Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
THEORY HOMEWORK SET #6 FOR MATH 550 (ORIENTATIONS) Problem #1: Suppose d : V ×V → R is a dot product on an n dimensional vector space V. Let ONB(V, d) be the set of all bases of V which are orthonormal relative to d. Let O(n) be the set of all A ∈ Rn×n such that A−1 = AT . (Such matrices are said to be orthogonal.) • Show that if (v1 , . . . , vn ) ∈ ONB(V, d) and A ∈ O(n) then (v1 , . . . , vn )A ∈ ONB(V, d). • Show that if (v1 , . . . , vn ) ∈ ONB(V, d) and A, B ∈ O(n) then (v1 , . . . , vn )(AB) = [(v1 , . . . , vn )A]B. • Show that if (v1 , . . . , vn ) ∈ ONB(V, d) and (w1 , . . . , wn ) ∈ ONB(V, d) then there is exactly one A ∈ O(n) such that (v1 , . . . , vn )A = (w1 , . . . , wn ). These are the rules governing how change of basis matrices act on orthonormal bases of V. Problem #2: Suppose V = R1 and d = · is the standard dot product on R1 , i.e. simply multiplication of two real numbers. Show that ONB(R1 , ·) = {(1), (−1)}. Show that for any 1-dimensional vector space V and any dot product d on it that ONB(V, d) = {(v), (−v)}, where d(v, v) = 1. (v) is a list with one vector in it. Problem #3: • Show that if A ∈ O(n) then det(A) is either 1 or −1. (Hint: det(AT ) = det(A) and det(AB) = det(A) det(B).) • Show that if A ∈ O(2) and A = (a1 , a2 ) then (a1 , a2 ) ∈ ONB(R2 , ·), where · is the standard dot product on R2 . Definition. Suppose d : V × V → R is a dot product on an n dimensional vector space V. An orientation on (V, d) is a mapping κ : ONB(V, d) → {1, −1} such that for every (v1 , . . . , vn ) ∈ ONB(V, d) and every A ∈ O(n) we have κ((v1 , . . . , vn )A) = κ((v1 , . . . , vn )) det(A). Problem #4: Show that there are only two possible orientations of (R1 , ·) and write them both down. Recall from problem #2 that if V is a 1-dimensional vector space and d is any dot product on it then ONB(V, d) = {(v), (−v)}, where d(v, v) = 1. Write down the two possible orientations on (V, d). For this reason we think of an orientation on a 1-dimensional vector space as being a choice of a positive 1 2 direction on that space (i.e. the line), i.e. if κ((v)) = 1 then v points in the positive direction on V. Problem #5: Suppose · denotes the standard dot product on R2 . Show that if κ1 , κ2 : ONB(R2 , ·) → {1, −1} are orientations and κ , ê2 )) = 1 and κ2 ((ê2 , ê1 )) = 1 then κ1 6= κ2 . (Hint: A = µ1 ((ê1¶ 0 1 ∈ O(2) and (ê1 , ê2 )A = (ê2 , ê1 ).) More generally show that if 1 0 V is a 2-dimensional vector space and d is any dot product on V then there are exactly two distinct orientations on (V, d). Problem #6: Suppose · denotes the standard dot product on R3 . Show that if κ1 , κ2 : ONB(R3 , ·) → {1, −1} are orientations and κ1 ((ê1 , ê2 , ê3 )) = 1 and κ2 ((ê2 , ê1 , ê3 )) = 1 then κ1 6= κ2 . On (R3 , ·) there is one orientation κ1 which assigns the value 1 to orthonormal bases which satisfy the ‘right-hand-rule’ and another orientation κ2 which assigns the value 1 to orthonormal bases which satisfy the ‘lefthand-rule’. κ1 is usually called the standard orientation of R3 . More generally show that if V is a 3-dimensional vector space and d is any dot product on V then there are exactly two distinct orientations on (V, d). Problem #7: Suppose κ1 is the standard orientation on (R3 , ·) and V ⊂ R3 is a 2-dimensional subspace of R3 (i.e. a plane through 0). V inherits a dot product · from R3 . Let n be a unit vector in R3 which is perpendicular to V. Define κn : ONB(V, ·) → {1, −1} by the rule: κn ((v1 , v2 )) = κ1 ((n, v1 , v2 )) for all (v1 , v2 ) ∈ ONB(V, ·). Show that κn is an orientation on (V, ·). Show that both orientations on (V, ·) arise from this construction: one is κn and the other is κ−n . For this reason an orientation of a 2-dimensional subspace of R3 is identified with a choice of the positive side of V, i.e. a unit vector n normal to V pointing toward the positive side of V. Remark: The trivial vector space V = {0} has dimension 0. A basis of V is an empty list of vectors; there is only one such empty list, which we denote by (). There is only one dot product on V, namely the one where d(0, 0) = 0. Let · denote this trivial dot product on V. So ONB(V, ·) = {()}, a set with one element, the empty list (). So an orientation of (V, ·) is a mapping κ : {()} → {1, −1}, i.e. an assignment of either κ(()) = 1 or κ(()) = −1. Hence orientation of a 0-dimensional vector space amounts to labelling it with either the number 1 or the number −1.