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# Big Thm2

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• 1.6D
• ### Big Thm2

1. 1. THE BIG THEOREM PART II
2. 2. EQUIVALENT STATEMENTS <ul><li>A is invertible </li></ul><ul><li>AX=0 has only the trivial solution </li></ul><ul><li>The reduced-row echelon for of A is I n </li></ul><ul><li>A is expressible as a product of elementary matrices </li></ul>We now add the following: e) AX=B is consistent f) AX=B has exactly one solution
3. 3. A X = B A -1 A -1 X = A -1 B If A is invertible then AX=B has exactly one solution. Proof (a  f) If A is invertible then A -1 exists. Then the following is true: I X = A -1 B (A -1 A) X = A -1 B <ul><li>A -1 B is clearly one solution. Is there another? </li></ul>
4. 4. If A is invertible then AX=B has exactly one solution. Proof (a  f) Let X 1 = A -1 B and assume X 2 is a second solution s.t. X 1  X 2 . Then both satisfy the equation AX=B so… AX 1 =B and AX 2 =B AX 1 - AX 2 = B – B = 0 A X 1 - A X 2 = 0 A (X 1 - X 2 ) = 0 A -1 A -1 I (X 1 - X 2 ) = 0 X 1 - X 2 = 0 This contradicts our assumption that X 1  X 2  X = A -1 B is the only solution.   X 1 = X 2
5. 5. Proof (f  e) If AX=B has exactly one solution then AX=B is consistent. If AX=B has exactly one solution, then by definition , the system is consistent. 
6. 6. Proof (e  a) If AX=B is consistent then A is invertible. If AX=B is consistent for every n x 1 matrix B, then it is consistent for each of the following choices of B:  B 1  B 2  B 3  B n
7. 7. Proof (e  a) - continued If AX=B is consistent then A is invertible. Let X 1 , X 2 , …, X n represent the respective solutions to each of the previous equations so that: Let C be the matrix formed by adjoining the columns of X 1 , X 2 , …, X n so that:
8. 8. Proof (e  a) - continued If AX=B is consistent then A is invertible.
9. 9. Proof (e  a) - continued Thus the matrix C exists s.t. AC = I. This implies that C = A -1 .  A is invertible  If AX=B is consistent then A is invertible. q.e.d.