Download THEORY HOMEWORK SET #6 FOR MATH 550 (ORIENTATIONS

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.