Lecture 9: Vector Algebra

Similar documents
Math 3191 Applied Linear Algebra

Math 2331 Linear Algebra

We showed that adding a vector to a basis produces a linearly dependent set of vectors; more is true.

Study Guide for Linear Algebra Exam 2

Math 2331 Linear Algebra

6.4 BASIS AND DIMENSION (Review) DEF 1 Vectors v 1, v 2,, v k in a vector space V are said to form a basis for V if. (a) v 1,, v k span V and

1 Last time: inverses

Vector Spaces 4.3 LINEARLY INDEPENDENT SETS; BASES Pearson Education, Inc.

MAT 242 CHAPTER 4: SUBSPACES OF R n

MATH 304 Linear Algebra Lecture 20: The Gram-Schmidt process (continued). Eigenvalues and eigenvectors.

Review Notes for Linear Algebra True or False Last Updated: February 22, 2010

Linear Combination. v = a 1 v 1 + a 2 v a k v k

MATH 2331 Linear Algebra. Section 2.1 Matrix Operations. Definition: A : m n, B : n p. Example: Compute AB, if possible.

Math 3191 Applied Linear Algebra

ICS 6N Computational Linear Algebra Vector Equations

Math 4377/6308 Advanced Linear Algebra

Vector Spaces. (1) Every vector space V has a zero vector 0 V

Chapter 3. Vector spaces

Math 4377/6308 Advanced Linear Algebra

Worksheet for Lecture 15 (due October 23) Section 4.3 Linearly Independent Sets; Bases

2018 Fall 2210Q Section 013 Midterm Exam II Solution

OHSX XM511 Linear Algebra: Multiple Choice Exercises for Chapter 2

MATH 2331 Linear Algebra. Section 1.1 Systems of Linear Equations. Finding the solution to a set of two equations in two variables: Example 1: Solve:

Chapter 3. More about Vector Spaces Linear Independence, Basis and Dimension. Contents. 1 Linear Combinations, Span

MATH 323 Linear Algebra Lecture 12: Basis of a vector space (continued). Rank and nullity of a matrix.

Worksheet for Lecture 15 (due October 23) Section 4.3 Linearly Independent Sets; Bases

Lecture 03. Math 22 Summer 2017 Section 2 June 26, 2017

Math Linear algebra, Spring Semester Dan Abramovich

Vector space and subspace

Chapter 1. Vectors, Matrices, and Linear Spaces

Objective: Introduction of vector spaces, subspaces, and bases. Linear Algebra: Section

2. Every linear system with the same number of equations as unknowns has a unique solution.

(i) [7 points] Compute the determinant of the following matrix using cofactor expansion.

Math 3C Lecture 25. John Douglas Moore

Abstract Vector Spaces

Lecture 6: Spanning Set & Linear Independency

Linear Algebra Review: Linear Independence. IE418 Integer Programming. Linear Algebra Review: Subspaces. Linear Algebra Review: Affine Independence

Lecture 3: Linear Algebra Review, Part II

6. Orthogonality and Least-Squares

Chapter 3. Directions: For questions 1-11 mark each statement True or False. Justify each answer.

Linear independence, span, basis, dimension - and their connection with linear systems

Span and Linear Independence

Assignment 1 Math 5341 Linear Algebra Review. Give complete answers to each of the following questions. Show all of your work.

Span & Linear Independence (Pop Quiz)

Sept. 26, 2013 Math 3312 sec 003 Fall 2013

Solutions of Linear system, vector and matrix equation

Dr. Abdulla Eid. Section 4.2 Subspaces. Dr. Abdulla Eid. MATHS 211: Linear Algebra. College of Science

DS-GA 1002 Lecture notes 10 November 23, Linear models

Lecture 22: Section 4.7

Solutions to Section 2.9 Homework Problems Problems 1 5, 7, 9, 10 15, (odd), and 38. S. F. Ellermeyer June 21, 2002

LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK)

Chapter 2. General Vector Spaces. 2.1 Real Vector Spaces

Introduction to Mathematical Programming IE406. Lecture 3. Dr. Ted Ralphs

Worksheet for Lecture 25 Section 6.4 Gram-Schmidt Process

1 Last time: determinants

Vector Spaces 4.4 Spanning and Independence

Math 3191 Applied Linear Algebra

1111: Linear Algebra I

Math 314H EXAM I. 1. (28 points) The row reduced echelon form of the augmented matrix for the system. is the matrix

Row Space, Column Space, and Nullspace

Lecture: Linear algebra. 4. Solutions of linear equation systems The fundamental theorem of linear algebra

Lecture 13: Orthogonal projections and least squares (Section ) Thang Huynh, UC San Diego 2/9/2018

Math 54 HW 4 solutions

Linear Algebra MATH20F Midterm 1

Criteria for Determining If A Subset is a Subspace

Overview. Motivation for the inner product. Question. Definition

March 27 Math 3260 sec. 56 Spring 2018

(II.B) Basis and dimension

MATH 1553 PRACTICE FINAL EXAMINATION

Chapter 1: Systems of Linear Equations

The definition of a vector space (V, +, )

NAME MATH 304 Examination 2 Page 1

Solutions: We leave the conversione between relation form and span form for the reader to verify. x 1 + 2x 2 + 3x 3 = 0

Math 102, Winter 2009, Homework 7

AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda 4. BASES AND DIMENSION

Exam 2 Solutions. (a) Is W closed under addition? Why or why not? W is not closed under addition. For example,

Math 54. Selected Solutions for Week 5

The set of all solutions to the homogeneous equation Ax = 0 is a subspace of R n if A is m n.

1. Determine by inspection which of the following sets of vectors is linearly independent. 3 3.

MATH 20F: LINEAR ALGEBRA LECTURE B00 (T. KEMP)

YORK UNIVERSITY. Faculty of Science Department of Mathematics and Statistics MATH M Test #1. July 11, 2013 Solutions

LINEAR ALGEBRA SUMMARY SHEET.

Math 550 Notes. Chapter 2. Jesse Crawford. Department of Mathematics Tarleton State University. Fall 2010

Chapter 2 Subspaces of R n and Their Dimensions

MTH 35, SPRING 2017 NIKOS APOSTOLAKIS

Abstract Vector Spaces and Concrete Examples

MATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian.

Linear equations in linear algebra

EXAM. Exam #2. Math 2360 Summer II, 2000 Morning Class. Nov. 15, 2000 ANSWERS

Online Exercises for Linear Algebra XM511

5.) For each of the given sets of vectors, determine whether or not the set spans R 3. Give reasons for your answers.

Homework 5. (due Wednesday 8 th Nov midnight)

Orthogonal Projection. Hung-yi Lee

VECTOR SPACES & SUBSPACES

7. Dimension and Structure.

Kevin James. MTHSC 3110 Section 4.3 Linear Independence in Vector Sp

Math 18, Linear Algebra, Lecture C00, Spring 2017 Review and Practice Problems for Final Exam

Math 3191 Applied Linear Algebra

Math 4A Notes. Written by Victoria Kala Last updated June 11, 2017

Vector Spaces and Linear Maps

Transcription:

Lecture 9: Vector Algebra Linear combination of vectors Geometric interpretation Interpreting as Matrix-Vector Multiplication Span of a set of vectors Vector Spaces and Subspaces Linearly Independent/Dependent set of vectors and bases Spanning set Dimension 1

Linear Combination of Vectors Let v = v 1 v 2 v 3. v p R p and c 1, c 2,, c p be scalars then vector y defined by: is called a linear combination of v 1, v 2,. v p with weights c 1, c 2, c p. The weights in a linear combination can be any real numbers including zero. And y is said to be linearly dependent on v 1, v 2,. v p.

Linear Dependence Definition: v 1, v 2, vn are linearly dependent if vi can be written down as a linear combination of rest of the prceeding vectors for any i. v i = c 1 v 1 + c 2 v 2 + c 3 v 3 +.. c i 1 v i 1 for any i = 1,2.. r A zero vector is linearly dependent on any set of N-dimensional vectors where x 1, x 2, are all 0.

Linear Dependence If there is a solution for the vector equation In other words, there exists values for a 1, a 2, a n such that the above equation is satisfied, then b is linearly dependent on x 1, x 2, x n.

Linear Independence If there is no value for a 1, a 2, exists such that the below equation is satisfied then b is linearly independent of the set of vectors v 1, v 2, v n. In other words, x 1 a 1 + x 2 a 2 + x n a n b for any value of x 1, x 2, or x 1 a 1 + x 2 a 2 + x n a n b 0 means b is linearly independent of the set of vectors a 1, a 2, a n In general, the set of vectors {a 1, a 2, a n, b} is linearly independent if x 1 a 1 + x 2 a 2 + x n a n + x n+1 b = 0, only if x 1, x 2, are all 0.

Linear Independence

Linearly Independent/Dependent Set A set of vectors that are linearly independent is called a Linearly Independent Set. If at least one vector in a set of vectors is linearly dependent on other vectors, then that set is called a Linearly Dependent Set. A set of vectors with a zero vector has to be a Linearly Dependent Set..

Examples of Linear Dependence A vector and its scalar multiple are linear dependent. (By definition.) x 1 a = b A vector that is a linear combination of a set of vectors is linearly dependent on those vectors. (By definition.) 8

Examples of Linear Dependence Let x1 = 1 2 3, x2 = 2 4 5 and b = [4 8 11] Then the rank of the matrix M is M = 1 2 3 2 4 5 4 8 11 - the number of linearly independent vectors

Span of a set of vectors A vector that is in the span of a set of vectors is linearly dependent on those vectors. (By definition.)

Span of a set of vectors Example 1:

Span of a set of vectors Example 2: 0 P 2 v 2 v 1 P 1

Same R 2 plane can be spanned by other vectors too Let a= 2 1 and b= 1 2. Any point in R 2 can be represented as a linear combination of the a and b axes. P arbitrary = c 1 a + c 2 b Example: a-coordinate P 1 = 2 0 = 4 2 a + b 3 3 P 2 = 1 2 = 4 a + 5 b 3 3 a and b vectors SPAN the R 2 plane. b-coordinate 5 3 b P 2 4 3 a b a P 1 4 3 a 2 3 b 13

A GEOMETRIC DESCRIPTION OF SPAN V Span of a set of vectors (geometric description) Let v be a nonzero vector in R 3. Then Span v is the set of all scalar multiples of v, which is the set of points on the line in R 3 through v and 0. See the figure below

Span A GEOMETRIC of a set of DESCRIPTION vectors (geometric OF SPAN U, Vdescription) If u and v are nonzero vectors in R 3, with v not a multiple of u, then Span u, v is the plane in R 3 that contains u, v, and 0. In particular, Span u, v contains the line in R 3 through u and 0 and the line through v and 0. See the figure below.

How many minimum number of vectors are necessary to span a space? ONE non-zero vector is required and sufficient to span a 1D space (in other words, a line). The dimension of the vectors has to be 1 or more. TWO linearly independent vectors are required and sufficient to span a 2D plane such as the XY plane. The dimension of the vectors has to be 2 or more. THREE linearly independent vectors are required and sufficient to span a 3D space (such as XYZ volume). The dimension of the vectors has to be 3 or more. N linearly independent vectors are required and sufficient to span an N-D space. The dimension of the vectors has to be N or more. The span of these vectors is called the vector space (or subspace, if it is a subset of a vector space). (Later we will see that a vector space is more generic than a span. But in this course, we will only see vector spaces that are spans of a sets of vectors.) 16

Span of a set of vectors Maximal subset consisting of linearly independent vectors is called the basis of the span of x 1, x 2, x 3, x n And the number of elements in the basis is called the dimension of the span.

Spanning Set A basis is an efficient spanning set that contains no unnecessary vectors. A basis can be constructed from a spanning set by discarding unneeded vectors Basis construction

Spanning Set v 4 =2v 1 v 2. So, by Spanning Set Theorem, {v 1, v 2, v 3 } span the same subspace as {v 1, v 2, v 3, v 4 }. We also see that {v 1, v 2, v 3 } are linearly independent. So that is the basis of W.

Vector Spaces Span of vectors also form a vector space. Now we will look more into the definition of vector spaces and their properties.

Vector Spaces 7) c u + v = cu + cv for any scalar c

Vector Spaces

Vector Subspace If a subset of a vector space also forms a vector space, then that subset is called a vector subspace. Hence, every vector subspace is also a vector space. And also every vector space is also a vector subspace (of itself and possibly of larger spaces).

Vector Subspace

Vector Subspace Example

Vector Subspace Example

Vector Basis

Vector Basis Example: No, because the Span{v 1, v 2 } H. Span{v 1, v 2 } is the entire XY plane, not just the vectors of the form [s,s,0].

Spanning Set The following three sets in R3 show how linearly independent set can be enlarged to a basis and how further enlargement destroys the linear independence of the set.

Dimension of a Vector Space We already discussed Dimension of a vector: Number of components of the vector Dimension of a vector space (or subspace): Minimum number of linearly independent vectors required to span that space. New definition Dimension of a vector space (or subspace): Number of vectors in its basis. What is Basis of vector space again? A Linearly Independent Set (of vectors) whose span is the vector space. This implies: Basis has the minimum number of linearly independent vectors that spans the space. Note: Basis is not unique. But all bases that span the same space has the same number of vectors. 30

Matrix equation Solving a system of m equation and n unknown in Algebra can be converted in matrixvector multiplication form. a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 A is a 3 3 matrix, x and b are 3 1 column vectors. A and b are known, where x is unknown. This is a matrix equation. Ax = b, a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 x 1 x 2 x 3 = b 1 b 2 b 3

Matrix equation as a vector equation Matrix equation can be interpreted as a vector equation. The left side of equation (Ax) can be written as: a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 x 1 x 2 x 3 = a 11 x 1 + a 12 x 2 + a 13 x 3 a 21 x 1 + a 22 x 2 + a 23 x 3 a 31 x 1 + a 32 x 2 + a 33 x 3 col 1 col 2 col 3 x 1 col 1 x 2 col 2 x 3 col 3 = x 1 a 11 a 21 a 31 + x 2 a 12 a 22 a 32 + x 3 a 13 a 23 a 33 Represented as a linear combination of column vectors

Matrix equation as a vector equation Now we have a vector equation, stating the same problem as Ax = b x 1 a 11 a 21 a 31 + x 2 a 12 a 22 a 32 + x 3 a 13 a 23 a 33 = b 1 b 2 b 3 So, we can represent any Ax = b equation as a vector equation. If b is in the span of column vectors, the system has either ONE or INFINTE solution. If b is NOT in the span of column vectors, the system has NO solution. But, we can find projection of b on the vector space, as best estimate.

Matrix equation (ONE solution) Example: y 2x + y = 5 4x 2y = 2 Row representation of above equation: 1 3 2 1 4 2 x y = 5 2 x

Vector equation (ONE solution) Span of col 1 Vector equation representation of previous example: dim 2 col 1 x 2 4 + y 1 2 = 5 2 col 2 b dim 1 b is in the span of col 1 and col 2. The only solution for this equation is x = 1, and y = 3 Span of col 2

Matrix equation (INFINITE solution) Example: 2x + y = 3 4x + 2y = 6 y Row representation of above equation: 2 1 x 4 2 y = 3 6 Both equations are the same line All points on the line are solutions x

Vector equation (INFINITE solution) Vector equation representation of previous example: x 2 4 + y 1 2 = 3 6 dim 2 b Span of col 1 and Span of col 2 col 1 and col 2 are linearly dependent b is in the span of col 1 and col 2 2x + y 1 2 = 3 6 Any combination of x and y that satisfies 2x + y = 3 is a solution. col 2 col 1 dim 1

Matrix equation (NO solution) Example: 2x + y = 3 4x + 2y = 4 y Row representation of above equation: 2 1 x 4 2 y = 3 4 The lines are parallel. There is no solution. x

Vector equation (NO solution) Vector equation representation of previous example: dim 2 Span of col 1 and Span of col 2 x 2 4 + y 1 2 = 3 4 col 1 and col 2 are linearly dependent col 2 col 1 b b is NOT in the span of col 1 and col 2 There is no solution dim 1

Vector equation (NO solution) dim 2 Projection of b Hence, in order to solve this we project b onto the span of col 1 and col 2 using Least Squares col 2 col 1 b dim 1