Matrix equation Ax = b

Transcription

1 Fall 2017 Matrix equation Ax = b Authors: Alexander Knop Institute: UC San Diego

2 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. Let u R 2 be a nonzero vector. Alexander Knop 2

3 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. Let u R 2 be a nonzero vector. Then Span {u} is a line. Alexander Knop 2

4 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. u Alexander Knop 3

5 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. u 2u Alexander Knop 3

6 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. u 2u u Alexander Knop 3

7 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. 1 0 Let u = 0 and v = Alexander Knop 4

8 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. 1 0 Let u = 0 and v = Consider Span {u, v}, Alexander Knop 4

9 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. 1 0 Let u = 0 and v = Consider Span {u, v}, it contains all the following vectors: xu + yv Alexander Knop 4

10 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. 1 0 Let u = 0 and v = Consider Span {u, v}, it contains all the following vectors: 1 0 xu + yv = x 0 + y Alexander Knop 4

11 Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,..., v l }. 1 0 Let u = 0 and v = Consider Span {u, v}, it contains all the following vectors: 1 0 x xu + yv = x 0 + y 1 = y Alexander Knop 4

12 Previously On Math 18 Note that, the set Span {a 1,..., a l } consists all the following vectors x 1 a x l a l. Alexander Knop 5

13 Previously On Math 18 Note that, the set Span {a 1,..., a l } consists all the following vectors x 1 a x l a l. REMARK Asking if b Span {a 1,..., a l } is equivalent to asking if an equation x 1 a x l a l = b has a solution. Alexander Knop 5

14 A Product of a Matrix and a Column DEFINITION If A is a n m matrix, with columns a 1,, a m and x R n, then the product of A and x denoted as Ax is Ax = x 1 a x n a n. Alexander Knop 6

15 A Product of a Matrix and a Column DEFINITION If A is a n m matrix, with columns a 1,, a m and x R n, then the product of A and x denoted as Ax is Ax = x 1 a x n a n. EXAMPLE [ ] Alexander Knop 6

16 A Product of a Matrix and a Column DEFINITION If A is a n m matrix, with columns a 1,, a m and x R n, then the product of A and x denoted as Ax is Ax = x 1 a x n a n. EXAMPLE [ ] 4 [ ] = [ ] [ ] [ 42 ] 3 Alexander Knop 6

17 A Product of a Matrix and a Column DEFINITION If A is a n m matrix, with columns a 1,, a m and x R n, then the product of A and x denoted as Ax is Ax = x 1 a x n a n. EXAMPLE [ ] 4 [ ] = [ ] [ ] [ 42 ] = 3 [ ] 4 17 Alexander Knop 6

18 Solutions of Matrix Equations THEOREM If A is a matrix with columns a 1,..., a n R m, then the matrix equation Ax = b has the same solution set as the vector equation x 1 a x n a n = b which in turn has the same solution as a system of linear equations with augmented matrix is [ a1 a 2... a n b ]. Alexander Knop 7

19 Solutions of Matrix Equations [ ] x 1 x 2 x 3 x 4 [ ] 4 = 17 Alexander Knop 8

20 Solutions of Matrix Equations [ ] has the same solution as [ ] [ ] 1 1 x 1 + x x 1 x 2 x 3 x 4 [ ] 4 = 17 [ ] [ ] [ ] x 3 + x 3 4 = 2 17 Alexander Knop 8

21 Solutions of Matrix Equations [ ] has the same solution as [ ] [ ] 1 1 x 1 + x x 1 x 2 x 3 x 4 [ ] 4 = 17 [ ] [ ] [ ] x 3 + x 3 4 = 2 17 which in turn has the same solution as a system of linear equations with augmented matrix is [ ] Alexander Knop 8

22 Existence of Solutions THEOREM The equation Ax = b has a solution iff b is a linear combination of the columns of A. Alexander Knop 9

23 Existence of Solutions THEOREM The equation Ax = b has a solution iff b is a linear combination of the columns of A. Does for all b R m the equation Ax = b has solution? Alexander Knop 9

24 Existence of Solutions Does for all b R m the equation Ax = b has solution? EXAMPLE Let A = and b = b 1 b b 3 Alexander Knop 10

25 Existence of Solutions Does for all b R m the equation Ax = b has solution? EXAMPLE Let A = and b = b 1 b b 3 Let us row reduce the corresponded augmented matrix b b b b 2 + 4b b b 3 + 3b b b 2 + 4b b 3 + 3b (b 2 + 4b 1 ) Alexander Knop 10

26 Existence of Solutions Does for all b R m the equation Ax = b has solution? EXAMPLE Let A = and b = b 1 b b 3 Hence, not for any b there is a solution. Alexander Knop 10

27 Existence of Solutions THEOREM Let A be a m n matrix. Then the following statements are logically equivalent. 1 For each b R m, the equation Ax = b has a solution. 2 Each b R m is a linear combination of the columns of A. 3 The columns of A span R m (Span {a 1,..., a n } = R m where a 1,, a n are columns of A). 4 A has a pivot position in every row. Alexander Knop 11

28 Existence of Solutions PROOF. Let U be an echelon form of A. Alexander Knop 12

29 Existence of Solutions PROOF. Let U be an echelon form of A. Given b we can row reduce [ A b ] [ U d ] for some d R m. Alexander Knop 12

30 Existence of Solutions PROOF. Let U be an echelon form of A. Given b we can row reduce [ A b ] [ U d ] for some d R m. Let us assume property 4 holds. Each row of U contains a pivot point and there can be no pivot in the column d. Alexander Knop 12

31 Existence of Solutions PROOF. Let U be an echelon form of A. Given b we can row reduce [ A b ] [ U d ] for some d R m. Let us assume property 4 holds. Each row of U contains a pivot point and there can be no pivot in the column d. Hence, system Ax = b has a solution for any b. Alexander Knop 12

32 Existence of Solutions PROOF. Let U be an echelon form of A. Given b we can row reduce [ A b ] [ U d ] for some d R m. Let us assume property 4 holds. Each row of U contains a pivot point and there can be no pivot in the column d. Hence, system Ax = b has a solution for any b. If property 4 false, then the last column of U is a zero column. Let d ba a column with 1 in the last entry. Alexander Knop 12

33 Existence of Solutions PROOF. Let U be an echelon form of A. Given b we can row reduce [ A b ] [ U d ] for some d R m. Let us assume property 4 holds. Each row of U contains a pivot point and there can be no pivot in the column d. Hence, system Ax = b has a solution for any b. If property 4 false, then the last column of U is a zero column. Let d ba a column with 1 in the last entry. Then [ U d ] is inconsistent. Alexander Knop 12

34 Existence of Solutions PROOF. Let U be an echelon form of A. Given b we can row reduce [ A b ] [ U d ] for some d R m. Let us assume property 4 holds. Each row of U contains a pivot point and there can be no pivot in the column d. Hence, system Ax = b has a solution for any b. If property 4 false, then the last column of U is a zero column. Let d ba a column with 1 in the last entry. Then [ U d ] is inconsistent. Transform this system back to Ax = b, this system is also inconsistent. Alexander Knop 12