MTH 464: Computational Linear Algebra

Transcription

1 MTH 464: Computational Linear Algebra Lecture Outlines Exam 1 Material Dr. M. Beauregard Department of Mathematics & Statistics Stephen F. Austin State University January 9, 2018 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Contents I 1 Lecture 01 2 Lecture 02 3 Lecture 03 4 Lecture 04 5 Lecture 05 6 Lecture 06 7 Lecture 07 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

2 Lecture 01 Overview Goal for today: Understand the basics of linear systems [section 1.1] Outline: 1 Linear Systems 2 Matrix notation and augmented form 3 Solutions 4 Solving Linear Systems Assignment (1.1): Read: sections 1.1 and 1.2 Work: section 1.1 (p. 10) #3, 5, 9, 13, 15, 17, 22, 23, 25 Extra practice: remaining odds (1 31) Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 01 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

3 Lecture 01 Linear Systems What is a linear system? Examples Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 01 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

4 Lecture 01 Matrix and Vector Notation Definition An m n matrix is a rectangular layout of numbers having m rows and n columns. For a matrix A the notation a ij will refer to the entry or number in the row i and column j of A. Definition A column vector of size m is a m 1 matrix. We denote a vector using a bold lower case letter or as an arrow above a lower case letter. For v or v the notation v i will refer to the entry or number in the i position. Remarks: We call an m n matrix of all zeros the zero matrix and is written as 0. We call a m dimensional vector of all zeros the zero vector and is written as 0. Sage makes it simple to evaluate and manipulate vectors and Linear matrices. Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 01 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

5 Coefficient Matrix Lecture 01 In a linear system we notice that what matter what the variable names are rather it is their coefficients. This motivates the creation of the coefficient matrix. Definition Consider the linear system a 11 x a 1n x n = b 1 a 21 x a 2n x n = b 2. =. a m1 x a mn x n = b m. The coefficient matrix for the linear system is defined as the matrix A with entries a ij. The solution vector is x with entries x i. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 01 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

6 Lecture 01 Augmented Matrix Definition Suppose we have a system of m equations with n variables with coefficient matrix A and the vector of constants b. Then the augmented matrix of the system of equations is the m (n + 1) matrix whose first n columns are the columns of A and last column is the column vector b. We denote this matrix as [ A b ]. Remarks: The solution set to the augmented problem is identical to the solution set of the linear system. It is far easier to manipulate the augmented problem since we are just focusing on the numbers. It is important to remember the augmented problem is equivalent to the linear system. Column j refers to the coefficients of x j for each equation. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 01 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

7 Lecture 01 Solutions Definition (page 3) A solution of a linear system in variables x 1, x 2,..., x n is a list (s 1, s 2,..., s n ) of corresponding values which together satisfy all of the equations. Claim (page 4) Any linear system has: exactly one solution, infinitely many solutions, or no solution. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 01 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

8 Solving Linear Systems Lecture 01 Overview: 1 Represent the system as an augmented matrix [ A b ] 2 Reduce it to an equivalent system (with same solution set) in simpler form using elementary row operations (EROs): Replacement: add a multiple of one row to another (replacing it) Interchange: swap two rows Scaling: multiply one row by a nonzero constant 3 Read off the solutions from the resulting simpler form Note: The simpler form is Reduced Echelon Form (section 1.2) Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 01 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

9 Lecture 02 Overview Goal for today: Outline: Understand the definition of consistent equations and to identify pivot positions [section 1.2] 1 Echelon forms 2 Pivots 3 Definition of consistency Assignment (1.2): Read: sections 1.3 Work: section 1.2 (p. 21) #1, 3, 9, 11, 14, 15, 17, 22, 23, 25 Extra practice: #5, 7, 13, 19, 21, 29 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 02 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

10 Echelon Forms Definition (page 13) Lecture 02 A matrix is in echelon form (EF) if and only if: every nonzero row is above all zero rows each leading entry is right of the leading entry in the previous row It is in reduced echelon form (REF) if also: each leading entry is a 1 each leading 1 is the only nozero entry in its column Definition (page 13) Two matrices are row-equivalent if and only if one can be row-reduced to the other. Notation: A B Theorem 1 (page 13) Each matrix is row-equivalent to exactly one matrix in REF. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 02 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

11 Lecture 02 Row Reduction Definition (page 14) A pivot position in a matrix is the location of a leading entry in the REF of that matrix. Method: Row-Reduction (pages 15 17) To reduce a matrix to REF, use EROs to put zeros: 1 below the pivot positions (working left to right) 2 above the pivot positions (working right to left) Remarks: Also known as Gaussian Elimination This is the fastest method for solving linear systems By hand: use 1 s as pivots where possible (to avoid fractions) On computer: use largest available pivot (to minimize roundoff) Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 02 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

12 Lecture 02 Solutions from REF Convention (page 19) From REF, solve for the variables corresponding to the pivots (basic variables) in terms of the rest, called free variables. Definition A linear system is consistent if there exists a solution. Theorem 2 (page 21) A linear system is consistent if and only if the REF of its augmented matrix does not have a row of the form [ b ] with b 0 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 02 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

13 Lecture 02 Free Variables Remarks: The number of pivots is enough denoted as r. This number enables one to prove many properties of a matrix or about the solution set of a linear system involving this matrix. The appearance of free variables is not bad, per se. It suggests a loss of uniqueness of a solution, if it exists, in a linear system. Much on more on this later. Your Mantra: If you need to know only one thing, then know that pivots are king! Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 02 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

14 Lecture 02 Examples Example 3 Consider the matrix A = If A is an augmented matrix then is there a solution? Is it unique? 2 If A is a coefficient matrix then given any 3 1 vector of constants, b, is there a solution? If so, is it unique? 3 Answer the same questions for the matrix: A = Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 02 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

15 Lecture 03 Overview Goal for today: Understand the vector equation x 1 a x n a n = b [section 1.3] Outline: 1 Vectors 2 Linear Combinations 3 Span Assignment (1.3): Read: section 1.4 Work: section 1.3 (p. 32) #3, 8, 9, 12, 13, 16, 19, 23, 24, 25 Extra practice: #1, 15, 17, 21 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 03 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

16 Lecture 03 Vectors Definition (page 27) A vector is a one-column matrix. The set of all vectors with n real entries is denoted by R n and called n-dimensional Euclidean space. Geometric interpretation: points and directions Operations: sums and scalar multiples Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 03 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

17 Interpretation of Vectors Lecture 03 There are two common ways of interpreting vectors: 1 Forces - the vector has a magnitude or length and is pointing in a particular direction. 2 Displacements - the vector stores the change in an object s coordinate position. For instance if you are at the point (2, 1) and move one unit along the line with slope 3/4 then where would your new position be? x new = x start + x = y new = y start + y = Hence we have moved one unit step along v = ( 3 5, 4 ). 5 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 03 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

18 Lecture 03 Algebraic Properties of Vectors For all vectors u, v, w in R n and all scalars c and d: (i) u + v = v + u (ii) (u + v) + w = u + (v + w) (iii) u + 0 = 0 + u = u (iv) u + ( u) = 0 (v) c(u + v) = cu + cv (vi) (c + d)u = cu + du (vii) c(du) = (cd)u (viii) 1u = u Remark: There is no such thing as vector multiplication. Then what is the dot product? Cross product? Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 03 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

19 Lecture 03 Linear Combinations Definition (page 27) Given vectors v 1,..., v p in R n and scalars c 1,..., c p, the vector y = c 1 v c p v p is a linear combination of vectors v 1,..., v p with weights c 1,..., c p. Fact (page 29 part of Theorem 3) The vector equation x 1 a 1 + x 2 a x n a n = b has the same solution set as the linear system with augmented matrix [ a1 a 2... a n b ] Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 03 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

20 Lecture 03 Geometric Interpretation of Linear Combinations d c b u v 2v a w -2v -v -u x y Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 ] Lecture 03 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

21 Lecture 03 Span (as a noun) Definition (page 30) Given vectors v 1,..., v p in R n, the span of these vectors is the set of all linear combinations of them. In symbols: Span { v 1,..., v p } = { c 1 v c p v p : c 1,..., c p R } Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 03 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

22 Lecture 04 Overview Goal for today: Understand the matrix equation Ax = b [section 1.4] Outline: 1 Span (revisited) 2 Matrix Equation 3 Equivalence 4 Linear Systems and Span Assignment (1.4): Read: section 1.5 Work: section 1.4 (p. 40) #3, 5, 7, 9, 11, 14, 17, 19, 23 Extra practice: #1, 13, 15, 21, 25, 27 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 04 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

23 Lecture 04 Span (revisited) Span (as a noun): Definition (page 30) Given vectors v 1,..., v p in R m, the span of (or subset generated by) these vectors is the set of all linear combinations of them. In symbols: Span { v 1,..., v p } = { c 1 v c p v p : c 1,..., c p R } Span (as a verb): Definition (page 37) A set of vectors v 1,..., v p in R m spans (or generates) R m if and only if every vector in R m is a linear combination of v 1,..., v p : that is, if and only if Span { v 1,..., v p } = R m Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 04 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

24 Lecture 04 Matrix Equation Definition (page 35) If A is an m n matrix with columns a 1,..., a n, and x R n then the matrix-vector product, is n Ax = x i a i Notice the matrix-vector product is a vector in R m. Remarks: Why is the matrix vector product generate a vector in R m? Why does x have to be of size n? Given b R m then Ax = b is called a matrix equation. i=1 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 04 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

25 Lecture 04 Equivalence A linear system of equations can be written in three alternative forms: 1 (Transformation) A matrix equation: Ax = b. 2 (Coordinates) A vector equation: x 1 a x n a n = b. 3 (Matrices) An augmented problem: [ A b ]. Remarks: The solution set is identical for each situation. The interpretation of the solution is different in each context. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 04 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

26 Lecture 04 Span and Existence of Solutions Theorem 4 (page 37) Let A be an m n matrix. The following are equivalent: For each b in R m, the matrix equation Ax = b has a solution. Each b in R m is a linear combination of the columns of A. The columns of A span R m. A has a pivot position in every row. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 04 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

27 Lecture 05 Overview Goal for today: Outline: Learn to find solution sets in parametric vector form [section 1.5] 1 Span (yep, again!) 2 Overview 3 Homogeneous Systems 4 Nonhomogeneous Systems Assignment (1.5): Read: section 1.6 Work: section 1.5 (p. 47) #5, 8, 12, 13, 17, 20(think about y=mx+b), 22, 24, 29, 30 Extra practice: Remaining problems in Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

28 Lecture 05 Span (one more time... ) Definition (page 30) Given vectors v 1,..., v p in R m, the span of (or subset generated by) these vectors is the set of all linear combinations of them. Definition (page 37) A set of vectors v 1,..., v p in R m spans (or generates) R m if and only if every vector in R m is a linear combination of v 1,..., v p. Theorem 4 (page 37) Let A be an m n matrix. The following are equivalent: For each b in R m, the matrix equation Ax = b has a solution. Each b in R m is a linear combination of the columns of A. The columns of A span R m. A has a pivot position in every row. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

29 Lecture 05 Overview A system of linear equations is equivalent to: An augmented problem: A matrix equation: [ A b ] Ax = b Solution Strategy: 1 Represent linear system or matrix equation as an augmented problem. 2 Reduce the matrix to an echelon form.rref is preferred but not necessary. 3 The number of pivots and locations indicate if a solution exists and if it is unique or not. Answering this question is often called the fundamental question of linear algebra. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

30 Lecture 05 Example 5 Consider the system of equations: x 1 + x 3 = 5 2x 2 + x 2 + 3x 3 = 0 1 Matrix Equation: ( ) x x x 3 = ( ) Augmented Problem: ( ) Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

31 Lecture 05 Example-Continued ( ) ( ) How many pivots? What are the pivot columns? How many solutions? Solve for x 1 and x 2 in terms of the free variable x 3. x = 5 x 3 10 x 3 x 3 It is preferred to write it in parametric vector form, that is, 5 1 x = x Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

32 Homogeneous Systems Lecture 05 Definition (page 43) A linear system Ax = b is: homogeneous if b = 0 nonhomogeneous if b 0 Fact (page 43) A homogeneous system Ax = 0 always has the solution x = 0, which is called the trivial solution. Fact (page 43) A homogeneous system Ax = 0 has a nontrivial solution (x 0) if and only if the system has at least one free variable. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

33 Finding the full solution Lecture 05 Method: 1 Reduce the augmented matrix to reduced echelon form 2 Express each basic variable in terms of the free variables 3 Write the solution x as a vector in terms of the free variables 4 Express the solution as combination of vectors with free variables factored out. This is called parametric vector form, that is, x = p + m α i v i, i=1 where Ax = p, α i are the free variables with coefficient vectors v i, and m is the number of free variables. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

34 Lecture 05 Exercises Consider the question: Is b Span { v 1, v 2, v 3, v 4 }? Rewrite this statement in an equivalent form (there are multiple!). True or False: 1 If b is in the span of the columns of a matrix A then there exists a solution to Ax = b. 2 If b = 2v 1 + 3v 3 then it is in the span of v 1, v 2, and v 3. 3 Let u be a nonzero vector. Then the Span{u} contains a line through u and the origin. 4 The set Span{u, v} can always be visualized as a plane through the origin. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 05 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

35 Lecture 06 Overview Goal for today: Learn to apply linear algebra to some problems [section 1.6] Outline: 1 Exchange Model 2 Network Flow Assignment (1.6): Read: section 1.7 Work: section 1.6 (p. 55) #3, 5, 13, 15 Extra practice: Remaining questions Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 06 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

36 Lecture 06 Exchange Model Example Suppose an economy has four sectors: Mining (M), Lumber (L), Energy (E), and Transportation (T): M sells 10% of its output to L, 60% to E, and retains the rest. L sells 15% to M, 50% to E, 20% to T, and retains the rest. E sells 20% to M, 15% to L, 20% to T, and retains the rest. T sells 20% to M, 10% to L, 50% to E, and retains the rest. (a) Construct the exchange table for this economy. (b) Find a set of equilibrium prices for the economy. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 06 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

37 Lecture 06 Balancing a chemical equation Example Aluminum oxide and carbon react to create elemental aluminum and carbon dioxide: α 1 Al 2 O 3 + α 2 C β 1 Al + β 2 CO 2 Determine the smallest natural numbers that satisfy the equation. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 06 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

38 Network Flow Example Lecture 06 Find the general flow pattern in the following network: 200 B x 1 x 2 A x 3 C x 4 x 5 D 60 What happens when x 4 = 0? What is the minimum value of x 1 then? Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 06 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

39 Overview Goal for today: Outline: Lecture 07 Further understand the concept of linear independence [section 1.7] 1 Concept and Definition 2 Matrix Formulation and connection to vector equation 3 Some Easy Cases 4 Alternate View Assignment (1.7): Review - Supplementary exercises (pg. 89) Work (DUE NEXT CLASS PERIOD): section 1.7 (p. 60) #2, 5, 12, 18, 20, 27, 28 Extra practice: #7, 9, 13, 25, 29 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 07 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

40 Lecture 07 What is Linear Independence? Concept: Linear independence Vectors are linearly independent iff they are in independent directions Definition (page 56) A set of vectors { v 1,..., v p } is: linearly independent if and only if the vector equation x 1 v x p v p = 0 has only the trivial solution x 1 = 0,..., x p = 0 linearly dependent if and only if there exist weights c 1,..., c p not all zero such that c 1 v c p v p = 0 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 07 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

41 Lecture 07 Matrix Formulation Fact (page 57) The columns of a matrix A are linearly independent if and only if the equation Ax = 0 has only the trivial solution x = 0. Some easy cases: Any set which includes the zero vector is linearly dependent Any set containing one (nonzero) vector is linearly independent Any set containing two vectors is linearly independent if and only if neither vector is a multiple of the other Any set containing more vectors than entries is linearly dependent: Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 07 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

42 Lecture 07 Alternate View Theorem 8 (page 59) If v 1,..., v p are p vectors in R n and p > n, then { v 1,..., v p } is linearly dependent. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 07 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86

43 Lecture 07 True/False If v 1,..., v 4 R 4 and v 3 = 2v 1 + v 2, then {v 1, v 2, v 3, v 4 } is a set of linearly independent vectors. If v 1 and v 2 are in R 4 and v 2 is not a scalar multiple of v 1, then {v 1, v 2 } is a set of linearly independent vectors. If v 1,..., v 4 are linear independent vectors in R 4 then {v 1, v 2, v 3 } is a set of linearly independent vectors. If v 1,..., v 4 are linear independent vectors in R 4 then the matrix [v 1 v 2 v 3 ] has 3 pivots. Let A be an m n matrix. If there exists a solution to Ax = b for every b R m then the columns of A are linearly independent. Linear Algebra (MTH 464) Lecture Outlines January 9, / 86 Lecture 07 Linear Algebra (MTH 464) Lecture Outlines January 9, / 86