Introduction to Linear Algebra V

Total Page:16

File Type:pdf, Size:1020Kb

Introduction to Linear Algebra V Introduction to Linear Algebra V Jack Xin (Lecture) and J. Ernie Esser (Lab) ∗ Abstract Eigenvalue, eigenvector, Hermitian matrices, orthogonality, orthonormal basis, singular value decomposition. 1 Eigenvalue and Eigenvector For an n × n matrix A, if A x = λ x; (1.1) has a nonzero solution x for some complex number λ, then x is eigenvector corresponding to eigenvalue λ. Equation (1.1) is same as saying x belongs to the null space of A − λI, or A − λI is singular or the so called characteristic equation holds: det(A − λI) ≡ p(λ) = 0; (1.2) p(λ) is a polynomial of degree n, hence n complex eigenvalues. In Matlab, eigenvalues and eigenvectors are given by [V,D]=eig(A), where columns of V are eigenvectors, D is a diagonal matrix with entries being eigenvalues. Matrix A is diagonalizable (A = V DV −1, D diagonal) if it has n linearly independent eigenvectors. A sufficient condition is that all n eigenvalues are distinct. 2 Hermitian Matrix For any complex valued matrix A, define AH = A¯T , where bar is complex conjugate. A is Hermitian if AH = A, for example: 3 2 − i A = 2 + i 4 ∗Department of Mathematics, UCI, Irvine, CA 92617. 1 A basic fact is that eigenvalues of a Hermitian matrix A are real, and eigenvectors of distinct eigenvalues are orthogonal. Two complex column vectors x and y of the same dimension are orthogonal if xH y = 0. The proof is short and given below. Consider eigenvalue equation: Ax = λx, and let α = xH Ax, then: α¯ = αH = (xH Ax)H = xH Ax = α; so α is real. On the other hand, α = λxH x, so λ is real. Let xi (i=1,2) be eigenvectors corresponding to distinct eigenvalues λi (i=1,2). We have the identities: H H H (Ax1) x1 = x1 Ax2 = λ2x1 x2; H H H H H H (Ax1) x2 = (x2 Ax1) = (λ1x2 x1) = λ1x1 x2; H so λ1 6= λ2 implies x1 x2 = 0. It follows that by choosing orthogonal basis for each eigenspace, Hermitian matrix A has n-orthonormal (orthogonal and of unit length) eigen-vectors, which become an orthogonal basis for Cn. Putting orthonomal eigenvectors as columns yield a matrix U so that U H U = I, which is called unitary matrix. If A is real, unitary matrix becomes orthogonal matrix U T U = I. Clearly a Hermitian matrix can be diagonalized by a unitary matrix (A = UDU H ). The necessary and sufficient condition for unitary diagonalization of a matrix is that it is normal, or satisfying the equation: AAH = AH A: This includes any skew-Hermitian matrix (AH = −A). 3 Orthogonal Basis n T In R , let v1, v2, ..., vn be n orthonormal column vectors, vi vj = δij (=1 if i=j, otherwise 0). Then each vector v has the representation: n X T v = cj vj; cj = v vj: j=1 2 Here cjvj is the projection of v onto vj. Pn If u = i=1 bjvj, then: n T X u v = bjcj; j=1 and n 2 T X 2 kuk2 = length of u squared = u u = bj ; j=1 which is called Parseval formula (generalization of Pythagorean theorem). An example of N-dimensional orthogonal basis is given by the discrete cosine transform: X = DCT ∗ x; where DCT is n × n orthogonal matrix: 1 DCT = w(k) cos π n − (k − 1)=N ; 2 k=1:N;n=1:N p w(1) = 1= N; w(k) = p2=N, if k ≥ 2. In Matlab, X = dct(x), DCT=dct(eye(n)), n ≥ 2. 4 Singular Value Decomposition (SVD) For a general real m × n matrix A, a factorization similar to orthogonal diagonalization of symmetric matrices (AT = A) is SVD. Suppose m ≥ n, then there are m × m orthogonal matrix U and n×n orthogonal matrix V , also non-negative numbers σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0, such that A = UΣV T ; (4.3) and Σ = [diag([σ1 σ2 ··· σn]); 0]; is m × n matrix, 0 is m-n zero rows of dimension n. The numbers σi's are called singular values. It follows from (4.3) that AT A = V ΣT ΣV T , T 2 Σ Σ = diag([σ1; σ2 ··· σn]); 3 2 T so σj (j=1:n) are real eigenvalues of A A, while columns of V are corresponding orthogonal eigenvectors. From AV = UΣ, we see that each column of U is uj = Avj/σj, j=1, 2, ..., r, where r is the rank of A (or the number of nonzero singular values). Check that uj's are orthonormal. Putting uj's (j=1:r) together gives part of the column vectors of U (the U1 in U = [U1 U2]), the other part U2 is the orthogonal complement. Since uj's (j=1:r) span than ? T the range of A (range(A)), U2 consist of orthonormal column vectors in range(A) = N(A ), the nullspace of AT . In Matlab, [U,S,V]=svd(A) gives the result (S in lieu of Σ). Keeping k < r of the T singular values gives the rank-k approximation of A, or A ≈ USkV , where Sk is obtained from S by zeroing out σj (j=k+1:r), so called low rank approximation, which is useful in image compression among other applications. The approximation is optimal in Frobenius sense (or in the sense of Euclidean, l2, norm of matrices). References [1] S. Leon, Linear Algebra with Applications, Pearson, Prentice-Hall, 2010. 4.
Recommended publications
  • Math 4571 (Advanced Linear Algebra) Lecture #27
    Math 4571 (Advanced Linear Algebra) Lecture #27 Applications of Diagonalization and the Jordan Canonical Form (Part 1): Spectral Mapping and the Cayley-Hamilton Theorem Transition Matrices and Markov Chains The Spectral Theorem for Hermitian Operators This material represents x4.4.1 + x4.4.4 +x4.4.5 from the course notes. Overview In this lecture and the next, we discuss a variety of applications of diagonalization and the Jordan canonical form. This lecture will discuss three essentially unrelated topics: A proof of the Cayley-Hamilton theorem for general matrices Transition matrices and Markov chains, used for modeling iterated changes in systems over time The spectral theorem for Hermitian operators, in which we establish that Hermitian operators (i.e., operators with T ∗ = T ) are diagonalizable In the next lecture, we will discuss another fundamental application: solving systems of linear differential equations. Cayley-Hamilton, I First, we establish the Cayley-Hamilton theorem for arbitrary matrices: Theorem (Cayley-Hamilton) If p(x) is the characteristic polynomial of a matrix A, then p(A) is the zero matrix 0. The same result holds for the characteristic polynomial of a linear operator T : V ! V on a finite-dimensional vector space. Cayley-Hamilton, II Proof: Since the characteristic polynomial of a matrix does not depend on the underlying field of coefficients, we may assume that the characteristic polynomial factors completely over the field (i.e., that all of the eigenvalues of A lie in the field) by replacing the field with its algebraic closure. Then by our results, the Jordan canonical form of A exists.
    [Show full text]
  • Lecture Notes: Qubit Representations and Rotations
    Phys 711 Topics in Particles & Fields | Spring 2013 | Lecture 1 | v0.3 Lecture notes: Qubit representations and rotations Jeffrey Yepez Department of Physics and Astronomy University of Hawai`i at Manoa Watanabe Hall, 2505 Correa Road Honolulu, Hawai`i 96822 E-mail: [email protected] www.phys.hawaii.edu/∼yepez (Dated: January 9, 2013) Contents mathematical object (an abstraction of a two-state quan- tum object) with a \one" state and a \zero" state: I. What is a qubit? 1 1 0 II. Time-dependent qubits states 2 jqi = αj0i + βj1i = α + β ; (1) 0 1 III. Qubit representations 2 A. Hilbert space representation 2 where α and β are complex numbers. These complex B. SU(2) and O(3) representations 2 numbers are called amplitudes. The basis states are or- IV. Rotation by similarity transformation 3 thonormal V. Rotation transformation in exponential form 5 h0j0i = h1j1i = 1 (2a) VI. Composition of qubit rotations 7 h0j1i = h1j0i = 0: (2b) A. Special case of equal angles 7 In general, the qubit jqi in (1) is said to be in a superpo- VII. Example composite rotation 7 sition state of the two logical basis states j0i and j1i. If References 9 α and β are complex, it would seem that a qubit should have four free real-valued parameters (two magnitudes and two phases): I. WHAT IS A QUBIT? iθ0 α φ0 e jqi = = iθ1 : (3) Let us begin by introducing some notation: β φ1 e 1 state (called \minus" on the Bloch sphere) Yet, for a qubit to contain only one classical bit of infor- 0 mation, the qubit need only be unimodular (normalized j1i = the alternate symbol is |−i 1 to unity) α∗α + β∗β = 1: (4) 0 state (called \plus" on the Bloch sphere) 1 Hence it lives on the complex unit circle, depicted on the j0i = the alternate symbol is j+i: 0 top of Figure 1.
    [Show full text]
  • A New Description of Space and Time Using Clifford Multivectors
    A new description of space and time using Clifford multivectors James M. Chappell† , Nicolangelo Iannella† , Azhar Iqbal† , Mark Chappell‡ , Derek Abbott† †School of Electrical and Electronic Engineering, University of Adelaide, South Australia 5005, Australia ‡Griffith Institute, Griffith University, Queensland 4122, Australia Abstract Following the development of the special theory of relativity in 1905, Minkowski pro- posed a unified space and time structure consisting of three space dimensions and one time dimension, with relativistic effects then being natural consequences of this space- time geometry. In this paper, we illustrate how Clifford’s geometric algebra that utilizes multivectors to represent spacetime, provides an elegant mathematical framework for the study of relativistic phenomena. We show, with several examples, how the application of geometric algebra leads to the correct relativistic description of the physical phenomena being considered. This approach not only provides a compact mathematical representa- tion to tackle such phenomena, but also suggests some novel insights into the nature of time. Keywords: Geometric algebra, Clifford space, Spacetime, Multivectors, Algebraic framework 1. Introduction The physical world, based on early investigations, was deemed to possess three inde- pendent freedoms of translation, referred to as the three dimensions of space. This naive conclusion is also supported by more sophisticated analysis such as the existence of only five regular polyhedra and the inverse square force laws. If we lived in a world with four spatial dimensions, for example, we would be able to construct six regular solids, and in arXiv:1205.5195v2 [math-ph] 11 Oct 2012 five dimensions and above we would find only three [1].
    [Show full text]
  • 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 ..
    [Show full text]
  • MATH 2370, Practice Problems
    MATH 2370, Practice Problems Kiumars Kaveh Problem: Prove that an n × n complex matrix A is diagonalizable if and only if there is a basis consisting of eigenvectors of A. Problem: Let A : V ! W be a one-to-one linear map between two finite dimensional vector spaces V and W . Show that the dual map A0 : W 0 ! V 0 is surjective. Problem: Determine if the curve 2 2 2 f(x; y) 2 R j x + y + xy = 10g is an ellipse or hyperbola or union of two lines. Problem: Show that if a nilpotent matrix is diagonalizable then it is the zero matrix. Problem: Let P be a permutation matrix. Show that P is diagonalizable. Show that if λ is an eigenvalue of P then for some integer m > 0 we have λm = 1 (i.e. λ is an m-th root of unity). Hint: Note that P m = I for some integer m > 0. Problem: Show that if λ is an eigenvector of an orthogonal matrix A then jλj = 1. n Problem: Take a vector v 2 R and let H be the hyperplane orthogonal n n to v. Let R : R ! R be the reflection with respect to a hyperplane H. Prove that R is a diagonalizable linear map. Problem: Prove that if λ1; λ2 are distinct eigenvalues of a complex matrix A then the intersection of the generalized eigenspaces Eλ1 and Eλ2 is zero (this is part of the Spectral Theorem). 1 Problem: Let H = (hij) be a 2 × 2 Hermitian matrix. Use the Min- imax Principle to show that if λ1 ≤ λ2 are the eigenvalues of H then λ1 ≤ h11 ≤ λ2.
    [Show full text]
  • Parametrization of 3×3 Unitary Matrices Based on Polarization
    Parametrization of 33 unitary matrices based on polarization algebra (May, 2018) José J. Gil Parametrization of 33 unitary matrices based on polarization algebra José J. Gil Universidad de Zaragoza. Pedro Cerbuna 12, 50009 Zaragoza Spain [email protected] Abstract A parametrization of 33 unitary matrices is presented. This mathematical approach is inspired by polarization algebra and is formulated through the identification of a set of three orthonormal three-dimensional Jones vectors representing respective pure polarization states. This approach leads to the representation of a 33 unitary matrix as an orthogonal similarity transformation of a particular type of unitary matrix that depends on six independent parameters, while the remaining three parameters correspond to the orthogonal matrix of the said transformation. The results obtained are applied to determine the structure of the second component of the characteristic decomposition of a 33 positive semidefinite Hermitian matrix. 1 Introduction In many branches of Mathematics, Physics and Engineering, 33 unitary matrices appear as key elements for solving a great variety of problems, and therefore, appropriate parameterizations in terms of minimum sets of nine independent parameters are required for the corresponding mathematical treatment. In this way, some interesting parametrizations have been obtained [1-8]. In particular, the Cabibbo-Kobayashi-Maskawa matrix (CKM matrix) [6,7], which represents information on the strength of flavour-changing weak decays and depends on four parameters, constitutes the core of a family of parametrizations of a 33 unitary matrix [8]. In this paper, a new general parametrization is presented, which is inspired by polarization algebra [9] through the structure of orthonormal sets of three-dimensional Jones vectors [10].
    [Show full text]
  • Lecture 4: April 8, 2021 1 Orthogonality and Orthonormality
    Mathematical Toolkit Spring 2021 Lecture 4: April 8, 2021 Lecturer: Avrim Blum (notes based on notes from Madhur Tulsiani) 1 Orthogonality and orthonormality Definition 1.1 Two vectors u, v in an inner product space are said to be orthogonal if hu, vi = 0. A set of vectors S ⊆ V is said to consist of mutually orthogonal vectors if hu, vi = 0 for all u 6= v, u, v 2 S. A set of S ⊆ V is said to be orthonormal if hu, vi = 0 for all u 6= v, u, v 2 S and kuk = 1 for all u 2 S. Proposition 1.2 A set S ⊆ V n f0V g consisting of mutually orthogonal vectors is linearly inde- pendent. Proposition 1.3 (Gram-Schmidt orthogonalization) Given a finite set fv1,..., vng of linearly independent vectors, there exists a set of orthonormal vectors fw1,..., wng such that Span (fw1,..., wng) = Span (fv1,..., vng) . Proof: By induction. The case with one vector is trivial. Given the statement for k vectors and orthonormal fw1,..., wkg such that Span (fw1,..., wkg) = Span (fv1,..., vkg) , define k u + u = v − hw , v i · w and w = k 1 . k+1 k+1 ∑ i k+1 i k+1 k k i=1 uk+1 We can now check that the set fw1,..., wk+1g satisfies the required conditions. Unit length is clear, so let’s check orthogonality: k uk+1, wj = vk+1, wj − ∑ hwi, vk+1i · wi, wj = vk+1, wj − wj, vk+1 = 0. i=1 Corollary 1.4 Every finite dimensional inner product space has an orthonormal basis.
    [Show full text]
  • 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.
    [Show full text]
  • 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.
    [Show full text]
  • Orthogonality Handout
    3.8 (SUPPLEMENT) | ORTHOGONALITY OF EIGENFUNCTIONS We now develop some properties of eigenfunctions, to be used in Chapter 9 for Fourier Series and Partial Differential Equations. 1. Definition of Orthogonality R b We say functions f(x) and g(x) are orthogonal on a < x < b if a f(x)g(x) dx = 0 . [Motivation: Let's approximate the integral with a Riemann sum, as follows. Take a large integer N, put h = (b − a)=N and partition the interval a < x < b by defining x1 = a + h; x2 = a + 2h; : : : ; xN = a + Nh = b. Then Z b f(x)g(x) dx ≈ f(x1)g(x1)h + ··· + f(xN )g(xN )h a = (uN · vN )h where uN = (f(x1); : : : ; f(xN )) and vN = (g(x1); : : : ; g(xN )) are vectors containing the values of f and g. The vectors uN and vN are said to be orthogonal (or perpendicular) if their dot product equals zero (uN ·vN = 0), and so when we let N ! 1 in the above formula it makes R b sense to say the functions f and g are orthogonal when the integral a f(x)g(x) dx equals zero.] R π 1 2 π Example. sin x and cos x are orthogonal on −π < x < π, since −π sin x cos x dx = 2 sin x −π = 0. 2. Integration Lemma Suppose functions Xn(x) and Xm(x) satisfy the differential equations 00 Xn + λnXn = 0; a < x < b; 00 Xm + λmXm = 0; a < x < b; for some numbers λn; λm. Then Z b 0 0 b (λn − λm) Xn(x)Xm(x) dx = [Xn(x)Xm(x) − Xn(x)Xm(x)]a: a Proof.
    [Show full text]
  • Inner Products and Orthogonality
    Advanced Linear Algebra – Week 5 Inner Products and Orthogonality This week we will learn about: • Inner products (and the dot product again), • The norm induced by the inner product, • The Cauchy–Schwarz and triangle inequalities, and • Orthogonality. Extra reading and watching: • Sections 1.3.4 and 1.4.1 in the textbook • Lecture videos 17, 18, 19, 20, 21, and 22 on YouTube • Inner product space at Wikipedia • Cauchy–Schwarz inequality at Wikipedia • Gram–Schmidt process at Wikipedia Extra textbook problems: ? 1.3.3, 1.3.4, 1.4.1 ?? 1.3.9, 1.3.10, 1.3.12, 1.3.13, 1.4.2, 1.4.5(a,d) ??? 1.3.11, 1.3.14, 1.3.15, 1.3.25, 1.4.16 A 1.3.18 1 Advanced Linear Algebra – Week 5 2 There are many times when we would like to be able to talk about the angle between vectors in a vector space V, and in particular orthogonality of vectors, just like we did in Rn in the previous course. This requires us to have a generalization of the dot product to arbitrary vector spaces. Definition 5.1 — Inner Product Suppose that F = R or F = C, and V is a vector space over F. Then an inner product on V is a function h·, ·i : V × V → F such that the following three properties hold for all c ∈ F and all v, w, x ∈ V: a) hv, wi = hw, vi (conjugate symmetry) b) hv, w + cxi = hv, wi + chv, xi (linearity in 2nd entry) c) hv, vi ≥ 0, with equality if and only if v = 0.
    [Show full text]
  • 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)...............................
    [Show full text]