Part 2. 1 Part 2. 2
Orthogonal Matrices
• If u and v are nonzero vectors then
u · v = kuk kvk cos(θ)
◦ £¥¤ ¢ is 0 if and only if cos(θ) = 0, i.e., θ = 90 . ¡
Hence, we say that two vectors u and v are perpendicular or orthogonal (in symbols u ⊥ v) if u · v = 0.
• A vector u is a unit vector if kuk = 1. Now rotate the vectors, i.e., apply T .
• Let T : R2 → R2 be rotation around the
£ ¥ origin by θ degrees. This is a linear
transformation. To see this, consider ¡£¢¥¤
the addition of u and v as in the §£¨ © following diagram. ¦
This diagram shows that T (u) + T (v) = T (u + v). It’s even easier to Part 2. 3 Part 2. 4
check that T (sv) = sT (v), so T is linear. • Recall the addition laws for sin and cos.
We need to find T (e1), T (e2). If we cos(A + B) = cos(A) cos(B) − sin(A) sin(B) rotate e by θ, we get (cos(θ), sin(θ)) by 1 cos(A − B) = cos(A) cos(B) + sin(A) sin(B) the unit circle definition of sin and cos.
sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
sin(A − B) = sin(A) cos(B) − cos(A) sin(B) £¥¤§¦¢¨ ©
Recall also that ¢¡ cos(−θ) = cos(θ) sin(−θ) = − sin(θ)
The vector w = (− sin(θ), cos(θ)) is a unit • Exercise: Rotating by θ and then by ϕ
vector and w · T (e1) = 0, so T (e2) must should be the same as rotating θ + ϕ. be either w or −w. Considering the sign Thus, we have of the first component shows that R(θ)R(ϕ) = R(θ + ϕ) = R(ϕ)R(θ). T (e2) = w. Use this and matrix multiplication to Thus, the matrix of T is derive the addition laws for sin and cos. −1 cos θ − sin(θ) What is R(θ) ?. R(θ) = . sin(θ) cos(θ) • Exercise: What is the matrix of reflection through the x-axis? Part 2. 5 Part 2. 6
• A matrix A is orthogonal if kAvk = kvk Hence, if A is orthogonal, we have for all vectors v. If A and B are 2Au · Av = kAuk2 + kAvk2 − kAu − Avk2 orthogonal, so is AB. If A is orthogonal, = kAuk2 + kAvk2 − kA(u − v)k2 so is A−1. Clearly I is orthogonal.. Rotation matrices are orthogonal. = kuk2 + kvk2 − ku − vk2 The set of orthogonal 2 × 2 matrices is = 2u · v denoted by O(2). so Au · Av = u · v. Classification of Orthogonal • If A is orthogonal, we must have Matrices 2 kAe1k = Ae1 · Ae1 = e1 · e1 = 1 2 • If A is orthogonal, then Au · Av = u · v (∗) kAe2k = Ae1 · Ae2 = e2 · e2 = 1 for all vectors u and v. (It follows that Ae1 · Ae2 = e1 · e2 = 0. A preserves the angles between vectors). To see this, recall that • Exercise: If A satisfies (∗), A is orthogonal. 2u · v = kuk2 + kv2k − ku − vk2. • Equations (∗) say that the columns of A are orthogonal unit vectors. Thus, if the first column of A is a 2 2 , a + b = 1 b Part 2. 7 Part 2. 8
the possibilities for the second column Call this matrix S(θ). What does this are represent geometrically? −b b , • We claim that there is a line L through a −a the origin so that A = S(θ) leaves every So A must look like one of the matrices point on L fixed. To see this, it will suffice to find a unit vector a −b a b , . cos(ϕ) b a b −a u = sin(ϕ)
• Since a2 + b2 = 1, we can find θ so that so that Au = u. If we do this, any other a = cos(θ) and b = sin(θ). Thus, in one vector along L will have the form tu for case, we get some scalar t, and we will have A(tu) = tAu = tu. Since we only need to a −b cos θ − sin(θ) consider each line through the origin A = = = R(θ), b a sin(θ) cos(θ) once, we can suppose 0 ≤ ϕ < 180◦. We may as well suppose that 0 ≤ θ < 360◦. so A is rotation by θ degrees. In the Thus, we are trying to find ϕ so that other case we have Au = u, i.e. a b cos θ sin(θ) cos θ sin(θ) cos(ϕ) cos(ϕ) A = = = b −a sin(θ) − cos(θ) sin(θ) − cos(θ) sin(ϕ) sin(ϕ) Part 2. 9 Part 2. 10
If we carry out the multiplication, this which is a unit vector orthogonal to u. becomes Calculate Av as follows cos(θ) cos(ϕ) + sin(θ) sin(ϕ) cos(ϕ) cos θ sin(θ) − sin(θ/2) = Av = sin(θ) cos(ϕ) − cos(θ) sin(ϕ) sin(ϕ) sin(θ) − cos(θ) cos(θ/2) Using the addition laws, this is the same − cos(θ) sin(θ/2) + sin(θ) cos(θ/2) = as − sin(θ) sin(θ/2) − cos(θ) cos(θ/2) cos(θ − ϕ) cos(ϕ) = sin(θ − θ/2) sin(θ − ϕ) sin(ϕ) = − cos(θ − θ/2) Because of our angle restrictions, we must have θ − ϕ = ϕ, so ϕ = θ/2. Thus, sin(θ/2) = Au = u for − cos(θ/2) cos(θ/2) = −v, u = sin(θ/2). So Av = −v. A little thought shows that A is reflection through the line L that Consider the vector passes through the origin and is parallel to u. Any vector w can be written as − sin(θ/2) v = , w = su + tv for scalars s and t and cos(θ/2) A(su + tv) = sAu + tAv = su − tv. Part 2. 11 Part 2. 12
Isometries of the Plane
£¥¤
¦¨§¥© • If a point has coordinates (x, y), the ¢¡
vector with components (x, y) is the
vector from the origin to the point. This is called the position vector of the point. Generally, we use the same
notation for the point and its position vector.
• Exercise: Verify the following • The distance between two points p and equations. q in R2 is given by d(p, q) = kp − qk.
R(θ)S(ϕ) = S(θ + ϕ) • A transformation T : R2 → R2 is called S(ϕ)R(θ) = S(ϕ − θ) an isometry if it is 1-1 and onto and d(T (p),T (q)) = d(p, q) for all points p and S(θ)S(ϕ) = R(θ − ϕ). q.
• One kind of isometry we have discovered is T (p) = Ap, where A is an orthogonal matrix.
• Another kind of isometry is translation.
Let v be a fixed vector and define Tv by Part 2. 13 Part 2. 14
Tv(p) = p + v. Translations are not linear. • If x, y and z are three noncollinear points and T is an isometry such that • The identity mapping id: 2 → 2 R R T (x) = x, T (y) = y and T (z) = z, then defined by id(p) = p is obviously an T = id. isometry. The composition of two To see this, suppose that is any point. isometries is an isometry. The inverse p Then is determined by the numbers of an isometry is an isometry. p d(x, p), d(y, p) and d(z, p). Since T is an • Let x, y and z be noncollinear points. isometry Then a point p is uniquely determined d(x, p) = d(T (x),T (p)) = d(x, T (p)). by the three numbers d(x, p), d(y, p) and Similarly, d(y, T (p)) = d(y, p) and d(z, p). d(z, T (p)) = d(z, p). Thus, we must have
T (p) = p.
Problem: If points , and form a
¢ • x y z
right triangle with the right angle at x, £
then T (x), T (y) and T (z) form a right ¡ triangle with the right angle at T (x). Part 2. 15 Part 2. 16
• Theorem If T is an isometry, there is a let M be the matrix of reflection vector v and an orthogonal matrix A so through the x axis. In either case M is 00 that T (p) = Ap + v for all p, i.e., T is the an orthogonal matrix and MT (e2) = e2. 000 000 composition of a rotation or reflection Let T = MRTvT . Then T (0) = 0, 000 000 000 at the origin and a translation. T (e1) = e1 and T (e2) = e2, so T must be the identity. If we set , is Proof: The points 0, e1 and e2 are A = MR A 000 noncollinear and form a right triangle, orthogonal and we have T = ATvT = id.
so the points T (0), T (e1) and T (e2) form Thus, for any point point p, −1 a right triangle. ATvT (p) = p. Multiplying by A we get T T (p) = A−1p. Apply T to both sides Let v = −T (0). Let Tv be translation by v −v 0 to get T (p) = T A−1p = A−1p − v. Since v. Then T = TvT is an isometry and −v −1 T 0(0) = 0. We must have A is orthogonal, the proof is 0 complete. d(0,T (e1)) = d(0, e1) = 1, so we can 0 rotate T (e1) around the origin so it • Every isometry can be written in the coincides with e1. Let R be the rotation form T (x) = Ax + u. To abbreviate this, matrix for this rotation. Then we’ll simply denote this isometry by 00 00 T = RTvT is an isometry with T (0) = 0 (A | u), so the rule is 00 00 and T (e1) = e1. Since d(T (e2), 0) = 1 00 (A | u)(x) = Ax + u and the points 0, e1 and T (e2) form a 00 right triangle, T (e2) must be either e2 In this notation, (I | 0) = id, (I | u) is
or −e2. If it’s e2 let M = I and if it’s −e2 translation by u, and (A | 0) is Part 2. 17 Part 2. 18
transformation x 7→ Ax. I. The identity. Given isometries (A | u) and (B | v), we II. A translation. have III. Rotation about some point (i.e., not (A | u)(B | v)(x) = (A | u)(Bx + v) necessarily the origin). = A(Bx + v) + u IV. Reflection about some line. = ABx + Av + u V. A glide reflection. = (AB | u + Av)(x), • We already know that (I | 0) is the so the rule for combining these symbols identity (Case I), (0 | u) is translation is (Case II) and that (A | 0) is either a rotation about the origin or a reflection (A | u)(B | v) = (AB | u + Av). about a line through the origin.
• Exercise: Verify that • Consider (A | w) where A is a rotation (A | u)−1 = (A−1 | −A−1u) matrix, A = R(θ). A Rotation leaves • A Glide Reflection is reflection in some fixed the rotation center, so we expect line followed by translation in a (A | w) to have a fixed point, i.e., a direction parallel to the line. point p so that Ap + w = p. Rearranging this equation gives • Theorem Every isometry falls into one w = p − Ap = (I − A)p, so p is the of the following mutually exclusive cases. Part 2. 19 Part 2. 20
solution, if any, of the equation Now that we know the fixed point p exists, we can write (∗) w = (I − A)p. (A | w) = (A | p − Ap). Then we have
The matrix I − A is given by (A | p − Ap)(x) = Ax + p − Ap = A(x − p) + p. 1 − cos(θ) sin(θ) If we think about this geometrically, this I − A = . − sin(θ) 1 − cos(θ) shows that (A | p − Ap) is rotation about the point p: We take the vector x − p so that points from p to x, take the arrow at the origin, rotate the arrow by A and det(A) = (1 − cos(θ))2 + sin2(θ) move the arrow back to p. This gives 2 2 = 1 − 2 cos θ + cos (θ) + sin (θ) us the rotation of x about p. = 2 − 2 cos θ. • Consider the case (A | w) where A is a This is nonzero as long as cos(θ) 6= 1. If reflection matrix. Recall that for a cos(θ) = 1, then A = I, and we’ve already reflection matrix, there are orthogonal dealt with that case. So, we can unit vectors u and v so that Au = u and assume det(I − A) 6= 0. Av = −v. Consider first the case where w = sv. Let p = sv/2. We claim that p is This means that I − A is invertible, so equation (∗) has the unique solution p = (I − A)−1w. Part 2. 21 Part 2. 22
fixed. so all the points on L are fixed. We can also write (A | sv)(p) = Ap + sv = A(sv/2) + sv (A | 2p)(x) = Ax + 2p = (s/2)Av + sv = Ax + p + p = (s/2)(−v) + sv = Ax − (−p) + p = (s − s/2)v = Ax − Ap + p = sv/2 = A(x − p) + p. = p. By the same reasoning as before, this Thus, sv = 2p and we can write shows that (A | 2p) is reflection through (A | w) = (A | 2p). Note Ap = −p. L. As t runs over all real numbers, the • Exercise: Show that (A | w) has fixed point p + tu run along the line, call it L, points if and only if w = sv. that passes through p and is parallel to u. For points on L, we have • Consider finally the case (A | w) where w (A | 2p)(p + tu) = A(p + tu) + 2p is not a multiple of v. Then we can write w = αu + βv. We then have = Ap + tu + 2p = −p + tu + 2p (A | w) = (A | αu + βv) = p + tu, = (I | αu)(A | βv) Part 2. 23
We know that (A | βv) is reflection about some line parallel to u and (I | αu) is translation in a direction parallel to u, so (A | w) is a glide reflection. A glide reflection has no fixed points.