Chapter 9. Vector Space

CHAPTER 9. VECTOR SPACE

Professor : Jorge M. SeminarioDate – Nov. 14th, 2011Group 3 - Pratik Darvekar Brian Zachary Harding

Min-Chi Hsieh

9.1 INTRODUCTION A vector is a quantity which has both a magnitude and

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

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

9.2 VECTORS; GEOMETRICAL REPRESENTATION9.2.1 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

F

G

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.

9.2 VECTORS; GEOMETRICAL REPRESENTATION9.2.2 Norm of a vector The norm, or magnitude, of a vector (a, b, c) is the

number defined by

ex : = (-1, 4, 1) =

X

Y

Z

(-1, 4 ,1)√18P

O


9.3 INTRODUCTION OF ANGLE AND DOT PRODUCT9.3.1 Dot Product

Let F = a1i + b1j + c1k , G = a2i + b2j + c2k

F· G = a1a2 + b1b2 + c1c2 = cosθ

F

G

θ

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.

9.3 INTRODUCTION OF ANGLE AND DOT PRODUCT9.3.2 Properties of Dot product

F· G = G· F(F + G)· H = F· H + G· Hα(F· G ) = (αF)· G = F· (αG)F· F = F· F = 0 if and only if F = 0


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 = 0ex: 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!


9.3 INTRODUCTION OF ANGLE AND DOT PRODUCT

Application for finding plant equationex: 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

X

(-6, 1, 1)

Z

N

Y

(x, y, z)


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


9.5 DOT PRODUCE, NORM AND ANGLE FOR N-SPACE9.5.1 Definition - NormThe 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


9.5 DOT PRODUCE, NORM AND ANGLE FOR N-SPACE9.5.3 Properties of Dot product - same as before

F· G = G· F(F + G)· H = F· H + G· Hα(F· G ) = (αF)· G = F· (αG)F· F = F· F = 0 if and only if F = 0


9.5 DOT PRODUCE, NORM AND ANGLE FOR N-SPACE9.5.4 Cauchy-Schwarz InequalityLet 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.

9.7 SPAN AND SUBSPACE9.7.1 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

𝑢= 𝛼1𝑢1 + ⋯ + 𝛼𝑘𝑢𝑘

[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.

9.7 SPAN AND SUBSPACE9.7.2 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.

9.7 SPAN AND SUBSPACE9.7.3 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 u1 because u2 is a scalar multiple of u1.


9.7 SPAN AND SUBSPACE9.7.4 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.


9.7 SPAN AND SUBSPACEex : 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

9.7 SPAN AND SUBSPACEex : 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

9.8: LINEAR DEPENDENCE9.8.1 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.

9.8: LINEAR DEPENDENCEex : Consider the 2-tuples, u1=(2, 1), u2=(-2, 1)

Now, 2α1-2α2=0 α1+α2=0Gauss elimination gives:

α1-α2=0 α1=0This gives only the trivial solution α1=α2=0This fits the governing definition so both of these are LI

9.8: LINEAR DEPENDENCE9.8.2 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 LIA 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.

9.8: LINEAR DEPENDENCEex: u1=(2, 1, 2), u2=(1, -1, 0) and u3=(-2, -1, 1)

2α1+α2-2α3=0 α1-α2-α3=0 2α1+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

9.9 BASES, EXPANSIONS, DIMENSION9.9.1 Definition : BasisThe 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=


9.9 BASES, EXPANSIONS, DIMENSION9.9.2 Test for basisThe 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.


9.9 BASES, EXPANSIONS, DIMENSIONex : 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

9.9 BASES, EXPANSIONS, DIMENSION9.9.3 Definition: dimensionIf 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.


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().

9.9 BASES, EXPANSIONS, DIMENSION


9.9 BASES, EXPANSIONS, DIMENSION9.9.4 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 .


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

9.9 BASES, EXPANSIONS, DIMENSION

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 =


9.10 BEST APPROXIMATIONex : 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

Chapter 9. Vector Space

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