11.3: Let 479 A = ( I 2 v/3 \/3\ 2 1 "2 I and B 0 - 1 (a) Prove that VA i...
- 6.11.4: For any real number , let A = cos* U sm< sin* cos* (a) Prove that L...
- 6.11.5: For any real number
- 6.11.6: Prove that the composite of any two rotations on R3 is a rotation o...
- 6.11.7: Given real numbers (p and ip, define matrices 1 0 0 A = | 0 cos
- 6.11.8: Prove Theorem 6.45 using the hints preceding the statement of the t...
- 6.11.9: Prove that no orthogonal operator can be both a rotation and a refl...
- 6.11.10: Prove that if V is a two- or three-dimensional real inner product s...
- 6.11.11: Give an example of an orthogonal operator that is neither a rcfiect...
- 6.11.12: Lt;t V be a finite-dimensional real inner product spae-e. Define T:...
- 6.11.13: Complete the proof e>f the lemma to Theorem 6.46 by shewing that W ...
- 6.11.14: Let T be an orthogonal [unitary] operator on a finite-dimensional r...
- 6.11.15: Let T be1 a linear e)perate>r on a finite-eliniensional vector spac...
- 6.11.16: Complete the proof of the1 corollary to The'orcm 6.47.
- 6.11.17: Let T be' a linear e)perate>r on an n-dimensional real inner produc...
- 6.11.18: Let V be a real inner product spae*? e)f elimension 2. For any x, y...
Solutions for Chapter 6.11: The Geometry of Orthogonal Operators
Full solutions for Linear Algebra | 4th Edition
Basis for V.
Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!
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.
Identity matrix I (or In).
Diagonal entries = 1, off-diagonal entries = 0.
Inverse matrix A-I.
Square matrix with A-I A = I and AA-l = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B-1 A-I and (A-I)T. Cofactor formula (A-l)ij = Cji! detA.
Kronecker product (tensor product) A ® B.
Blocks aij B, eigenvalues Ap(A)Aq(B).
Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
Length II x II.
Square root of x T x (Pythagoras in n dimensions).
Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.
Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q -1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •
Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.
Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).
The diagonal entry (first nonzero) at the time when a row is used in elimination.
Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.
Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.
Rayleigh quotient q (x) = X T Ax I x T x for symmetric A: Amin < q (x) < Amax.
Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).
Saddle point of I(x}, ... ,xn ).
A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.
Skew-symmetric matrix K.
The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.
Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.
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).