MTH 464: Computational Linear Algebra

Transcription

1 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, / 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, / 120

2 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, / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

3 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, / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

4 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 = , v 2 = , 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 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

5 Lecture 1 Null and Column Spaces Definition (pages ) 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, / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

6 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, / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

7 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 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, / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

8 Null Space - Examples Lecture 1 Determine the null space of the following matrices: [ ] [ ] ,, , Linear Algebra (MTH 464) Lecture Outlines March 2, / 120 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

9 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, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

10 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, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

11 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, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

12 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, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

13 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, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

14 Column Space Example Lecture 2 Determine the column space of A = A /3 10/ /3 26/ /3 10/ So the basis for the column space is? Linear Algebra (MTH 464) Lecture Outlines March 2, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

15 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, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

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

17 Lecture 2 Null Space Example - Continued So x = A 2x 2 + x 4 x 2 2x 4 x = x x Thus the Nul(A) = Span , 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, / 120 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

18 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, / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

19 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, / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

20 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 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, / 120 c 1. c p Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

21 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, / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

22 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, / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

23 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, / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

24 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, / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

25 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, / 120 Lecture 3 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

26 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, / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

27 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, / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

28 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, / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

29 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, / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

30 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, / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

31 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, / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

32 Chapter 2 Review Supplementary Exercises (True/False) - Page 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 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 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120 Chapter 2 Review Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

33 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, / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

34 Lecture 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, / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

35 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, / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

36 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 ( 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 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, / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

37 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: 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, / 120 Lecture 4 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

38 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, / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

39 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, / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

40 Lecture 5 Method): To compute det A by row reduction 1 Reduce A to echelon form B by elementary row operations: a 11 a a 1n b 11 b b 1n a 21 a a 2n EROs 0 b b 2n A = = B a n1 a n2... a nn 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, / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

41 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 ,864, ,124 n O(n!) O(n 3 ) To compute a determinant by cofactors on the MilkyWay-2 supercomputer (3.1M cores, 34 petaflops < flops per second): seconds years Note: determinant is a theoretical tool numerical value rarely needed. Linear Algebra (MTH 464) Lecture Outlines March 2, / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

42 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, / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

43 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, / 120 Lecture 5 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

44 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, / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

45 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, / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

46 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 determinants, so far more work than computing [ A I ] [ I A 1 ] Linear Algebra (MTH 464) Lectureby Outlines EROs March 2, / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

47 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, / 120 Lecture 6 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

48 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, / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

49 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, / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

50 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, / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

51 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, / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

52 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, / 120 Lecture 7 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

53 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

54 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

55 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

56 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

57 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

58 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

59 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120

60 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, / 120 Lecture 8 Linear Algebra (MTH 464) Lecture Outlines March 2, / 120