MTH 464: Computational Linear Algebra

MTH 464: Computational Linear Algebra Lecture Outlines Exam 2 Material Prof. M. Beauregard Department of Mathematics & Statistics Stephen F. Austin State University March 2, 2018 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 1 / 120 Contents I 1 Lecture 1 2 Lecture 2 3 Lecture 3 4 Chapter 2 Review 5 Lecture 4 6 Lecture 5 7 Lecture 6 8 Lecture 7 9 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 2 / 120

Lecture 1 Overview Goal for today: Understand the concept of a subspace in R n [section 2.8 part 1] Outline: 1 Subspaces 2 Null and Column Spaces Assignment (2.8.1): Read: section 2.8 (seriously, do it again) Work: section 2.8 (p. 153) #1, 3, 5, 7, 8, 10, 11, 12 (This is due next class period!) Extra practice: Remaining exercises in #2-14 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 3 / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 4 / 120

Subspaces Basic Idea: Lecture 1 A subspace of R n is a subset of R n (like a copy of R k for some k n). Definition (page 148) A subset H of R n is a subspace of R n if and only if: The zero vector 0 is in H H is closed under addition: for each u and v in H, the sum u + v is in H H is closed under scalar multiplication: for each u in H and each scalar c, the vector cu is in H Fact (Example 3, page 149) For any v 1,..., v p in R n, Span { v 1,..., v p } is a subspace of R n. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 5 / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 6 / 120

Subspaces - Examples Lecture 1 Are the following subspaces of R 2? Why or why not? 1 H = {(x, y) 0 x, y 2} 2 H = {(x, y) 0 x, y < } 3 H = {(x, y) < x <, y = x} H = Span{v 1, v 2, v 3 }, where v 1 = 1 2 3 1, v 2 = 1 0 0 1, v 3 = Is H a subspace of R 4? What dimension is the subspace? We ll talk more about this in the coming lectures, but since the vectors are linearly independent the dimension is 3. 1 0 3 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 7 / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 8 / 120

Lecture 1 Null and Column Spaces Definition (pages 149 150) Let A be an m n matrix with columns a 1,..., a n. The null space of A is the set of all solutions of Ax = 0: Nul A = { x R n : Ax = 0 } This is a subspace of R n (by Theorem 12, pg. 150). The column space of A is the span of the columns of A: Col A = Span { a 1,..., a n } This is a subspace of R m. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 9 / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 10 / 120

Lecture 1 Questions 1 Let A be a 4 7 matrix. Is the column space of A guaranteed to be a subspace of R 4 or R 7? What about the null space? 2 Let A be a 7 7 matrix. Is the column space of A guaranteed to be a subspace of R 4 or R 7? What about the null space? 3 (True or False) Given A = [ a 1 a 2 a 3 ]. Then the vector w = 2 a 1 6 a 3 is in the column space of A. 4 If A is a 4 4 matrix with 2 pivot columns then does the column space span R 4? Describe the null space. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 11 / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 12 / 120

Lecture 1 Determining the Null space Given a matrix A then the null space is determined by: 1 Reduce [ A 0 ] to reduced echelon form (or at least to the point you can identify the pivots, free variables, and back substitute) 2 Write the solution x in parametric vector form, that is, where α i are free variables. x = α 1 v 1 +... + α γ v γ 3 The null space is then the span of v 1,..., v γ. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 13 / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 14 / 120

Null Space - Examples Lecture 1 Determine the null space of the following matrices: [ ] [ ] 1 3 3 7 3 6,, 0 1 4 5 1 2 1 2 3 2 2 4 4 2 1 3, 4 4 8 8 1 1 0 0 3 3 0 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 15 / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 16 / 120

Lecture 2 Overview Goal for today: Outline: Understand the concept of a basis [section 2.8 part 2] 1 Null and Column Space [Review] 2 What is a Basis? 3 Basis for the Null Space 4 Basis for the Column Space Assignment (2.8.2): Read: section 2.9 Work: section 2.8 (p. 153) #15, 17, 21, 23, 25 (This is due next class period!) Extra practice: #19, 22, 24, 27, 29, 31 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 17 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 18 / 120

Lecture 2 Null Space Definition The null space of A is the set of all solutions of Ax = 0. Questions: 1 (True or False) Let A be a 4 7 matrix with 4 pivots. Then the null space is not empty. 2 (True or False) If x 1 and x 2 are in the null space of A then x 1 + 8x 2 is also in the null space of A. 3 How do we determine the null space? Reduce [ A 0 ] to reduced echelon form. Write the solution x in parametric vector form. The null space is then the span of the independent vectors. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 19 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 20 / 120

Lecture 2 Column Space Definition The column space of A is the span of the columns of A. Questions: 1 Let A be a 9 2 matrix. Is the column space of A guaranteed to be a subspace of R 2 or R 9? 2 Let A be a 5 3 matrix. Is the column space of A guaranteed to be a subspace of R 3 or R 5? 3 (True or False) If A is an n m matrix if Col(A) = R n then m n. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 21 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 22 / 120

What is a Basis? Idea: Lecture 2 A basis for a subspace H a small set of vectors which describes H. Definition (page 150) A basis for a subspace H of R n is a set of vectors in H which is linearly independent and spans H. Analogy: Minimal Toolkit job to do describe H tools vectors enough tools span H no duplicates linearly independent minimal toolkit basis Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 23 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 24 / 120

Lecture 2 Basis for the Column Space Recall: The column space of an m n matrix A = [ a 1 span of the columns of A:... a n ] is the Col A = Span { a 1,..., a n } This is a subspace of R m (by Example 3). Method (Example 7, page 151) To find a basis for the column space of a matrix A: 1 Reduce A to echelon form 2 Identify the pivot positions 3 The pivot columns of A form a basis for Col A (Theorem 13, p. 152) Warning: Use the pivot columns of A not of the echelon form. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 25 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 26 / 120

Column Space Example Lecture 2 Determine the column space of 3 6 1 1 A = 1 2 2 3 2 4 5 8 A 3 6 1 1 0 0 5/3 10/3 0 0 13/3 26/3 3 6 1 1 0 0 5/3 10/3 0 0 0 0 So the basis for the column space is? Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 27 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 28 / 120

Basis for the Null Space Recall: Lecture 2 The null space of an m n matrix A is the set of all solutions of Ax = 0: Nul A = { x R n : Ax = 0 } This is a subspace of R n (by Theorem 12). Method (Example 6, page 151) To find a basis for the null space of a matrix A: 1 Reduce [ A 0 ] to reduced echelon form 2 Write the solution x in parametric vector form 3 The vectors which multiply the free variables form a basis for Nul A Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 29 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 30 / 120

Null Space Example Lecture 2 Determine the null space space of 3 6 1 1 A = 1 2 2 3 2 4 5 8 We found this matrix has the reduced form of 3 6 1 1 A 0 0 5/3 10/3 0 0 0 0 Reducing this further yields, A 1 2 0 1 0 0 1 2 0 0 0 0 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 31 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 32 / 120

Lecture 2 Null Space Example - Continued So x = A 2x 2 + x 4 x 2 2x 4 x 4 1 2 0 1 0 0 1 2 0 0 0 0 = x 2 2 1 0 0 + x 4 1 0 2 1 Thus the Nul(A) = Span 2 1 0 0, 1 0 2 1 Span{v 1, v 2 } The vectors v 1 and v 2 form a basis for the null space of A. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 33 / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 34 / 120

Lecture 3 Overview Goal for today: Understand coordinates, dimension, and rank [section 2.9] Outline: 1 Coordinates 2 Dimension and Rank 3 Three Important Theorems Assignment: Read: 3.1 Work: section 2.9 (p. 159) #1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 35 / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 36 / 120

Lecture 3 Review 1 A basis B is a minimal set of vectors that span a subspace H 2 A minimal set of vectors means that each basis vector is linearly independent 3 We can determine the basis for the column space of a matrix A 4 We can determine the basis for the null space of a matrix A Key Idea: A basis B can be used to describe other vectors in a subspace H. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 37 / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 38 / 120

Coordinates Definition (page 156) Lecture 3 Suppose B = { b 1,..., b p } is a basis for a subspace H of R n. The coordinates of a vector x in H are the weights c 1,..., c p in the linear combination x = c 1 b 1 + + c p b p. The corresponding vector [x] B = is called the coordinate vector of x relative to the basis B. Remarks: Coordinates are unique (given x and B) [section 4.4] Change of basis (B to C): [x] C = P C B [x] B [section 4.7] Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 39 / 120 c 1. c p Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 40 / 120

Lecture 3 Example [ 1 Given a B = {b 1, b 2 }, where b 1 = 1 the Euclidean basis, determine [x] B. ] [ ] [ ] 2 4, b 2 =. Let x = in 3 5 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 41 / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 42 / 120

Lecture 3 Dimension and Rank Claim If a subspace H has a basis consisting of p vectors, then every basis for H consists of p vectors (see Exercises 27 and 28 of section 2.9). Definition (page 157) The dimension of a subspace H of R n, denoted by dim H, is the number of vectors in any basis for H. The dimension of the zero subspace { 0 } is the number 0. Definition (page 157) The rank of a matrix A of R n, denoted by rank A, is the dimension of the column space: rank A = dim(col A). Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 43 / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 44 / 120

Lecture 3 Three Important Theorems The Rank Theorem (Theorem 14, page 158) If a matrix A has n columns, then: rank A + dim(nul A) = n. The Basis Theorem (Theorem 15, page 158) If H is a subspace of R n with dim H = p, then: Any linearly independent set of p vectors in H spans H. Any set of p vectors in H which spans H is linearly independent. Any set of more than p vectors in H is linearly dependent. Any set of fewer than p vectors in H does not span H. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 45 / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 46 / 120

Lecture 3 The Invertible Matrix Theorem continued (page 158) Let A be an n n matrix. The following are equivalent: a. A is invertible.. e. The columns of A form a linearly independent set.. h. The columns of A span R n. m. The columns of A form a basis for R n n. Col A = R n o. dim Col A = n p. rank A = n q. Nul A = { 0 } r. dim Nul A = 0 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 47 / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 48 / 120

Lecture 3 True/False Questions 1 A 4 4 matrix A has a rank of 4. Then the columns are linearly independent. 2 A 4 4 matrix A has a rank of 4. Then the columns form a basis for R 4. 3 A n n matrix A has a rank of n. Let B be the basis to the column space of A. Then any vector x in R n can also be written as a B-coordinate vector. 4 A 5 5 matrix A has null space with dimension 2. Then the dimension of the column space is also 2. 5 Given a n m matrix A. Then number of pivots enables an observer to objectively determine the dimensions of the null and column space. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 49 / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 50 / 120

Chapter 2 Review Goal for today: Review the fundamental theorems of chapter 2 and its applications. Outline: 1 Three fundamental theorems 2 Supplementary questions Reading assignment: Read: 3.1 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 51 / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 52 / 120

Chapter 2 Review Rank Theorem The Rank Theorem (Theorem 14, page 156) If a matrix A has n columns, then: rank A + dim(nul A) = n. A big remark: Given Ax = b. Then b is either: 1 In the span of Col A a solution exists. 2 In the span of Nul A no solution exists. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 53 / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 54 / 120

Chapter 2 Review Basis Theorem The Basis Theorem (Theorem 15, page 156) If H is a subspace of R n with dim H = p, then: Any linearly independent set of p vectors in H spans H. Any set of p vectors in H which spans H is linearly independent. Any set of more than p vectors in H is linearly dependent. Any set of fewer than p vectors in H does not span H. A not so big remark: This theorem brings together ideas/theorems from section 1.7 where linear independence was introduced. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 55 / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 56 / 120

Chapter 2 Review The Grand Poopa The Invertible Matrix Theorem continued (page 156) Let A be an n n matrix. The following are equivalent: a. A is invertible. b. A is row-equivalent to the identity matrix. c. A has n pivot positions. d. The equation Ax = 0 has only the trivial solution. e. The columns of A form a linearly independent set. f. The linear transformation x Ax is one-to-one. g. The equation Ax = b has at least one solution for each b in R n. h. The columns of A span R n. i. The linear transformation x Ax is onto (i.e, maps R n onto R m ). j. There is an n n matrix C such that CA = I. k. There is an n n matrix D such that AD = I. l. A T is invertible. m. The columns of A form a basis for R n n. Col A = R n o. dim Col A = n p. rank A = n q. Nul A = { 0 } r. dim Nul A = 0 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 57 / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 58 / 120

Chapter 2 Review Supplementary Exercises (True/False) - Page 160 Matrix multiplication: 1 If A and B are m n matrix then both AB and A B. 2 If AB = C and C has 2 columns, then A has 2 columns. 3 Left-multiplying a matrix B by a diagonal matrix A, with nonzero entries on the diagonal, scales the rows of B. 4 If BC = BD, then C = D. 5 If AC = 0, then either A = 0 or C = 0. 6 If A and B are n n matrices, then (A + B)(A B) = A 2 B 2. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 59 / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 60 / 120

Chapter 2 Review Supplementary Exercises (True/False) - Page 160 Elementary Matrices: 1 An elementary n n matrix has either n or n + 1 nonzero entries. 2 The transpose of an elementary matrix is an elementary matrix. 3 An elementary matrix must be square. Hint: Are elementary matrices invertible? 4 Every square matrix is a product of elementary matrices. Hint: Is every square matrix equivalent to the identity matrix? Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 61 / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 62 / 120

Chapter 2 Review Supplementary Exercises (True/False) - Page 160 1 If A is a 3 3 matrix with three pivot positions, there exist elementary matrices E 1,..., E p such that E p E p 1 E 1 = I. 2 If AB = I, then A is invertible. Hint: What happens if A is not square? 3 If A and B are n n, square and invertible matrices, then AB is invertible and (AB) 1 = A 1 B 1. 4 If AB = BA and if A is invertible, then A 1 B = BA 1. 5 If A is invertible and if r 0 then (ra) 1 = ra 1. 1 6 If A is a 3 3 matrix and the equation Ax = 0 has a unique 0 solution, then A is invertible. 7 If A is a 3 3 invertible matrix then the solution to Ax = determines the first column of A 1. 1 0 0 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 63 / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 64 / 120

Lecture 4 Overview Goal for today: Learn to compute determinants by cofactors [section 3.1] Outline: 1 Determinants 2 Cofactor Expansion 3 Triangular Matrices Assignment (3.1): Read: section 3.2 Work: section 3.1 (p. 169) #3, 9, 14, 19, 21, 23, 37, 39 Extra practice: #1, 5, 7, 11, 13 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 65 / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 66 / 120

Lecture 4 2 2 Matrices Question: What condition on the entries of the matrix invertible? [ a b c d ] guarantee that it is Answer: When the determinant is not zero, that is, when ad bc 0. Idea: Can we extend this to n n matrices? Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 67 / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 68 / 120

Lecture 4 Determinants Concept: The determinant of a square matrix A is a scalar which tells whether or not A is invertible: A is invertible if and only if det A 0. Easy cases: n = 1: A = [ a ] det A = a n = 2: A = [ ] a b c d det A = ad bc Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 69 / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 70 / 120

Cofactor Expansion Lecture 4 Definition (page 167) The determinant of an n n matrix A = [a i,j ] with n > 1 is det A = a 11 det A 11 a 12 det A 12 + + ( 1) 1+n a 1n det A 1n where A ij is the (n 1) (n 1) matrix obtained by removing row i and column j from A, called the (i, j) minor matrix. The (i, j) cofactor of A is the scalar C ij = ( 1) i+j det A ij : det A = a 11 C 11 + a 12 C 12 + + a 1n C 1n Theorem 1 (page 168): can expand by cofactors of any row: det A = a i1 C i1 + a i2 C i2 + + a in C in (any i) any column: det A = a 1j C 1j + a 2j C 2j + + a nj C nj (any j) Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 71 / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 72 / 120

Lecture 4 Triangular Matrices Recall: A matrix A = [a i,j ] is triangular iff either: lower-triangular iff a i,j = 0 for i < j: upper-triangular iff a i,j = 0 for i > j: 0... 0... 0..... 0...... 0........ 0 0... Theorem 2 (page 169) If A is a triangular matrix, then det A is the product of the entries on the main diagonal of A. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 73 / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 74 / 120

Lecture 5 Overview Goal for today: Learn to compute determinants by row reduction [section 3.2] Outline: 1 Determinants by Row Reduction 2 Cofactors vs. Row Reduction 3 More Properties of Determinants Assignment (3.2): Read: section 3.3 Work: section 3.2 (p. 177) #7, 12, 15, 17, 19, 21, 27, 28, 29, 39 Extra practice: #1, 3, 5, 9, 11, 13, 37 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 75 / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 76 / 120

Fact Lecture 5 The determinant of an elementary matrix E is: Operation Effect Records Replacement: R i + cr j R i det E = c Records Interchange: R i R j det E = 1 Records Scaling: kr i R i det E = k Theorem 3 (page 171) Let A be a square matrix and let B be obtained from A by one elementary row operation. Then: Operation Effect Replacement: R i + cr j R i det B = det A Interchange: R i R j det B = det A Scaling: kr i R i det B = k det A This motivates an alternative way to determine a determinant! Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 77 / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 78 / 120

Lecture 5 Method): To compute det A by row reduction 1 Reduce A to echelon form B by elementary row operations: a 11 a 12... a 1n b 11 b 12... b 1n a 21 a 22... a 2n EROs 0 b 22... b 2n A =............ = B a n1 a n2... a nn 0 0... b nn and keep track of the elementary row operations used. 2 Compute det B = b 11 b 22 b nn. 3 Adjust det B by the effect of the EROs to obtain det A. Remarks: Be careful if you scale or do a row exchange as this influences the value of the determinant. Can combine row reduction and cofactors Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 79 / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 80 / 120

Lecture 5 Cofactors vs. Row Reduction Claim: Row reduction is much faster than cofactors Work for an n n determinant in flops (floating-point operations): n by cofactors by row reduction 3 14 15 5 324 74 10 9,864,100 624 50 8 10 64 82,124 n O(n!) O(n 3 ) To compute a 50 50 determinant by cofactors on the MilkyWay-2 supercomputer (3.1M cores, 34 petaflops < 10 17 flops per second): 10 48 seconds 10 42 years Note: determinant is a theoretical tool numerical value rarely needed. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 81 / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 82 / 120

Lecture 5 More Properties of Determinants Theorem 4 (page 173) A square matrix A is invertible if and only if det A 0. Theorem 5 (page 174) For a square matrix A: det A T = det A Theorem 6 (page 174) For n n matrices A and B: det(ab) = (det A)(det B) Consequences: det(a k ) = (det A) k for k = 1, 2, 3,... If A is invertible then det(a 1 ) = 1 det A [see 3.2 #31] Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 83 / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 84 / 120

Lecture 5 True/False Questions 1 Consider the LU-factorization of A. The det A = det L det U. 2 Matlab uses a cofactor expansion to determine the determinant of a matrix. 3 A row exchange influences the determinant of a matrix. 4 det(a + B) = det A + det B. 5 The determinant of matrix can be determined by calculating the product of the diagonal entries. 6 If det A = 0 and A is an n n matrix then A has at most n 1 pivots. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 85 / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 86 / 120

Lecture 6 Overview Goal for today: Learn some applications of determinants [section 3.3] Outline: 1 Cramer s Rule 2 Inverse by Cofactors 3 Area and Volume Assignment (3.3): Read: section 5.1 (skip chapter 4) Work: section 3.3 (p. 186) #3, 7, 11, 19, 22, 23, 27 Extra practice: #1, 9, 13, 21, 25, 29 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 87 / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 88 / 120

Lecture 6 Cramer s Rule Theorem 7 (page 179) Let A be an n n invertible matrix. For each vector b R n, the unique solution x of Ax = b has entries x i = det A i(b), i = 1,..., n, det A where A i (b) is the matrix obtained from A by replacing column i by b. Remarks: Useful when A contains a parameter (e.g., in Laplace transforms) Requires the work of n + 1 determinants of size n n, and thus requires more calculation that standard Gaussian elimination. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 89 / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 90 / 120

Lecture 6 Inverse by Cofactors Recall that the cofactor is C ij = ( 1) i+j det A ij Theorem 8 (page 181) If A is an n n invertible matrix, then A 1 = 1 det A C 11. C n1... C 1n.... C nn T where C ij is the (i, j) cofactor of A. Remarks: Useful when A contains parameters or only certain entries needed The matrix of cofactors (transposed) is called the adjugate of A Requires the work of n 2 + 1 determinants, so far more work than computing [ A I ] [ I A 1 ] Linear Algebra (MTH 464) Lectureby Outlines EROs March 2, 2018 91 / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 92 / 120

Area and Volume Theorem 9 (page 182) Lecture 6 If A is a 2 2 matrix, then the area of the parallelogram determined by the columns of A is det A. If A is a 3 3 matrix, then the volume of the parallelepiped determined by the columns of A is det A. Theorem 10 (page 184) If T : R 2 R 2 is determined by a 2 2 matrix A and S is a parallelogram in R 2, then { area of T (S) } = det A { area of S } If T : R 3 R 3 is determined by a 3 3 matrix A and S is a parallelepiped in R 3, then { volume of T (S) } = det A { volume of S } Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 93 / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 94 / 120

Lecture 7 Overview Goal for today: Outline: Understand the concepts of eigenvalues and eigenvectors [section 5.1] 1 Eigenvalues and Eigenvectors 2 Verifying Eigenvectors and Eigenvalues 3 Finding Eigenvectors and Eigenspaces Assignment (5.1): Read: section 5.2 Work: section 5.1 (p. 273) #3, 7, 9, 13, 14, 15, 17, 19, 21, 33 Extra practice: #1, 5, 11, 13, 22, 23, 25 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 95 / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 96 / 120

Lecture 7 Motivating Example Consider vectors in R 2 and the matrix: ( 5 3 A = 3 5 ). Consider the set of vectors x such that x = 1. We say that A transforms x into a new vector in R 2 through x Ax Let s examine how A transforms any members of this set. What is special about these vectors? v 1 = [ ] 2/2 2/2 v 2 = [ ] 2/2 2/2 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 97 / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 98 / 120

Lecture 7 Eigenvalues and Eigenvectors Idea: Describe the action of (multiplying by) a matrix: eigenvectors special directions eigenvalues magnification factors Definition Let A be an n n matrix. A scalar λ is an eigenvalue of A if and only if Ax = λx for some x 0. Any such vector x is called an eigenvector of A corresponding to λ. Notes: Eigenvectors are not unique (nonzero multiples are eigenvectors). Eigenvectors cannot be zero (by definition) Eigenvalues may be zero (if and only if A is singular) Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 99 / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 100 / 120

Lecture 7 Verifying Eigenvectors and Eigenvalues To verify that a vector x is an eigenvector of A: Just multiply and see whether Ax = λx for some scalar λ C To verify that a scalar λ is an eigenvalue of A: The matrix A λi must be singular, since: Ax = λx (x 0) Ax λx = 0 (x 0) Ax λix = 0 (x 0) (A λi)x = 0 (x 0) Reduce A λi to echelon form: must have a row of zeros What about the det(a λi)? Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 101 / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 102 / 120

Lecture 7 Finding Eigenvectors and Eigenspaces Definition The eigenspace corresponding to an eigenvalue λ of an n n matrix A is the null space of the matrix A λi; this may be written as Notes: { x R n : Ax = λx }. The eigenspace consists of all corresponding eigenvectors (and 0) Finding eigenvectors finding a basis for the eigenspace To do so, reduce [ A λi 0 ] to REF and solve for eigenvectors x Write the solutions x in parametric vector form (PVF) The fixed vectors in the PVF form a basis for the eigenspace. The dimension of a particular eigenspace is the geometric multiplicity of the associated eigenvalue. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 103 / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 104 / 120

Lecture 8 Overview Goal for today: Learn to find eigenvalues and eigenvectors [section 5.2] Outline: 1 Review 2 Finding Eigenvalues 3 Similarity Assignment (5.2): Read: section 5.3 Work: section 5.2 (p. 281) #3, 7, 11, 14, 15, 21, 22, 23 Extra practice: #1, 5, 9, 13, 17, 19 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 105 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 106 / 120

Lecture 8 Review Definition Let A be an n n matrix. A scalar λ is an eigenvalue of A if and only if Ax = λx for some x 0. Any such vector x is called an eigenvector of A corresponding to λ. Definition The eigenspace corresponding to an eigenvalue λ of A is: Nul(A λi) = { x R n : (A λi)x = 0 } = { x R n : Ax = λx } Finding basis for eigenspace of a particular λ 1 Row-reduce [ A λi 0 ] to reduced echelon form 2 Write corresponding solutions x in parametric vector form Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 107 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 108 / 120

Lecture 8 Review 1 (T/F) Eigenvectors are unique. 2 (T/F) If x is an eigenvector then so is αx for α R. 3 (T/F) Given y if Ay = βy then β is an eigenvalue of the matrix A. 4 (T/F) Given y if Ay = βy for a scalar β then y is an eigenvector of the matrix A. 5 (T/F) The basis for the null space of A λi is the eigenspace for this particular λ. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 109 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 110 / 120

Lecture 8 Determining Eigenvalues Fact A scalar λ is an eigenvalue of an n n matrix A if and only if λ satisfies the characteristic equation det(a λi) = 0. Notes: The function p(λ) = det(a λi) is a polynomial of degree n in λ, called the characteristic polynomial Its zeros [i.e., roots of p(λ) = 0] are the eigenvalues of A The algebraic multiplicity of an eigenvalue is its multiplicity as a root of p The geometric multiplicity of an eigenvalue is the number of linearly independent vectors associated with it. In other words, it is the dimension of the eigenspace. There are n eigenvalues (counting multiplicities, possibly complex) of every n n matrix. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 111 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 112 / 120

Lecture 8 Review of our Methodology Method: To find the eigenvalues of a matrix A 1 Find p(λ) = det(a λi) [probably by cofactor expansion, since we have parameters] 2 Find the roots of p(λ) = 0: by factoring, synthetic division, etc. Method: To find the eigenvectors of a matrix A 1 For each eigenvalue determine the null space of A λi. That is, find a nontrivial solution to (A λi)v = 0 2 Each independent null vector of A λi is an eigenvector. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 113 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 114 / 120

Lecture 8 Two Theorems Theorem 2 (page 271) If v 1,..., v r are eigenvectors corresponding to distinct eigenvalues λ 1,..., λ r, then { v 1,..., v r } is linearly independent. Theorem 3 (page 277, Invertible Matrix Theorem continued!) Let A be an n n matrix. Then the following are equivalent: a. A is invertible. s. The number 0 is not an eigenvalue of A. t. The determinant of A is not zero. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 115 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 116 / 120

Lecture 8 Similarity Definition Let A and B be n n matrices. The matrix A is similar to the matrix B if and only if there is an invertible matrix P such that PB = AP or equivalently, B = P 1 AP Notes: Similarity is an equivalence relation because it is: Reflexive: A A [where here denotes is similar to ] Symmetric: if A B, then B A Transitive: if A B, and B C, then A C Similar matrices represent the same linear transformation with respect to different bases (see section 5.4) Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 117 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 118 / 120

Lecture 8 An Important Result for Similar Matrices Theorem 4 (page 279) If n n matrices A and B are similar, then they have the same characteristic polynomial, and therefore the same eigenvalues (with the same multiplicities). Remark: The above theorem says NOTHING about eigenvectors. Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 119 / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, 2018 120 / 120