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College Algebra Exam Review 360

College Algebra Exam Review 360 - n It is possible for a...

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370 8. MODULES Since f w 1 ;:::;w m g spans M , we can also write each v j as an R –linear combinations of the elements w i , v j D b 1;j w 1 C b 2;j w 2 C ±±± C b n;j w n : Let B denote the m –by– n matrix B D .b i;j / . The n relations above can be written as a single matrix equation: Œw 1 ;:::;w m ŁB D Œv 1 ;:::;v n Ł: (8.4.2) Combining ( 8.4.1 ) and ( 8.4.2 ), we have Œv 1 ;:::;v n ŁAB D Œw 1 ;:::;w m ŁB D Œv 1 ;:::;v n Ł; or Œv 1 ;:::;v n Ł.AB ² E n / D 0; where E n denotes the n –by– n identity matrix. Because of the linear inde- pendence of the v j , we must have AB D E n . Now, if m < n , we augment A by appending n ² m columns of zeros to obtain an n –by– n matrix A 0 . Likewise, we augment B by adding n ² m rows of zeros to obtain an n by– n matrix B 0 . We have A 0 B 0 D AB D E n . Taking determinants, we obtain 1 D det .E n / D det .A 0 B/ D det .A 0 / det .B 0 / . But det .A 0 / D 0 , since the matrix A 0 has a column of zeros. This contradiction shows that m ³ n . In particular, any two basis have the same cardinality.
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Unformatted text preview: n It is possible for a free module over an non-commutative ring with identity to have two bases of different cardinalities. See Exercise 8.4.4 . Definition 8.4.2. Let R be a commutative ring with identity element. The rank of a finitely generated free R –module is the cardinality of any basis. Remark 8.4.3. The zero module over R is free of rank zero. The empty set is a basis. This is not just a convention; it follows from the definitions. For the rest of this section, R denotes a principal ideal domain. Lemma 8.4.4. Let F be a free module of finite rank n over a principal idea domain R . Any submodule of F has a generating set with no more than n elements...
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