Math 3191 Applied Linear Algebra

Transcription

1 Math 191 Applied Linear Algebra Lecture 9: Characterizations of Invertible Matrices Stephen Billups University of Colorado at Denver Math 191Applied Linear Algebra p.1/

2 Announcements Review for Exam 1 posted. (Exam next Tuesday). HWK 5 (due Oct. ) posted. Math 191Applied Linear Algebra p./

3 Tonight Finish Sec.., Inverse of a Matrix. Sec.., Characterizations of Invertible Matrices. Math 191Applied Linear Algebra p./

4 Sec.: The Inverse of a Matrix Key Concepts: Definition of an inverse. Uniqueness of inverse. Terminology: Invertible=nonsingular, non-invertible=singular. Formula for inverse of matrix. Solving linear equations using the inverse. Properties of inverses: (inverses of inverses, products, transposes). Elementary matrices. Algorithm for computing inverses using row operations. Math 191Applied Linear Algebra p./

5 Theorem Inverse of by matrix. Let A = a b c d 5. If ad bc 0, then A is invertible and A 1 = 1 ad bc d b c a 5. If ad bc = 0, then A is not invertible. Example: A = 1 5 A 1 = = 1 / 1/ 5 Check: AA 1 = / 1/ 5 = 5. Math 191Applied Linear Algebra p.5/

6 Solving Linear Equations with Inverse Assume A is any invertible matrix and we wish to solve Ax = b. Then Ax = b and so Ix = or x =. Is the Solution Unique? Suppose w is also a solution to Ax = b. Then Aw = b and Aw = b which means w =A 1 b. So, w =A 1 b, which is in fact the same solution. Math 191Applied Linear Algebra p./

7 Theorem 5 We have proved the following result: Theorem 5 If A is an invertible n n matrix, then for each b in R n, the equation Ax = b has the unique solution x = A 1 b. EXAMPLE: A = Solution: A 1 = [ x =A 1 b = [ [ ] 5 5, b = [ 1 ] ] [ ] ] =. [ = 5 [ ] ]. Math 191Applied Linear Algebra p./

8 Theorem Suppose A and B are invertible. Then the following results hold: a. A 1 is invertible and `A 1 1 = A (i.e. A is the inverse of A 1 ). b. AB is invertible and (AB) 1 = B 1 A 1 c. A T is invertible and `A T 1 `A 1 T = Partial proof of part b: (AB) `B 1 A 1 = A ( ) A 1 = A ( ) A 1 = =. Similarly, one can show that `B 1 A 1 (AB) = I. Math 191Applied Linear Algebra p.8/

9 More Matrices... Theorem, part b can be generalized to three or more invertible matrices: (ABC) 1 =. Proof: (ABC) 1 = ((AB) C) 1 = C 1 ( ) 1 = C 1 ( ) =. Math 191Applied Linear Algebra p.9/

10 Elementary Matrices To calculate inverses of larger than matrices, we first need to look at elementary matrices. Definiton: An elementary matrix is one that is obtained by performing a single elementary row operation on an identity matrix. EX: Let E 1 = 0 0 5, E = and E are elementary matrices. Why? 5, E = E 1, E, Math 191Applied Linear Algebra p.10/

11 Multiplying by an Elementary Matrix Let A = E 1 A = a b c d e f g h i a b c d e f g h i = Describe the effect multiplying by E 1 had on A. Math 191Applied Linear Algebra p.11/

12 Effect of Multiplying by Elementary Matrix If an elementary row operation is performed on an m n matrix A, the resulting matrix can be written as EA, where the m m matrix E is created by performing the same row operations on I m. Examples: E A = a b c d e f g h i 5 = a b c g h i d e f 5 E A = a b c d e f g h i 5 = a b c d e f a + g b + h c + i 5 Math 191Applied Linear Algebra p.1/

13 Inverses of Elementary Matrices Elementary matrices are invertible because row operations are reversible. To determine the inverse of an elementary matrix E, determine the elementary row operation needed to transform E back into I and apply this operation to I to find the inverse. For example, E = E = E 1 = E 1 = Math 191Applied Linear Algebra p.1/

14 Inverting A with Elementary Matrices Example: Let A = E 1 A = E (E 1 A) = E (E E 1 A) = Then 5 = = = Math 191Applied Linear Algebra p.1/

15 So Then multiplying on the right by A 1, we get E E E 1 A = I. So E E E 1 A = I. E E E 1 I = A 1 E E E 1 I = I = = = Math 191Applied Linear Algebra p.15/

16 The elementary row operations that row reduce A to I n are the same elementary row operations that transform I n into A 1. Theorem An n n matrix A is invertible if and only if A is row equivalent to I n, and in this case, any sequence of elementary row operations that reduces A to I n will also transform I n to A 1. Math 191Applied Linear Algebra p.1/

17 Algorithm for finding A 1 Place A and I side-by-side to form an augmented matrix [A I]. Then perform row operations on this matrix (which will produce identical operations on A and I). So by Theorem :[A I] will row reduce to ˆI A 1 (or A is not invertible). EXAMPLE: Find the inverse of A = , if it exists. Solution:[A I] = So A 1 = Math 191Applied Linear Algebra p.1/

18 Bad Example Find the inverse of A = Solution: [A I] = What went wrong? A is. Math 191Applied Linear Algebra p.18/

19 Order of multiplication is important! EXAMPLE Suppose A,B,C, and D are invertible n n matrices and A = B(D I n )C. Solve for D in terms of A, B, C and D. Solution: A = B(D I n )C D I n = B 1 AC 1 D I n + = B 1 AC 1 + D = Math 191Applied Linear Algebra p.19/

20 Sec.. Characterizations of Invertible Matrices Overview: The Invertible Matrix Theorem. Math 191Applied Linear Algebra p.0/

21 Theorem 8 (The Invertible Matrix Theorem) Let A be a square n n matrix. The the following statements are equivalent (i.e., for a given A, they are either all true or all false). a. A is an invertible matrix. b. A is row equivalent to I n. 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 maps R n onto R n. j. There is an n n matrix C such that CA = I n. k. There is an n n matrix D such that AD = I n. l. A T is an invertible matrix. Math 191Applied Linear Algebra p.1/

22 Comments about Proof There are many theorems in the mathematical literature that say The Following Are Equivalent. To prove that two statements are equivalent, (e.g., to show (a) (b), you need to prove two things: (a) (b), and (b) (a). To prove that three statements are equivalent, you could show (a) (b), (b) (c), and (c) (b) ( implications to prove); But that is more work than necessary. It is sufficient to show a cycle of implications: (a) (b) (c) (a). ( implications to prove). Alternatively, you could show: (a) (b), and (b) (c). ( implications). Both of these strategies are used in the proof of Theorem 8. Math 191Applied Linear Algebra p./

23 Example Use the Invertible Matrix Theorem to determine if A is invertible, where 1 0 A = Solution A = }{{} pivots positions Circle correct conclusion: Matrix A is / is not invertible. Math 191Applied Linear Algebra p./

24 Example Suppose H is a 5 5 matrix and suppose there is a vector v in R 5 which is not a linear combination of the columns of H. What can you say about the number of solutions to Hx = 0? Solution Since v in R 5 is not a linear combination of the columns of H, the columns of H do not R 5. So by the Invertible Matrix Theorem, Hx = 0 has. Math 191Applied Linear Algebra p./

25 Invertible Linear Transformations For an invertible matrix A, A 1 Ax = x for all x in R n and Pictures: AA 1 x = x for all x in R n. Math 191Applied Linear Algebra p.5/

26 Invertible Linear Transformations A linear transformation T : R n R n is said to be invertible if there exists a function S : R n R n such that S(T (x)) = x for all x in R n and T (S(x)) = x for all x in R n. Math 191Applied Linear Algebra p./

27 Theorem 9 Let T : R n R n be a linear transformation and let A be the standard matrix for T. Then T is invertible if and only if A is an invertible matrix. In that case, the linear transformation S given by S(x) = A 1 x is the unique function satisfying S(T (x)) = x for all x in R n and T (S(x)) = x for all x in R n. Math 191Applied Linear Algebra p./