1 MATH 304 Linear Algebra Lecture 24: Scalar product.
2 Vectors: geometric approach B A B A A vector is represented by a directed segment. Directed segment is drawn as an arrow. Different arrows represent the same vector if they are of the same length and direction.
3 Vectors: geometric approach v B A v B A AB denotes the vector represented by the arrow with tip at B and tail at A. AA is called the zero vector and denoted 0.
4 Vectors: geometric approach v B A v B A If v = AB then BA is called the negative vector of v and denoted v.
5 Vector addition Given vectors a and b, their sum a + b is defined by the rule AB + BC = AC. That is, choose points A, B, C so that AB = a and BC = b. Then a + b = AC. A a B b a + b A a C B b C a + b
6 The difference of the two vectors is defined as a b = a + ( b). a b a b
7 Properties of vector addition: a + b = b + a (commutative law) (a + b) + c = a + (b + c) (associative law) a + 0 = 0 + a = a a + ( a) = ( a) + a = 0 Let AB = a. Then a + 0 = AB + BB = AB = a, a + ( a) = AB + BA = AA = 0. Let AB = a, BC = b, and CD = c. Then (a +b) +c = ( AB + BC) + CD = AC + CD = AD, a + (b +c) = AB + ( BC + CD) = AB + BD = AD.
8 Parallelogram rule Let AB = a, BC = b, AB = b, and B C = a. Then a + b = AC, b + a = AC. B b C C a a A b B Wrong picture!
9 Parallelogram rule Let AB = a, BC = b, AB = b, and B C = a. Then a + b = AC, b + a = AC. B b C = C a a + b a A b B Right picture!
10 Scalar multiplication Let v be a vector and r R. By definition, rv is a vector whose magnitude is r times the magnitude of v. The direction of rv coincides with that of v if r > 0. If r < 0 then the directions of rv and v are opposite. v 2v 3v
11 Scalar multiplication Let v be a vector and r R. By definition, rv is a vector whose magnitude is r times the magnitude of v. The direction of rv coincides with that of v if r > 0. If r < 0 then the directions of rv and v are opposite. Properties of scalar multiplication: r(a + b) = ra + rb (distributive law #1) (r + s)a = r a + sa (distributive law #2) r(sa) = (rs)a 1a = a (associative law)
12 Beyond linearity: length of a vector The length (or the magnitude) of a vector AB is the length of the representing segment AB. The length of a vector v is denoted v or v. Properties of vector length: x 0, x = 0 only if x = 0 rx = r x (positivity) (homogeneity) x + y x + y (triangle inequality) x y x + y
13 Beyond linearity: angle between vectors Given nonzero vectors x and y, let A, B, and C be points such that AB = x and AC = y. Then BAC is called the angle between x and y. The vectors x and y are called orthogonal (denoted x y) if the angle between them equals 90 o. y C A θ x B
14 x x + y y Pythagorean Theorem: x y = x + y 2 = x 2 + y 2 3-dimensional Pythagorean Theorem: If vectors x,y,z are pairwise orthogonal then x + y + z 2 = x 2 + y 2 + z 2
15 C y x y A θ x B Law of cosines: x y 2 = x 2 + y 2 2 x y cos θ
16 y x y x x + y x y Parallelogram Identity: x + y 2 + x y 2 = 2 x y 2
17 Beyond linearity: dot product The dot product of vectors x and y is x y = x y cos θ, where θ is the angle between x and y. The dot product is also called the scalar product. Alternative notation: (x,y) or x,y. The vectors x and y are orthogonal if and only if x y = 0. Relations between lengths and dot products: x = x x x y x y x y 2 = x 2 + y 2 2x y
18 Vectors: algebraic approach An n-dimensional coordinate vector is an element of R n, i.e., an ordered n-tuple (x 1, x 2,...,x n ) of real numbers. Let a = (a 1, a 2,...,a n ) and b = (b 1, b 2,...,b n ) be vectors, and r R be a scalar. Then, by definition, a + b = (a 1 + b 1, a 2 + b 2,...,a n + b n ), ra = (ra 1, ra 2,...,ra n ), 0 = (0, 0,...,0), b = ( b 1, b 2,..., b n ), a b = a + ( b) = (a 1 b 1, a 2 b 2,...,a n b n ).
19 Properties of vector addition and scalar multiplication: a + b = b + a (a + b) + c = a + (b + c) a + 0 = 0 + a = a a + ( a) = ( a) + a = 0 r(a + b) = ra + rb (r + s)a = ra + sa r(sa) = (rs)a 1a = a
20 Cartesian coordinates: geometric meets algebraic ( 3, 2) ( 3, 2) (2,1) (2,1) Once we specify an origin O, each point A is associated a position vector OA. Conversely, every vector has a unique representative with tail at O. Cartesian coordinates allow us to identify a line, a plane, and space with R, R 2, and R 3, respectively.
21 Length and distance Definition. The length of a vector v = (v 1, v 2,...,v n ) R n is v = v v v2 n. The distance between vectors/points x and y is y x. Properties of length: x 0, x = 0 only if x = 0 rx = r x x + y x + y (positivity) (homogeneity) (triangle inequality)
22 Scalar product Definition. The scalar product of vectors x = (x 1, x 2,...,x n ) and y = (y 1, y 2,...,y n ) is n x y = x 1 y 1 + x 2 y x n y n = x k y k. k=1 Properties of scalar product: x x 0, x x = 0 only if x = 0 (positivity) x y = y x (symmetry) (x + y) z = x z + y z (distributive law) (rx) y = r(x y) (homogeneity)
23 Relations between lengths and scalar products: x = x x x y x y (Cauchy-Schwarz inequality) x y 2 = x 2 + y 2 2x y By the Cauchy-Schwarz inequality, for any nonzero vectors x,y R n we have cos θ = x y for some 0 θ π. x y θ is called the angle between the vectors x and y. The vectors x and y are said to be orthogonal (denoted x y) if x y = 0 (i.e., if θ = 90 o ).
24 Problem. Find the angle θ between vectors x = (2, 1) and y = (3, 1). x y = 5, x = 5, y = 10. cos θ = x y x y = 5 = 1 = θ = 45 o Problem. Find the angle φ between vectors v = ( 2, 1, 3) and w = (4, 5, 1). v w = 0 = v w = φ = 90 o
25 Orthogonal projection Let x,y R n, with y 0. Then there exists a unique decomposition x = p +o such that p is parallel to y and o is orthogonal to y. o x y p p = orthogonal projection of x onto y
26 Orthogonal projection Let x,y R n, with y 0. Then there exists a unique decomposition x = p +o such that p is parallel to y and o is orthogonal to y. Namely, p = αu, where u is the unit vector of the same direction as y, and α = x u. Indeed, p u = (αu) u = α(u u) = α u 2 = α = x u. Hence o u = (x p) u = x u p u = 0 = o u = o y. p is called the vector projection of x onto y, α = ± p is called the scalar projection of x onto y. u = y y, α = x y y, p = x y y y y.
27 Problem. Find the distance from the point x = (3, 1) to the line spanned by y = (2, 1). Consider the decomposition x = p + o, where p is parallel to y while o y. The required distance is the length of the orthogonal component o. p = x y y y y = 5 (2, 1) = (2, 1), 5 o = x p = (3, 1) (2, 1) = (1, 2), o = 5. Problem. Find the point on the line y = x that is closest to the point (3, 4). The required point is the projection p of v = (3, 4) on the vector w = (1, 1) spanning the line y = x. p = v w w w w = 1 ( (1, 1) = 1 2 2, 1 ) 2
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