Loading in 5 sec....

Chapter 9. Vector SpacePowerPoint Presentation

Chapter 9. Vector Space

- 361 Views
- Uploaded on
- Presentation posted in: General

Chapter 9. Vector Space

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 9. Vector Space

Professor : Jorge M. Seminario

Date – Nov. 14th, 2011

Group 3 - Pratik Darvekar

Brian Zachary Harding

Min-Chi Hsieh

- 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

- 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

9.2.1 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.

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

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.

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.

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.

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

- 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.

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.

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.

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.

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.

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.

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.

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.

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

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

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.

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

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.

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

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.

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.

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

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.

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().

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 .

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

- 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.

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