m235notes11 - Notes 11 Dimension Rank Nullity theorem...

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Unformatted text preview: Notes 11: Dimension, Rank Nullity theorem Lecture Oct , 2011 Definition 1. Let V be a subspace of R n . A basis of V is a subset S of V provided • the set S spans V, and • the set S is independent. Definition 2. Let V be a subspace of R n . The dimension of V is the number of elements in a basis of V. Example 1 . The set { e 1 = (1 , 0) ,e 2 = (0 , 1) } is a basis of R 2 . The dimension of R 2 is 2 . Example 2 . The set { u = (1 , 2) ,v = (- 2 , 3) } is a basis of R 2 . To show this is the case we have to show two things. • Every equation of the form x 1 2 + y- 2 3 = a b in unknows x,y can be solved. This says that our set spans R 2 . • The only relations of the form au + bv = 0 are the ones with a = b = 0 . This says that the set is independent. Example 3 . Let e i denote the element of R n with all zero entries except in the i-th position where a 1 occurs. Then S = { e 1 ,e 2 , ··· ,e n } is a basis of R n . The dimension of R n is n. Example 4 . We show how to find a basis of the kernel of a matrix. We examine our algorithm for finding the kernel of a matrix. For example if we start with the matrix M = 1 1 1 1 1 1 0- 1 2- 2 3 2 1 4 . We find that its RREF is 1 0- 1 2- 2 0 1 2- 1 3 0 0 . From this we see that the kernal is all the linear combinations a 1- 2 1 + b - 2 1 1...
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m235notes11 - Notes 11 Dimension Rank Nullity theorem...

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