Geometrically, W is a plane through the origin.ģ38 CHAPTER 6 Orthogonality and Least SquaresĢ7. Subspace of 3, because W is the null space of the 1 3 matrix. Theorem 2 in Chapter 4 may be used to show that W is a
.jpg "algebra lineal david c lay solucionario")
If a = 0Īnd b = 0, then H = 2 since the equation 0x + 0y = 0 places noĢ6. Is still a basis for H since a = 0 and b 0.
ALGEBRA LINEAL DAVID C LAY SOLUCIONARIO FREE
A natural choice for a basis for H in this case isī 0, y = 0 and x is a free variable. If a 0, then x = (b/a)y with y aįree variable, and H is a line through the That are orthogonal to is the subspace of vectors whoseĮntries satisfy ax + by = 0. If and only if all the numbers are themselves zero.Ģ3. U is the sum of the squares of the entries in u, u u0. Theorems 3(c) and 2(d), respectively, from Section 2.1. )T Tc c c c u v u v u v u v The second and third equalities used
U w v w The second and third equalities used Theorems 3(b) andĢ(c), respectively, from Section 2.1. Theorem 1(b): ( ) ( ) ( )T T T T T u v w u v w u v w u w v w See the defintion of orthogonal complement. A unit vector in the direction of the given vector isĨ/ 3 8/ 3 4 / 51 12 2 3/ 5100 / 9(8 / 3) 2 A unit vector in the direction of the given vector isġ2. A unit vector in the direction of the given vector isġ1. A unit vector in the direction of the given vector isġ0. The optional material on angles is not used later. Only for Supplementary Exercise 13 at the end of the chapter and in Theorem 3 is an important general fact, but is needed Orthogonality and orthogonal complements, which are essential for Notes: The first half of this section is computational and isĮasily learned. Exercises 27–31 concern facts used later. Theorem 3 is an important general fact, but is needed only for Supplementary Exercise 13 at the end of the chapter and in Section 7.4. The second half concerns the concepts of orthogonality and orthogonal complements, which are essential for later work. Since the set is 2 Spanp q u u v v is a subset of H + K.335 6.1 SOLUTIONS Notes : The first half of this section is computational and is easily learned. Note that u and v are on the line L, but u + v is not. Since H is not closed under scalar multiplication, H is not a subspace of 2.
5 = u and c = 4, then u is in H but c u is not in H. Example: If 1 7 − = − u and 2 3 = v, then u and v are in W but u + v is not in W. If x y = u is in W, then the vector x cx c c y cy = u is in W because 2 ( )( ) ( ) 0 cx cy c xy = ≥ since xy ≥ 0. Example: If 2 2 = u and c = –1, then u is in V but c u is not in V. Since a sum of nonnegative numbers is nonnegative, the vector u + v has nonnegative entries. If u and v are in V, then their entries are nonnegative.
Other vectors do appear later in the chapter: the space of signals is used in Section 4.8, and the spaces n of polynomials are used in many sections of Chapters 4 and 6. The exercises in this section (and the next few sections) emphasize n, to give students time to absorb the abstract concepts.
ALGEBRA LINEAL DAVID C LAY SOLUCIONARIO HOW TO
Students should be taught how to check the closure axioms. Theorem 1 provides the main homework tool in this section for showing that a set is a subspace. 185 4.1 SOLUTIONS Notes : This section is designed to avoid the standard exercises in which a student is asked to check ten axioms on an array of sets.
0 kommentar(er)
