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Unformatted text preview: LECTURE 1 I. Column space, nullspace, solutions to the Basic Problem Let A be a m n matrix, and ~ y a vector in R m . Recall the fact: Theorem: A~x = ~ y has a solution if and only if ~ y is in the column space R ( A ) of A . Now lets prove a new theorem which talks about when such a solution is unique. Suppose ~x,~x are two solutions to A~x = ~ y . Then A ( ~x ~x ) = A~x A~x = ~ y ~ y = ~ . So ~x ~x is in the nullspace N ( A ) of A . In fact, we can say more. Theorem: Let ~x be a solution to A~x = ~ y . Then the set of all solutions is the set of vectors which can be written ~x + ~n for some ~n N ( A ). Proof omittedits not much more than the above argument. We also get the corollary: Theorem: Suppose ~ y is in the column space of A . A~x = ~ y has a unique solution if and only if N ( A ) = ~ 0. Ex: Take V = 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 . Now N ( V ) is the set ~x : x 1 + x 2 + x 3 = 0 R ( V ) is the set span { 1 / 3 1 / 3 1 / 3 } . In particular, the equation V ~x = ~ y , if it has a solution, will never have a unique one. One minute contemplation: in view of the above theorem, how do we think of the set of solutions of A~x = 1 1 1 ? 1 By contrast, take D = 2 3 5 . Then ask what people think the nullspace and column space are. N ( D ) is the zero space; R ( D ) = R 3 . So, for every ~ y , the equation D~x = ~ y has a unique solution (which we already knew, because D is invertible.) II. Rank Lets understand the averaging example a little more closely. If we try to solve 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 1 / 3 x 1 x 2 x 3 = y 1 y 2 y 3...
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 Winter '09
 IgorDolgachev
 Linear Algebra, Algebra

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