8. Orthogonality
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The Grassmann Manifold
The Grassmann Manifold 1. For vector spaces V and W denote by L(V; W ) the vector space of linear maps from V to W . Thus L(Rk; Rn) may be identified with the space Rk£n of k £ n matrices. An injective linear map u : Rk ! V is called a k-frame in V . The set k n GFk;n = fu 2 L(R ; R ): rank(u) = kg of k-frames in Rn is called the Stiefel manifold. Note that the special case k = n is the general linear group: k k GLk = fa 2 L(R ; R ) : det(a) 6= 0g: The set of all k-dimensional (vector) subspaces ¸ ½ Rn is called the Grassmann n manifold of k-planes in R and denoted by GRk;n or sometimes GRk;n(R) or n GRk(R ). Let k ¼ : GFk;n ! GRk;n; ¼(u) = u(R ) denote the map which assigns to each k-frame u the subspace u(Rk) it spans. ¡1 For ¸ 2 GRk;n the fiber (preimage) ¼ (¸) consists of those k-frames which form a basis for the subspace ¸, i.e. for any u 2 ¼¡1(¸) we have ¡1 ¼ (¸) = fu ± a : a 2 GLkg: Hence we can (and will) view GRk;n as the orbit space of the group action GFk;n £ GLk ! GFk;n :(u; a) 7! u ± a: The exercises below will prove the following n£k Theorem 2. The Stiefel manifold GFk;n is an open subset of the set R of all n £ k matrices. There is a unique differentiable structure on the Grassmann manifold GRk;n such that the map ¼ is a submersion. -
Orthogonal Bases and the -So in Section 4.8 We Discussed the Problem of Finding the Orthogonal Projection P
Orthogonal Bases and the -So In Section 4.8 we discussed the problemR. of finding the orthogonal projection p the vector b into the V of , . , suhspace the vectors , v2 ho If v1 v,, form a for V, and the in x n matrix A has these basis vectors as its column vectors. ilt the orthogonal projection p is given by p = Ax where x is the (unique) solution of the normal system ATAx = A7b. The formula for p takes an especially simple and attractive Form when the ba vectors , . .. , v1 v are mutually orthogonal. DEFINITION Orthogonal Basis An orthogonal basis for the suhspacc V of R” is a basis consisting of vectors , ,v,, that are mutually orthogonal, so that v v = 0 if I j. If in additii these basis vectors are unit vectors, so that v1 . = I for i = 1. 2 n, thct the orthogonal basis is called an orthonormal basis. Example 1 The vectors = (1, 1,0), v2 = (1, —1,2), v3 = (—1,1,1) form an orthogonal basis for . We “normalize” ‘ R3 can this orthogonal basis 1w viding each basis vector by its length: If w1=—- (1=1,2,3), lvii 4.9 Orthogonal Bases and the Gram-Schmidt Algorithm 295 then the vectors /1 I 1 /1 1 ‘\ 1 2” / I w1 0) W2 = —— _z) W3 for . form an orthonormal basis R3 , . ..., v, of the m x ii matrix A Now suppose that the column vectors v v2 form an orthogonal basis for the suhspacc V of R’. Then V}.VI 0 0 v2.v .. -
2 Hilbert Spaces You Should Have Seen Some Examples Last Semester
2 Hilbert spaces You should have seen some examples last semester. The simplest (finite-dimensional) ex- C • A complex Hilbert space H is a complete normed space over whose norm is derived from an ample is Cn with its standard inner product. It’s worth recalling from linear algebra that if V is inner product. That is, we assume that there is a sesquilinear form ( , ): H H C, linear in · · × → an n-dimensional (complex) vector space, then from any set of n linearly independent vectors we the first variable and conjugate linear in the second, such that can manufacture an orthonormal basis e1, e2,..., en using the Gram-Schmidt process. In terms of this basis we can write any v V in the form (f ,д) = (д, f ), ∈ v = a e , a = (v, e ) (f , f ) 0 f H, and (f , f ) = 0 = f = 0. i i i i ≥ ∀ ∈ ⇒ ∑ The norm and inner product are related by which can be derived by taking the inner product of the equation v = aiei with ei. We also have n ∑ (f , f ) = f 2. ∥ ∥ v 2 = a 2. ∥ ∥ | i | i=1 We will always assume that H is separable (has a countable dense subset). ∑ Standard infinite-dimensional examples are l2(N) or l2(Z), the space of square-summable As usual for a normed space, the distance on H is given byd(f ,д) = f д = (f д, f д). • • ∥ − ∥ − − sequences, and L2(Ω) where Ω is a measurable subset of Rn. The Cauchy-Schwarz and triangle inequalities, • √ (f ,д) f д , f + д f + д , | | ≤ ∥ ∥∥ ∥ ∥ ∥ ≤ ∥ ∥ ∥ ∥ 2.1 Orthogonality can be derived fairly easily from the inner product. -
Orthogonal Complements (Revised Version)
Orthogonal Complements (Revised Version) Math 108A: May 19, 2010 John Douglas Moore 1 The dot product You will recall that the dot product was discussed in earlier calculus courses. If n x = (x1: : : : ; xn) and y = (y1: : : : ; yn) are elements of R , we define their dot product by x · y = x1y1 + ··· + xnyn: The dot product satisfies several key axioms: 1. it is symmetric: x · y = y · x; 2. it is bilinear: (ax + x0) · y = a(x · y) + x0 · y; 3. and it is positive-definite: x · x ≥ 0 and x · x = 0 if and only if x = 0. The dot product is an example of an inner product on the vector space V = Rn over R; inner products will be treated thoroughly in Chapter 6 of [1]. Recall that the length of an element x 2 Rn is defined by p jxj = x · x: Note that the length of an element x 2 Rn is always nonnegative. Cauchy-Schwarz Theorem. If x 6= 0 and y 6= 0, then x · y −1 ≤ ≤ 1: (1) jxjjyj Sketch of proof: If v is any element of Rn, then v · v ≥ 0. Hence (x(y · y) − y(x · y)) · (x(y · y) − y(x · y)) ≥ 0: Expanding using the axioms for dot product yields (x · x)(y · y)2 − 2(x · y)2(y · y) + (x · y)2(y · y) ≥ 0 or (x · x)(y · y)2 ≥ (x · y)2(y · y): 1 Dividing by y · y, we obtain (x · y)2 jxj2jyj2 ≥ (x · y)2 or ≤ 1; jxj2jyj2 and (1) follows by taking the square root. -
Math 2331 – Linear Algebra 6.2 Orthogonal Sets
6.2 Orthogonal Sets Math 2331 { Linear Algebra 6.2 Orthogonal Sets Jiwen He Department of Mathematics, University of Houston [email protected] math.uh.edu/∼jiwenhe/math2331 Jiwen He, University of Houston Math 2331, Linear Algebra 1 / 12 6.2 Orthogonal Sets Orthogonal Sets Basis Projection Orthonormal Matrix 6.2 Orthogonal Sets Orthogonal Sets: Examples Orthogonal Sets: Theorem Orthogonal Basis: Examples Orthogonal Basis: Theorem Orthogonal Projections Orthonormal Sets Orthonormal Matrix: Examples Orthonormal Matrix: Theorems Jiwen He, University of Houston Math 2331, Linear Algebra 2 / 12 6.2 Orthogonal Sets Orthogonal Sets Basis Projection Orthonormal Matrix Orthogonal Sets Orthogonal Sets n A set of vectors fu1; u2;:::; upg in R is called an orthogonal set if ui · uj = 0 whenever i 6= j. Example 82 3 2 3 2 39 < 1 1 0 = Is 4 −1 5 ; 4 1 5 ; 4 0 5 an orthogonal set? : 0 0 1 ; Solution: Label the vectors u1; u2; and u3 respectively. Then u1 · u2 = u1 · u3 = u2 · u3 = Therefore, fu1; u2; u3g is an orthogonal set. Jiwen He, University of Houston Math 2331, Linear Algebra 3 / 12 6.2 Orthogonal Sets Orthogonal Sets Basis Projection Orthonormal Matrix Orthogonal Sets: Theorem Theorem (4) Suppose S = fu1; u2;:::; upg is an orthogonal set of nonzero n vectors in R and W =spanfu1; u2;:::; upg. Then S is a linearly independent set and is therefore a basis for W . Partial Proof: Suppose c1u1 + c2u2 + ··· + cpup = 0 (c1u1 + c2u2 + ··· + cpup) · = 0· (c1u1) · u1 + (c2u2) · u1 + ··· + (cpup) · u1 = 0 c1 (u1 · u1) + c2 (u2 · u1) + ··· + cp (up · u1) = 0 c1 (u1 · u1) = 0 Since u1 6= 0, u1 · u1 > 0 which means c1 = : In a similar manner, c2,:::,cp can be shown to by all 0. -
CLIFFORD ALGEBRAS Property, Then There Is a Unique Isomorphism (V ) (V ) Intertwining the Two Inclusions of V
CHAPTER 2 Clifford algebras 1. Exterior algebras 1.1. Definition. For any vector space V over a field K, let T (V ) = k k k Z T (V ) be the tensor algebra, with T (V ) = V V the k-fold tensor∈ product. The quotient of T (V ) by the two-sided⊗···⊗ ideal (V ) generated byL all v w + w v is the exterior algebra, denoted (V ).I The product in (V ) is usually⊗ denoted⊗ α α , although we will frequently∧ omit the wedge ∧ 1 ∧ 2 sign and just write α1α2. Since (V ) is a graded ideal, the exterior algebra inherits a grading I (V )= k(V ) ∧ ∧ k Z M∈ where k(V ) is the image of T k(V ) under the quotient map. Clearly, 0(V )∧ = K and 1(V ) = V so that we can think of V as a subspace of ∧(V ). We may thus∧ think of (V ) as the associative algebra linearly gener- ated∧ by V , subject to the relations∧ vw + wv = 0. We will write φ = k if φ k(V ). The exterior algebra is commutative | | ∈∧ (in the graded sense). That is, for φ k1 (V ) and φ k2 (V ), 1 ∈∧ 2 ∈∧ [φ , φ ] := φ φ + ( 1)k1k2 φ φ = 0. 1 2 1 2 − 2 1 k If V has finite dimension, with basis e1,...,en, the space (V ) has basis ∧ e = e e I i1 · · · ik for all ordered subsets I = i1,...,ik of 1,...,n . (If k = 0, we put { } k { n } e = 1.) In particular, we see that dim (V )= k , and ∅ ∧ n n dim (V )= = 2n. -
LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 9 SOLUTIONS In
MATH 110: LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 9 SOLUTIONS In the following V will denote a finite-dimensional vector space over R that is also an inner product space with inner product denoted by h ·; · i. The norm induced by h ·; · i will be denoted k · k. 1. Let S be a subset (not necessarily a subspace) of V . We define the orthogonal annihilator of S, denoted S?, to be the set S? = fv 2 V j hv; wi = 0 for all w 2 Sg: (a) Show that S? is always a subspace of V . ? Solution. Let v1; v2 2 S . Then hv1; wi = 0 and hv2; wi = 0 for all w 2 S. So for any α; β 2 R, hαv1 + βv2; wi = αhv1; wi + βhv2; wi = 0 ? for all w 2 S. Hence αv1 + βv2 2 S . (b) Show that S ⊆ (S?)?. Solution. Let w 2 S. For any v 2 S?, we have hv; wi = 0 by definition of S?. Since this is true for all v 2 S? and hw; vi = hv; wi, we see that hw; vi = 0 for all v 2 S?, ie. w 2 S?. (c) Show that span(S) ⊆ (S?)?. Solution. Since (S?)? is a subspace by (a) (it is the orthogonal annihilator of S?) and S ⊆ (S?)? by (b), we have span(S) ⊆ (S?)?. ? ? (d) Show that if S1 and S2 are subsets of V and S1 ⊆ S2, then S2 ⊆ S1 . ? Solution. Let v 2 S2 . Then hv; wi = 0 for all w 2 S2 and so for all w 2 S1 (since ? S1 ⊆ S2). -
Geometric (Clifford) Algebra and Its Applications
Geometric (Clifford) algebra and its applications Douglas Lundholm F01, KTH January 23, 2006∗ Abstract In this Master of Science Thesis I introduce geometric algebra both from the traditional geometric setting of vector spaces, and also from a more combinatorial view which simplifies common relations and opera- tions. This view enables us to define Clifford algebras with scalars in arbitrary rings and provides new suggestions for an infinite-dimensional approach. Furthermore, I give a quick review of classic results regarding geo- metric algebras, such as their classification in terms of matrix algebras, the connection to orthogonal and Spin groups, and their representation theory. A number of lower-dimensional examples are worked out in a sys- tematic way using so called norm functions, while general applications of representation theory include normed division algebras and vector fields on spheres. I also consider examples in relativistic physics, where reformulations in terms of geometric algebra give rise to both computational and conceptual simplifications. arXiv:math/0605280v1 [math.RA] 10 May 2006 ∗Corrected May 2, 2006. Contents 1 Introduction 1 2 Foundations 3 2.1 Geometric algebra ( , q)...................... 3 2.2 Combinatorial CliffordG V algebra l(X,R,r)............. 6 2.3 Standardoperations .........................C 9 2.4 Vectorspacegeometry . 13 2.5 Linearfunctions ........................... 16 2.6 Infinite-dimensional Clifford algebra . 19 3 Isomorphisms 23 4 Groups 28 4.1 Group actions on .......................... 28 4.2 TheLipschitzgroupG ......................... 30 4.3 PropertiesofPinandSpingroups . 31 4.4 Spinors ................................ 34 5 A study of lower-dimensional algebras 36 5.1 (R1) ................................. 36 G 5.2 (R0,1) =∼ C -Thecomplexnumbers . 36 5.3 G(R0,0,1)............................... -
Orthogonal Bases
Orthogonal bases • Recall: Suppose that v1 , , vn are nonzero and (pairwise) orthogonal. Then v1 , , vn are independent. Definition 1. A basis v1 , , vn of a vector space V is an orthogonal basis if the vectors are (pairwise) orthogonal. 1 1 0 3 Example 2. Are the vectors − 1 , 1 , 0 an orthogonal basis for R ? 0 0 1 Solution. Note that we do not need to check that the three vectors are independent. That follows from their orthogonality. Example 3. Suppose v1 , , vn is an orthogonal basis of V, and that w is in V. Find c1 , , cn such that w = c1 v1 + + cnvn. Solution. Take the dot product of v1 with both sides: If v1 , , vn is an orthogonal basis of V, and w is in V, then w · v j w = c1 v1 + + cnvn with cj = . v j · v j Armin Straub 1 [email protected] 3 1 1 0 Example 4. Express 7 in terms of the basis − 1 , 1 , 0 . 4 0 0 1 Solution. Definition 5. A basis v1 , , vn of a vector space V is an orthonormal basis if the vectors are orthogonal and have length 1 . If v1 , , vn is an orthonormal basis of V, and w is in V, then w = c1 v1 + + cnvn with cj = v j · w. 1 1 0 Example 6. Is the basis − 1 , 1 , 0 orthonormal? If not, normalize the vectors 0 0 1 to produce an orthonormal basis. Solution. Armin Straub 2 [email protected] Orthogonal projections Definition 7. The orthogonal projection of vector x onto vector y is x · y xˆ = y. -
A Guided Tour to the Plane-Based Geometric Algebra PGA
A Guided Tour to the Plane-Based Geometric Algebra PGA Leo Dorst University of Amsterdam Version 1.15{ July 6, 2020 Planes are the primitive elements for the constructions of objects and oper- ators in Euclidean geometry. Triangulated meshes are built from them, and reflections in multiple planes are a mathematically pure way to construct Euclidean motions. A geometric algebra based on planes is therefore a natural choice to unify objects and operators for Euclidean geometry. The usual claims of `com- pleteness' of the GA approach leads us to hope that it might contain, in a single framework, all representations ever designed for Euclidean geometry - including normal vectors, directions as points at infinity, Pl¨ucker coordinates for lines, quaternions as 3D rotations around the origin, and dual quaternions for rigid body motions; and even spinors. This text provides a guided tour to this algebra of planes PGA. It indeed shows how all such computationally efficient methods are incorporated and related. We will see how the PGA elements naturally group into blocks of four coordinates in an implementation, and how this more complete under- standing of the embedding suggests some handy choices to avoid extraneous computations. In the unified PGA framework, one never switches between efficient representations for subtasks, and this obviously saves any time spent on data conversions. Relative to other treatments of PGA, this text is rather light on the mathematics. Where you see careful derivations, they involve the aspects of orientation and magnitude. These features have been neglected by authors focussing on the mathematical beauty of the projective nature of the algebra. -
Basics of Euclidean Geometry
This is page 162 Printer: Opaque this 6 Basics of Euclidean Geometry Rien n'est beau que le vrai. |Hermann Minkowski 6.1 Inner Products, Euclidean Spaces In a±ne geometry it is possible to deal with ratios of vectors and barycen- ters of points, but there is no way to express the notion of length of a line segment or to talk about orthogonality of vectors. A Euclidean structure allows us to deal with metric notions such as orthogonality and length (or distance). This chapter and the next two cover the bare bones of Euclidean ge- ometry. One of our main goals is to give the basic properties of the transformations that preserve the Euclidean structure, rotations and re- ections, since they play an important role in practice. As a±ne geometry is the study of properties invariant under bijective a±ne maps and projec- tive geometry is the study of properties invariant under bijective projective maps, Euclidean geometry is the study of properties invariant under certain a±ne maps called rigid motions. Rigid motions are the maps that preserve the distance between points. Such maps are, in fact, a±ne and bijective (at least in the ¯nite{dimensional case; see Lemma 7.4.3). They form a group Is(n) of a±ne maps whose corresponding linear maps form the group O(n) of orthogonal transformations. The subgroup SE(n) of Is(n) corresponds to the orientation{preserving rigid motions, and there is a corresponding 6.1. Inner Products, Euclidean Spaces 163 subgroup SO(n) of O(n), the group of rotations. -
The Orthogonal Projection and the Riesz Representation Theorem1
FORMALIZED MATHEMATICS Vol. 23, No. 3, Pages 243–252, 2015 DOI: 10.1515/forma-2015-0020 degruyter.com/view/j/forma The Orthogonal Projection and the Riesz Representation Theorem1 Keiko Narita Noboru Endou Hirosaki-city Gifu National College of Technology Aomori, Japan Gifu, Japan Yasunari Shidama Shinshu University Nagano, Japan Summary. In this article, the orthogonal projection and the Riesz re- presentation theorem are mainly formalized. In the first section, we defined the norm of elements on real Hilbert spaces, and defined Mizar functor RUSp2RNSp, real normed spaces as real Hilbert spaces. By this definition, we regarded sequ- ences of real Hilbert spaces as sequences of real normed spaces, and proved some properties of real Hilbert spaces. Furthermore, we defined the continuity and the Lipschitz the continuity of functionals on real Hilbert spaces. Referring to the article [15], we also defined some definitions on real Hilbert spaces and proved some theorems for defining dual spaces of real Hilbert spaces. As to the properties of all definitions, we proved that they are equivalent pro- perties of functionals on real normed spaces. In Sec. 2, by the definitions [11], we showed properties of the orthogonal complement. Then we proved theorems on the orthogonal decomposition of elements of real Hilbert spaces. They are the last two theorems of existence and uniqueness. In the third and final section, we defined the kernel of linear functionals on real Hilbert spaces. By the last three theorems, we showed the Riesz representation theorem, existence, uniqu- eness, and the property of the norm of bounded linear functionals on real Hilbert spaces.