1 / 32

Chapter 9. Vector Space

Chapter 9. Vector Space. Professor : Jorge M. Seminario Date – Nov. 14 th , 2011 Group 3 - Pratik Darvekar Brian Zachary Harding Min-Chi Hsieh. 9.1 Introduction.

kalona
Download Presentation

Chapter 9. Vector Space

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 9. Vector Space Professor : Jorge M. Seminario Date – Nov. 14th, 2011 Group 3 - Pratik Darvekar Brian Zachary Harding Min-Chi Hsieh

  2. 9.1 Introduction • A vector is a quantity which has both amagnitudeand a direction. Vectors arise naturally as physical quantities. • Examples of vectors are displacement, velocity, acceleration, force and electric field. Some physical quantities cannot be added in the simple way described for scalars. [1]http://www.physchem.co.za/index.htm

  3. 9.1 Introduction • ex: Walk 4m in a northerly direction and then 3m in an easterly direction, how far would you be from your starting point? • The answer is clearly NOT 7 m! One could calculate the distance using the theorem of Pythagoras, • You could have reached the same final position by walking 5 m in the direction 36.9° east of north. • These quantities are called displacements. Displacement is an example of a vector quantity. [1]http://www.physchem.co.za/index.htm

  4. 9.2 Vectors; Geometrical Representation 9.2.1 Algebra of Vectors • F + G = G + F • (F + G) + H = F + (G + H) • F + 0 = F • α(F + G) = αF + αG • (αβ)F = α(βF) • (α + β)F = αF + βF G F F + G F G [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.1, pp. 204-208.

  5. 9.2 Vectors; Geometrical Representation 9.2.2 Norm of a vector • The norm, or magnitude, of a vector (a, b, c) is the number defined by ex : = (-1, 4, 1) = Z P (-1, 4 ,1) Y O [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.1, pp. 201-203. X

  6. 9.3 Introduction of Angle and Dot Product 9.3.1 Dot Product • Let F = a1i + b1j + c1k , G = a2i + b2j + c2k F· G = a1a2 + b1b2 + c1c2 = cosθ G θ F Gcosθ [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.2, pp. 209.

  7. 9.3 Introduction of Angle and Dot Product 9.3.2 Properties of Dot product • F· G = G· F • (F + G)· H = F· H + G· H • α(F· G ) = (αF)· G = F· (αG) • F· F = • F· F = 0 if and only if F = 0 [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.2, pp. 210-211.

  8. 9.3 Introduction of Angle and Dot Product 9.3.3 Orthogonal Vectors • Vector F and G are orthogonal if and only if F· G = 0 ex: F = -4i + j + 2k, G = 2i + 4k, H = 6i – j – 2k F· G = (-4)(2) + (1)(0) + (2)(4) = 0 F· H = (-4)(6) + (1)(-1) + (2)(-2) = -29 ≠ 0 G· H = (2)(6) + (0)(-1) + (4)(-2) = 4 ≠ 0 only F & G are orthogonal, others are not! [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.2, pp. 213-215.

  9. 9.3 Introduction of Angle and Dot Product Application for finding plant equation ex: Suppose we want the equation of a plant Π containing the point (-6, 1, 1) and perpendicular to the vector N = -2i + 4j + k First we suggest a point (x, y, z) is on Π, so the vector from (-6, 1, 1) to (x, y, z) must be orthogonal to N. this means that ((x+6)i + (y-1)j + (z-1)k)· N = 0 ((x+6)i + (y-1)j + (z-1)k)· (-2i + 4j + k) = 0 -2(x+6) + 4(y-1) + (z-1) = 0 -2x + 4y + z = 17 Z (-6, 1, 1) (x, y, z) N [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.2, pp. 213-215. Y X

  10. 9.4&9.6 n-Space& Generalized Vector Space • If n is a positive integer, an n-vector is an n-tuple (x1, x2,…, xn), with each coordinate xj a real number. The set of all n-vector is denoted Rn. 9.4.1 Algebra of Vectors - same as before • F + G = G + F • (F + G) + H = F + (G + H) • F + 0 = F • α(F + G) = αF + αG • (αβ)F = α(βF) • (α + β)F = αF + βF [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.4, pp. 222-223.

  11. 9.5 Dot Produce, Norm and Angle for n-Space 9.5.1 Definition - Norm The Norm of F = (x1, x2,…,xn) 9.5.2 Definition - Dot Product We let F = (x1, x2,…,xn), G = (y1, y2,…,yn), we can get F· G = x1y1 + x2y2 + … + xnyn [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.4, pp. 222-223.

  12. 9.5 Dot Produce, Norm and Angle for n-Space 9.5.3 Properties of Dot product - same as before • F· G = G· F • (F + G)· H = F· H + G· H • α(F· G ) = (αF)· G = F· (αG) • F· F = • F· F = 0 if and only if F = 0 [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.4, pp. 222-223.

  13. 9.5 Dot Produce, Norm and Angle for n-Space 9.5.4 Cauchy-Schwarz Inequality Let F and G be vectors, then Proof: from dot product we know cosθ = since so then equivalent to the Cauchy-Schwarz inequality! [1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch.5.4, pp. 224.

  14. 9.7 Span and Subspace 9.7.1 Definition If you have a vector or a group of vectors, the set of all linear combinations of these vectors is called the span. Symbolically this is written as: where α1,… αk are scalars [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.

  15. 9.7 Span and Subspace 9.7.2 Single Vector • When this definition is applied to a single vector, the span is a line extending infinitely from the vector • ex: The span of the vector u1=(4,7) is the set of all scalar multiples of (4,7). 2u1=(8,14) 0u1=(0,0) -5u1=(-20,-35) [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.

  16. 9.7 Span and Subspace 9.7.3 Two Vectors • Similar to the span of a single vector, the span of two vectors is all linear combinations of both vectors. ex: u1=(4,7), u2=(8,14). The span of these two vectors is the same as the span of just u1because u2 is a scalar multiple of u1. [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.

  17. 9.7 Span and Subspace 9.7.4 Subspace • Up to this point, all of the vectors have been in R2. That is to say that the spans of all of the vectors have been subspaces of R2. • The governing definition of subspace is: If a subset T of a vector space S is itself a vector space, then T is a subspace of S. • Using this definition a subspace is either all or part of a vector space, meaning R2 is a subspace of R2. [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.

  18. 9.7 Span and Subspace ex : If the span of u1=(2,3) u2=(2,4) all or part of R2. To determine the span of (u1, u2) let v=(v1, v2) be any vector in R2 and try to express v=(α1u1+α2u2) (v1, v2)=α1(2,3)+α2(2,4) v1=2α1+2α2 v2=3α1+4α2 Now, use Gauss elimination: α1= 2v1-v2 This system is consistent, meaning that you can solve α1 for every v

  19. 9.7 Span and Subspace ex : In R3 do a similar exercise u1 = (2, 1, 2), u2 = (-2, 1, 1) and v = (α1u1+α2u2) Gauss elimination: 2α1-2α2=v1 α1+α2=v2 2α1+α2=v3 0=v3-(1/4)v1-(3/2) v2 This means that the span is constrained by this equation. Any two can be chosen arbitrarily and the last is set. The solution is a plane and is not all of R3, rather it is a subspace

  20. 9.8: Linear Dependence 9.8.1 Governing definition • A finite set of vectors is LI if and only if there exist scalar multipliers such that 0=α1u1+α2u2…αjuj is only true if all scalar multipliers are zero Independent Dependent [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch. 9.7 pp. 444-446. Print.

  21. 9.8: Linear Dependence ex : Consider the 2-tuples, u1=(2, 1), u2=(-2, 1) Now, 2α1-2α2=0α1+α2=0 Gauss elimination gives: α1-α2=0 α1=0 This gives only the trivial solution α1=α2=0 This fits the governing definition so both of these are LI

  22. 9.8: Linear Dependence 9.8.2 Some important theorems concerning linear independence • If a set of 2 vector is LD, then one vector must be a scalar multiple of the other • Every finite set of orthogonal vectors is LI • A set containing the zero vector is LD [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch. 9.7 pp. 446. Print.

  23. 9.8: Linear Dependence ex: u1=(2, 1, 2), u2=(1, -1, 0) and u3=(-2, -1, 1) 2α1+α2-2α3=0 α1-α2-α3=02α1+α3=0 Gauss elimination gives: 2α1+α2-2α3=0 -3α2=0 3α3=0 All α are equal to 0, so the set is LI Independent Dependent

  24. 9.9 Bases, Expansions, Dimension 9.9.1 Definition : Basis The finite set of vectors {e1,e2,…,ek} in a vector space S is a basis for S if each vector u in S can be can be expressed uniquely in the form- u=1e1+…+ek= [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch.9.9 pp. 470-484. Print.

  25. 9.9 Bases, Expansions, Dimension 9.9.2 Test for basis The finite set {e1,e2,…,ek} in a vector space S is a basis for S if and only if it spans S and is Linearly Independent. [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch.9.9 pp. 470-484. Print.

  26. 9.9 Bases, Expansions, Dimension ex : Show whether the vectors (3, 2) and (-1, -5) is a basis for R2 (3, 2) and (-1, -5) is a basis for R2 if α1 (3, 2)+(-1, -5) = (u1,u2) admits a unique solution for α1, α2 for any values of u1,u2. 3 α1 – α2= u1 2 α1 - 5α2= u2, solving we get,α1 = α2 = Thus we get unique α1 , α2 for every u1,u2. So it’s a basis for R2

  27. 9.9 Bases, Expansions, Dimension 9.9.3 Definition: dimension If the greatest number of LI vectors that can be found in a vector space S is k, where 1≤k≤∞, then S is k-dimensional. dim S = k • If a vector space S admits a basis consisting of k vectors, then S is k-dimensional. [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch.9.9 pp. 470-484. Print.

  28. 9.9 Bases, Expansions, Dimension 9.9.4 Orthogonal bases • For non-orthogonal bases, the expansion process of a given vector can be quite laborious. For example, if we seek to expand a given vector u in R8(8 dimensional space), then there will be 8 base vectors() and 8 expansion coefficients(), and these will be found by solving a system of 8 equations in 8 unknowns(). [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch.9.9 pp. 470-484. Print.

  29. 9.9 Bases, Expansions, Dimension 9.9.4 Orthogonal bases • Hence orthogonal bases are preferred, in which- . = 0 if i≠j. if {e1,e2,…,ek} are orthogonal bases, the expansion of any vector u is simply- u = 1+…+ k . [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch.9.9 pp. 470-484. Print.

  30. 9.9 Bases, Expansions, Dimension ex : Expand u = (4, 3, -3, 6) in terms of the orthogonal base vectors e1=(1, 0, 2, 0), e2=(0, 1, 0, 0), e3=(-2, 0, 1, -1) and e4=(-2, 0, 1, -1) of R4. Compute = -2, = 5, and so on, u = 1+…+ 4 , u = e1 + 3 e2 + e3 + e4

  31. 9.10 Best approximation • If the vectors {e1,e2,…,en } are orthonormal, but fall short of being a basis for S (i.e., N < dim S), and we still wish to expand a vector u in S which does not fall within the span {e1,e2,…,en }, in such a case the best approximation of u in terms of {e1,e2,…,en } is given by- u = [1]Greenberg, Michael D. "Vector Space: Span and Subspace." Advanced Engineering Mathematics 2nd Edition. Upper Saddle River, NJ: Prentice-Hall, 1998. ch.9.10 pp. 470-484. Print.

  32. 9.10 Best approximation ex : Let S be R2, N=1, e1= (12,5), and u= (1,1). Find the best approximation uc1e1. The best approximation of u in terms of {e1,e2,…,en } is given by u = 1= + = Hence u = e1

More Related