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Unformatted text preview: (Partial) Solutions to Homework 4 Problem 10.1: Let M be a square matrix. Explain why the following statements are equivalent: i. MX = V has a unique solution for every column vector V . ii. M is nonsingular. (Show that ( i ) ⇒ ( ii ) and ( ii ) ⇒ ( i ).) Answer: As suggested, we will prove this in two steps, first showing that ( i ) implies ( ii ) and then that ( ii ) implies ( i ). ( i ) = ⇒ ( ii ): Suppose that ( i ) is true; i.e., MX = V has a unique solution for every column vector V (and remember, for this problem, we can’t assume the results of later sections...so this is all we can assume is true, plus whatever theorems were covered in Lecture 10). Then since MX = V has a unique solu tion for every V , MX = 0 has a unique solution. By Theorem 10.1, M must be invertible. (There are many other ways to prove this; this is just one way. If you have a different proof and are wondering if it’s correct, ask me.) ( ii ) = ⇒ ( i ): Assume M is nonsingular. Then there exists an inverse of M , M 1 . Now, let V be any column vector. Then M 1 V is a solution to MX = V , since MM 1 V = IV = V . Morever, M 1 V is the only solu tion to MX = V , since if Y is any other solution, we must have MY = V , so M 1 MY = M 1 V , which implies Y = M 1 V . Therefore, there exists a unique solution to MX = V , for any V . (All the steps in here are necessary. Try to understand the general structure of the proof. In order to show ( ii ) = ⇒ ( i ), we start by assuming that (ii) is true. Using only this assumption, we can then show that (i) must be true. In order to show that (i) is true, we have to let V be an arbitrary vector, and then show two things: (a) there is a solution to MX = V and (b) there is only one solution to MX = V . This is what the above proof does.)....
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This note was uploaded on 03/23/2010 for the course MAT 022A taught by Professor Pon during the Spring '10 term at UC Davis.
 Spring '10
 Pon

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