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Geometric Algebra Crash Course, Part 2

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Geometric Algebra Crash Course, Part 2 Hochschule für Wintersemester 2014/2015 Wirtschaft und Recht Berlin Berlin School of Economics and Law Dr. Horn Mathematics for Business and Economics – LV-Nr. 200691.01 – Modern Linear Algebra (A Geometric Algebra crash course, Part II: solving systems of linear equations) Stand: 31. Dez. 2014 Teaching & learning contents according to the modular description of LV 200 691.01 Linear functions, multidimensional linear models, matrix algebra Systems of linear equations including methods for solving a system of linear equations and examples in business processes Most of this will be discussed in the standard language of the rather old- fashioned linear algebra or matrix algebra found in most textbooks of business mathematics or mathematical economics. But as it might be helpful to get an impression of some more interesting new approaches, we will talk about solving systems of linear equations in this second part (4 x 45 min.) of our short introduction to Geometric Algebra. Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 1 Repetition: Basics of Geometric Algebra 1 + 3 + 3 + 1 = 23 = 8 different base elements exist in three-dimensional space. One base scalar: 1 Three base vectors: x, y, z Three base bivectors: xy, yz, zx (sometimes called pseudovectors) One base trivector: xyz (sometimes called pseudoscalar) Base scalar and base vectors square to one: 2 2 2 2 1 = x = y = z = 1 Base bivectors and base trivector square to minus one: 2 2 2 2 (xy) = (yz) = (zx) = (xyz) = – 1 Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 2 Anti-Commutativity The order of vectors is important. It encodes information about the orientation of the re- sulting area elements. right-handed orientation (anticlockwise orientation) x y y positive orientation in a right- handed coordinate system x x left-handed orientation (clockwise orientation) y y x negative orientation in a right- handed coordinate system Base vectors anticommute. Thus the product of two base vectors follows Pauli algebra: xy = – yx yz = – zy zx = – xz Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 3 Scalars Scalars are geometric entities without direction. They can be expressed as a multiple of the base scalar: k = k 1 Vectors Vectors are oriented line segments. They can be expressed as linear combinations of the base vectors: r = x x + y y + z z Bivectors Bivectors are oriented area elements. They can be expressed as linear combinations of the base bivectors: A = Axy xy + Ayz yz + Azx zx Trivectors Trivectors are oriented volume elements. They can be expressed as a multiple of the base trivector: V = Vxyz xyz Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 4 Geometric Multiplication of Vectors The product of two vectors consists of a scalar term and a bivector term. They are called inner product (dot product) and outer product (exterior product or wedge product). r1 r2 = r1 r2 + r1 r2 The inner product of two vectors is a commuta- tive product as a reversion of the order of two vectors does not change it: 1 r r = r r = (r r + r r ) 1 2 2 1 2 1 1 2 2 The outer product of two vectors is an anti-com- mutative product as a reversion of the order of two vectors changes the sign of the outer pro- duct: r r = – r r = (r r – r r ) 2 1 1 2 2 1 1 2 This is the end of the repetition. More about the basics of Geometric Algebra can be found in the slides of the first part. Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 5 Systems of Two Linear Equations Let’s start with an example of a rather simple system of two linear equations: 2 x + y = 3 2 x + 4 y = 6 Of course this system of two linear equations can be solved algebraically: 2 x + y = 3 y = – 2 x + 3 2 x + 4 y = 6 substitution 2 x + 4 (– 2 x + 3) = 6 – 6 x + 12 = 6 x = 1 substitution y = – 2 x + 3 = 1 Check of the result: 2 x + y = 2 ∙ 1 + 1 = 3 The result is correct. 2 x + 4 y = 2 ∙ 1 + 4 ∙ 1 = 6 Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 6 Graphical Solutions of the System of Linear Equations There are two different strategies to solve this system of linear equations graphically. rows 2 x + y = 3 2 x + 4 y = 6 columns First strategy: Row picture The two rows 2 x + y = 3 and 2 x + 4 y = 6 are shown in a diagram. Second strategy: Column picture 2 1 3 The columns , , and of the 2 4 6 system of linear equations are shown in a diagram. Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 7 Row Picture The two rows 2 x + y = 3 and 2 x + 4 y = 6 are represented by the two straight lines y = – 2 x + 3 1 3 y = – x + 2 2 y slope x-intercept 3 1.5 1 x 1 The point of intersection (x, y) = (1, 1) of the two lines represents the solution x = 1 and y = 1 of the system of linear equations. Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 8 Column Picture The system of linear equations 2 x + y = 3 and 2 x + 4 y = 6 2 1 3 is now written in vector form x + y = 2 4 6 Solution strategy: We are looking for a linear combination of 2 1 vector a = and b = which gives 2 4 3 us vector c = . 6 y Graphical solution: 6 c b 2 a x 2 3 As just one vector a and one vector b is needed to get vector c, the solution equals x = 1 and y = 1. Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 9 Translating the Column Picture into Geometric Algebra (Pauli Algebra) 2 x + y = 3 2 x x + y x = 3 x 2 x + 4 y = 6 2 x y + 4 y y = 6 y 2 a = a = 2 + 2 x y 2 1 b = b = + 4 x y 4 3 6 c = c = 3 + 6 x y y 6 b c 2 a x 2 3 Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 10 Translating the Column Picture into Geometric Algebra (Pauli Algebra) 2 x + y = 3 Two equations 2 x + 4 y = 6 2 1 3 2 4 6 x + y = a x + b y = c Only one equation 2 x x + y x = 3 x 2 x y + 4 y y = 6 y y 6 b c 2 a x 2 3 a x + b y = c Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 11 Conceptional Core of Transforming Algebraic Problems into Geometric Situations 2 x + y = 3 Two equations 2 x + 4 y = 6 2 x x + y x = 3 x This is only 2 x y + 4 y y = 6 y one equation! By adding directional information, we con- dense the two original equations into only one final equation: 2 x x + y x + 2 x y + 4 y y = 3 x + 6 y (2 x + 2 y) x + (x + 4 y) y = 3 x + 6 y a x + b y = c See for example relativity: The four Maxwell equations can be written as one equation in Geometric Algebra. Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 12 (2 x + 2 y) x + (x + 4 y) y = 3 x + 6 y To recover the two original equations, just reflect the final equation at the x-axis (or at the y-axis) and add the results to the final equation: 1 [x (a x + b y) x + (a x + b y)] 2 1 = [(2 – 2 ) x + ( – 4 ) y 2 x y x y + (2 x + 2 y) x + (x + 4 y) y] = 2 x x + x y 1 [ c + c] 2 x x 1 = [3 x – 6 y + 3 x + 6 y] 2 = 3 x First original equation: 2 x x + y x = 3 x Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 13 But it is more interesting to look for a ma- thematical strategy to recover the columns (or vectors) as they will give us a simple solution of the system of linear equations. To find such a strategy, remember the characteristic features of inner and outer product: If the product of two vectors equals the inner product (the bivector terms cancel), the two vectors are parallel. 1 2 If two parallel vectors are multiplied, the outer product will disappear and the pro- duct of the two parallel vectors will equal the inner product. The outer product of a vector with itself equals zero: 2 2 a a = (a – a ) = 0 To get rid of vector a, we only have to find the outer product of a linear equation with a. Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M. HORN) 14 Solving a System of Two Linear Equations in Geometric Algebra The wedge product delivers a simple solu- tion of a system of linear equations: a x + b y = c Getting rid of vector a x: a (a x + b y) = a c (a a) x + (a b) y = a c (a b) y = a c This gives the solution of variable y: 1 y = (a c) a b Getting rid of vector b y: (a x + b y) b = c b (a b) x + (b b) y = c b (a b) x = c b This gives the solution of variable x: 1 x = (c b) a b Modern Linear Algebra: Solving Systems of Linear Equations (OHP Slides M.
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