Lecture 12: Orthonormal Matrices

Lecture 12: Orthonormal Matrices 12 Orthogonal Matrices In this lecture, we start formally studying the symmetry of shapes, combining group theory with linear algebra. The matrices considered will be over R, the feld of real numbers, rather than C. 12.1 Dot Products and Orthogonal Matrices Recall the following defnitions. Defnition 12.1 Pn Given column vectors x, y ∈ Rn , the dot product is defned as x · y = xT y = i=1 xi yi . The length of a √ vector v is |v| = v · v. The dot product is defned algebraically, but also carries geometric information about two vectors: x · y = |x||y| cos θ. Moreover, if x · y = 0, then x and y will be perpendicular vectors in Rn . To start out with, consider bases for which the pairwise dot products are as simple as possible. Defnition 12.2 √ A basis {v1 , · · · , vn } is called orthonormal if |vi | = 1 and vi · vj = 0 for i ̸= j. That is, since |vi | = vi · vi , vi · vj = δij , where δij denotes the Kronecker delta.a a The Kronecker delta δij is equal to 0 if i ̸= j and 1 if i = j. Now, Rn not√only has a vector space structure, but it also has some extra structure provided by the dot product. Since |v| = v · v, the dot product produces some notion of "length" or "distance." Guiding Question What kinds of matrices interact well with this notion of distance? Orthogonal matrices are those preserving the dot product. Defnition 12.3 A matrix A ∈ GLn (R) is orthogonal if Av · Aw = v · w for all vectors v and w. In particular, taking v = w means that lengths are preserved by orthogonal matrices. There are many equivalent characterizations for orthogonal matrices. Theorem 12.4 The following conditions are all equivalent: 1. The matrix A is orthogonal. 2. For all vectors v ∈ Rn , |Av| = |v|. That is, A preserves lengths. 3. For an n-dimensional matrix A, AT A = In . 4. The columns of A form an orthonormal basis.a a Since AT also satisfes the third condition, this means that the rows of A, which are the columns of AT , will also form an orthonormal basis. Proof. All the conditions will end up equivalent. Lecture 12: Orthonormal Matrices • Condition (1) implies (2). Because A preserves dot products, |Av| = also preserves lengths. √ Av · Av = √ v · v = |v|, and so A • Condition (2) implies (1) because
1� |Av + Aw|2 − |Av|2 − |Aw|2 2
1� = |v + w|2 − |v|2 − |w|2 2 = v · w. Av · Aw = Namely, dot products can be written in terms of lengths, and lengths can be written in terms of dot products, so preserving one is equivalent to preserving the other. • Condition (1) states that Av · Aw = v · w; unwinding the dot product in terms of matrix multiplication, this equation is v T AT Aw = v T w for all v, w ∈ Rn . Evidently, (3) implies (1), since if AT A = In , v T AT Aw = v T w. By calculation, it can be seen that for ei and ej the ith and jth standard basis vectors, eTi M ej = Mij , which is the (i, j)th component of the matrix M. If (1) is true, taking v = ei and w = ej over all i and j gives us that the (i, j)th component of AT A is 1 when i = j and 0 otherwise. • Condition (4) is equivalent to (3) from simply computing the matrix product: the (i, j)th entry of AT A is the dot product of the ith column of A with the jth column of A, which is 1 when i = j and 0 otherwise. Orthogonal matrices preserve lengths, as well as preserving angles up to sign. In general, a set of matrices satisfying some well-behaved properties of a set of matrices generally form a subgroup, and this principle does hold true in the case of orthogonal matrices. Proposition 12.5 The orthogonal matrices form a subgroup On of GLn . Proof. Using condition (3), if for two orthogonal matrices A and B, AT A = B T B = In , it is clear that (AB)T AB = B T AT AB = B T B = In . The other subgroup properties are not difcult to verify. 12.2 The Special Orthogonal Group Given an orthogonal matrix A, AT A = In , and so det(AT A) = det(AT ) det(A) = det(A)2 = det(In ) = 1. As a result, det(A) = ±1. The determinant is a homomorphism from det : GLn −→ R, and the restriction to On is a homomorphism det : On −→ {±1}. The kernel forms a subgroup of On . Defnition 12.6 (Special Orthogonal Group) The orthogonal matrices with determinant 1 form a subgroup SOn ⊂ On ⊂ GLn called the special orthogonal group. Because the determinant is surjective41 , the kernel, SOn , is an index 2 subgroup inside of On . The two cosets are SOn itself and all the matrices with determinant −1. To gain some intuition for orthogonal matrices, we will look at some examples! For n = 1, the orthogonal group has two elements, [1] and [−1], which is not too interesting. 12.3 Orthogonal Matrices in Two Dimensions What are the orthogonal matrices in two dimensions? 41 For example, the identity matrix is always orthogonal and has determinant 1, and the diagonal matrix with −1 in the frst row and column and 1 down the rest of the diagonal is also orthogonal and has determinant −1. Lecture 12: Orthonormal Matrices Example 12.7 (O2 ) 2 Describing an element of O2 is equivalent to writing down an orthonormal   basis {v1 , v2 } of R . Evidently, cos θ v1 must be a unit vector, which can always be described as v1 = for some angle θ. Then v2 must sin θ     − sin θ sin θ also have length 1 and be perpendicular to v1 . There are two choices, v2 = or . This cos θ − cos θ characterizes all 2×2 orthogonal matrices:     cos θ − sin θ cos θ sin θ O2 = , . sin θ cos θ sin θ − cos θ In particular, the frst type of matrix has determinant 1, and forms the subgroup SOn , and the second has determinant −1 and forms its the non-trivial coset. Geometrically, the frst type   of matrix in O2 are rotations cos θ sin θ by θ around the origin. The matrices of the second type, A = , have characteristic polynomial sin θ − cos θ pA (t) = t2 − 1 = (t + 1)(t − 1). Thus, they have distinct eigenvalues ±1, in contrast to rotation matrices, which do not have any real eigenvalues. Because the eigenvalues are distinct, there is an eigenbasis {v⃗+ , v⃗− }. Lecture 12: Orthonormal Matrices Theorem 12.8 The matrices of the second type are refections across a line through the origin at an angle of θ/2. Proof. Consider the line L = Span(v⃗+ ); since v⃗+ is an eigenvector with eigenvalue 1, A fxes this line. Notice that v⃗+ · v⃗− = Av⃗+ · Av⃗− = v⃗+ · (−v⃗− ), where the frst equality comes from the fact that A is orthogonal, and the second comes from the eigenvalues 1 and −1 of v+ and v− . The only possibility is v⃗+ · v⃗− = 0, so the two eigenvectors are orthogonal. Writing out any other vector in terms of the eigenvectors, Av is precisely the refection across L. As expected, rotations and refections preserve distance, and in fact they make up all the 2×2 orthogonal matrices. A fun fact that comes from this analysis is that the composition of two refections over diferent lines will be a rotation, since the product of determinants will be (−1) · (−1) = 1. Orthogonal matrices can be thought of either geometrically or algebraically! 12.4 Orthogonal Matrices in Three Dimensions In two dimensions, SO2 consists of rotation matrices. It turns out that in three dimensions, SO3 also consists of rotation matrices. Lecture 12: Orthonormal Matrices In particular, a rotation in R3 is characterized by the axis of the rotation, which is a unit vector ⃗u ∈ R3 , and the angle of the rotation, which is some θ ∈ R. The plane u⊥ = {v ∈ R3 : u · v = 0} consists of all the vectors in R3 that are perpendicular to R3 . Defnition 12.9 The rotation operator with spin labels u and θ is ρ(u,θ) , the linear operator ρ : R3 −→ R3 such that ρ(u) = u and ρ|u⊥ is the rotation by θ counterclockwise with respect to the direction that u points in.a a Since every vector in R3 is a linear combination of u and some vector in u⊥ , the rotation operator is described completely by these conditions. There is some redundancy in this description; for example, ρ(u,θ) = ρ(−u,−θ) . Theorem 12.10 The rotation operators are exactly SO3 . From geometric intuition, this result is not very surprising, since rotations preserve distance.42 Proof. First, we show that all the rotation matrices are in SO3 , and then we show that all matrices in SO3 are rotation matrices. • We frst show that all of these rotation matrices belong to SO3 . Let {v, w} be an orthonormal basis for the plane u⊥ , and let P be a 3 × 3 matrix with columns (u, v, w). Since v and w are orthogonal to each other, and u is orthogonal to both v and w, P ∈ O3 . Conjugating a rotation matrix by P demonstrates the action of ρ(u,θ) with respect to the basis (u, v, w). Since u is fxed by the rotation matrix, the frst column is (1, 0, 0)t , and since the plane u⊥ is being rotated by θ, the rest of the matrix M is given by the form of a 2×2 rotation matrix. That is,   1 0 0 P −1 ρ(u,θ) P = 0 cos θ − sin θ = M, 0 sin θ cos θ which is in SO3 . Since ρ(u,θ) = P M P −1 , and since P, P −1 ∈ O3 and M ∈ SO3 , the rotation matrix ρ(u,θ) is also in O3 . Taking the determinant of both sides43 demonstrates that ρ(u,θ) ∈ SO3 . • To show the other direction, an element A ∈ SO3 must be shown to be rotation around some axis u, which has to be some eigenvector with eigenvalue λ = 1. There exists such an eigenvector if and only if 1 is a root of the characteristic polynomial of A, which is precisely when det(I − A) = 0. 42 And orientation = det(P ) det(M ) det(P )−1 = det(M ) = 1 43 det(P M P −1 ) Lecture 12: Orthonormal Matrices Since det(AT ) = 1, det(A − I) = det(AT (A − I)). Using the fact that A is orthogonal, this is det(I − AT ). Taking the transpose, this is det(I − A). Since the matrices are 3×3, det(I − A) = (−1)3 det(A − I). Combining these, det(A − I) = det(AT (A − I)) = det(I − AT ) = det(I − A) = (−1)3 det(A − I), implying that det(A − I) = 0. Therefore, there does exist an eigenvector of eigenvalue 1 for A, which can be scaled to be a unit vector u. We extend u to an orthonormal basis P = (u, v, w) by picking an orthonormal basis for u⊥ . Consider taking A in this basis. The frst column is (1, 0, 0)t , since u is an eigenvector, and the frst row is (1, 0, 0) because the columns are orthogonal. Then, the bottom right submatrix is an element of SO2 by taking the determinant. So   1 0 0 P −1 AP = 0 cos − sin , 0 sin cos and we are done.