- 7.1.1: Find loo and I2 norms ofthe vectors. a. x = (3, -4. 0, |)' b. x = (...
- 7.1.2: Find Igo and I2 norms of the vectors. a. x = (2, -2, 1)' b. x = (-4...
- 7.1.3: Prove that the following sequences are convergent and find their li...
- 7.1.4: Prove that the following sequences are convergent and find their li...
- 7.1.5: Find the /ao norm ofthe matrices.
- 7.1.6: Find the norm ofthe matrices. 10 -I -1 11 a. b. d. b. 10 0 15 1 4 1...
- 7.1.7: The following linear systems Ax = b have x as the actual solution a...
- 7.1.8: The following linear systems Ax = Compute ||x xlloo and || Ax b| a....
- 7.1.9: a. Verify that the function || * II1, defined on R" by Ixlli = 51 l...
- 7.1.10: The matrix norm || ||i, defined by ||/t||i = max ||/lx||i, can be c...
- 7.1.11: Show by example that || ||oo, defined by || AHoo = max |a,:;|, does...
- 7.1.12: Show that || ||, defined by 11. Show by example that || ||oo, defin...
- 7.1.13: a. The Frobenius norm (which is not a natural norm) is defined for ...
- 7.1.14: In Exercise 13, the Frobenius norm of a matrix was defined. Show th...
- 7.1.15: Let S be a positive definite n x n matrix. For any x in E" define |...
- 7.1.16: Let S be a real and nonsingular matrix and let || || be any norm on...
- 7.1.17: Prove that if || II is a vector norm on R", then || A|| = maX||X||=...
- 7.1.18: The following excerpt from the Mathematics Magazine [Sz] gives an a...
- 7.1.19: Show that the Cauchy-Buniakowsky-Schwarz Inequality can be strength...
Solutions for Chapter 7.1: Norms of Vectors and Matrices
Full solutions for Numerical Analysis | 10th Edition
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].
Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and
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.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.
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.
Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.
Nullspace N (A)
= All solutions to Ax = O. Dimension n - r = (# columns) - rank.
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).
Permutation matrix P.
There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or -1) based on the number of row exchanges to reach I.
Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.
Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.
Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
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).
Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.
Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.
Solvable system Ax = b.
The right side b is in the column space of A.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.
Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.
Volume of box.
The rows (or the columns) of A generate a box with volume I det(A) I.