In this handout, we discuss orthogonal maps and their significance from a geometric standpoint. Preliminary results on the transpose The definition of an orthogonal matrix involves transpose, so we prove some facts about it first. Proposition 1. (a) If A is an ` × m matrix and B is an m × n matrix, then (AB)> = B>A>: (b) If A is an invertible n × n matrix, then A> is invertible and (A>)−1 = (A−1)>: Proof. (a) We compute the (i; j)-entries of both sides for 1 ≤ i ≤ n and 1 ≤ j ≤ `: m m m > X X > > X > > > > [(AB) ]ij = [AB]ji = AjkBki = [A ]kj[B ]ik = [B ]ik[A ]kj = [B A ]ij: k=1 k=1 k=1 (b) It suffices to show that A>(A−1)> = I. By (a), A>(A−1)> = (A−1A)> = I> = I: Orthogonal matrices Definition (Orthogonal matrix). An n × n matrix A is orthogonal if A−1 = A>. We will first show that \being orthogonal" is preserved by various matrix operations. Proposition 2. (a) If A is orthogonal, then so is A−1 = A>. (b) If A and B are orthogonal n × n matrices, then so is AB. Proof. (a) We have (A−1)> = (A>)> = A = (A−1)−1, so A−1 is orthogonal. (b) We have (AB)−1 = B−1A−1 = B>A> = (AB)>, so AB is orthogonal. The collection O(n) of n × n orthogonal matrices is the orthogonal group in dimension n. The above definition is often not how we identify orthogonal matrices, as it requires us to compute an n × n inverse. Instead, let A be an orthogonal matrix and suppose its columns are v1;:::; vn. Then we can compute 0 > 1 − v1 − 0 j j 1 > B . C v ··· v A A = @ . A @ 1 nA = (vi · vj)1≤i;j≤n; > j j − vn − so comparing with the identity matrix I, we obtain 1 Proposition 3. An n×n matrix A is orthogonal if and only if its columns form an orthonormal basis of Rn. By considering A>, we also have that A is orthogonal if and only if its rows (or rather, their transposes) form an orthonormal basis of Rn. The importance of orthogonal matrices from a geometric perspective is that they preserve dot products, and hence lengths and angles. Theorem 1. Let A 2 O(n) and x; y 2 Rn. Then (Ax) · (Ay) = x · y. Conversely, if A is an n × n matrix preserving dot products, then A 2 O(n). Proof. For the forward direction, (Ax) · (Ay) = (Ax)>(Ay) = x>A>Ay = x>y = x · y. For the reverse, if A is an n × n matrix preserving dot products, then by considering Aei for standard n basis vectors ei, the columns of A form an orthonormal basis of R , hence A 2 O(n). Corollary 1. 1. If A 2 O(n) and x 2 Rn, then kAxk = kxk. 2. If A 2 O(n) and x; y 2 Rn, then Ax ? Ay if and only if x ? y. Example: The two-dimensional orthogonal group O(2) a c Before continuing with general results, we describe the matrices in O(2). Let A = be an b d orthogonal 2 × 2 matrix. By Proposition 3, the equations a; b; c; d must satisfy are a2 + b2 = c2 + d2 = 1 and ac + bd = 0: From the last equation, (c; d) = (tb; −ta) for some t 2 R, so then 1 = c2 + d2 = t2(b2 + a2) = t2: This means that t = ±1, which gives, for a2 + b2 = 1, a b a −b A = or A = : b −a b a The solutions to a2 + b2 = 1 can be parametrised by a single real parameter θ via a = cos θ and b = sin θ, so in conclusion, cos θ − sin θ cos θ sin θ O(2) = [ : sin θ cos θ sin θ − cos θ These are matrices that we have seen before: the first set consists of counterclockwise rotations about the origin by angle θ, whereas the second set consists of reflections about lines through the origin making angles θ=2 with the positive x-axis. Algebraically, another way to separate these two sets is that the first set consists of matrices with determinant 1 and the second set consists of matrices with determinant −1. 2 The special orthogonal group The fact that the only determinants were ±1 in the two-dimensional case is not a coincidence. Proposition 4. If A 2 O(n), then det A = ±1. Proof. Since det A> = det A, we have 1 = det I = det(AA>) = (det A)(det A>) = (det A)2: Definition (Special orthogonal group). The subset SO(n) = fA 2 O(n) j det A = 1g is the special orthogonal group in dimension n. Example: The three-dimensional special orthogonal group SO(3) In three dimensions, the elements of the whole orthogonal group O(3) do not admit as simple a description as in two dimensions, but it turns out that there is a simple description of SO(3). Let A 2 SO(3), so that det A = 1. We analyse the complex eigenvalues of A. To do this, given a matrix M with complex number entries, define My to be its conjugate transpose, i.e. My is obtained from M by taking the transpose and taking the complex conjugate of every entry. If n y v 2 C isp a vector, then v v is a non-negative real number, so we can define the magnitude of v by kvk = vyv. The magnitude satisfies kcvk = jcjkvk for all v 2 Cn and c 2 C, and if v has all real entries, then the magnitude of v as a complex number vector is the same as the magnitude of v as a real number vector. Proposition 5. Let v 2 Cn and A 2 O(n; R). (Here we use the notation O(n; R) to emphasise that A has real entries.) Then kAvk = kvk. Proof. It suffices to compare square magnitudes. We have kAvk2 = (Av)y(Av) = vyAyAv: Since A is real, Ay = A>, so the middle product simplifies to I and we are left with vyv = kvk2. Corollary 2. Let λ 2 C be a complex eigenvalue of A 2 O(n; R). Then jλj = 1. Proof. Let v be an eigenvector of A with eigenvalue λ. Then kvk = kAvk = kλvk = jλjkvk; so jλj = 1 since kvk 6= 0. Returning to the specific case that A 2 SO(3; R), the eigenvalues satisfy det(A − λI) = 0, which upon expanding the left hand side is a cubic with real coefficients. This shows that A has at least one real eigenvalue, which by Corollary 2 must be ±1. Together with the fact that the product of the roots of the eigenvalue equation, counted with multiplicity, is det A = 1, we can show that A must have 1 as an eigenvalue. 3 • If A has only real roots, then each of the three roots is ±1 and their product is 1. They cannot all be −1, as the product would be −1, so at least one of the roots is 1. • If A has non-real complex roots, then since the coefficients are real, the two non-real roots must be a conjugate pair λ, λ¯. Their product is λλ¯ = jλj2 = 1 by Corollary 2, so the last root must be 1. Let u be an eigenvector of A with eigenvalue 1, and if necessary, rescale so that kuk = 1. Since u solves the linear system (A − I)u = 0, which has real coefficients, we can take u to have real coordinates, so although we passed to complex numbers above, we can now return to the 3 3 setting of R . Extend the singleton list (u) to an orthonormal basis B = (u; u2; u3) of R . If S is the matrix whose columns are the vectors of B, then the matrix of A with respect to B is B = S−1AS. Since the columns of S form an orthonormal basis, S is itself an orthogonal matrix, so by Proposition 2(b), B is orthogonal. Moreover, since Au = u, 01 ∗ ∗1 B = @0 ∗ ∗A : 0 ∗ ∗ The first column must be orthogonal to the second and third columns, so 01 0 01 B = @0 ∗ ∗A : 0 ∗ ∗ The second and third columns are an orthonormal list, so this together with det B = det A = 1 shows that the bottom 2 × 2 matrix is a two-dimensional rotation matrix. Hence A is a matrix of rotation about the u-axis, and the conclusion is that SO(3) is the collection of all rotations about lines through the origin. Reflections about hyperplanes A linear map is orthogonal if it preserves dot products, or equivalently, if its matrix (with respect to the standard basis) is an orthogonal matrix. A useful class of orthogonal maps is that of reflections. In 2 dimensions, the most important reflections are those about lines through the origin (dimension-1 subspaces), whereas in 3 dimensions, the most important reflections are those about planes through the origin (dimension-2 subspaces). To generalise, we define Definition. A (linear) hyperplane in Rn is a subspace of dimension n − 1. Equivalently, a linear hyperplane is a subspace of Rn defined by a single linear equation. Proposition 6. Let a1; : : : ; an be real numbers, not all zero. Then n V = fx 2 R j a1x1 + ··· + anxn = 0g is a linear hyperplane in Rn. Conversely, every linear hyperplane is of this form. > Proof. Given such an equation, let a = a1 ··· an . Then V is the orthogonal complement of span(a), hence has dimension n − 1.