College Algebra Exam Review 366

# College Algebra Exam Review 366 - 376 8 MODULES Recall that...

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Unformatted text preview: 376 8. MODULES Recall that any two bases of a ﬁnite dimensional vector space are related by an invertible change of basis matrix. The same is true for bases of free modules over a commutative ring with identity. Suppose that fv1 ; : : : ; vn g is a basis of the free module F . Let .w1 ; : : : ; wn / be another sequence of n module elements. Each wj has a unique expression as an R–linear combination of the basis elements vi , X wj D ci;j vi : i Let C denote the matrix C D .ci;j /. We can write the n equations above as the single matrix equation: Œv1 ; : : : ; vn  C D Œw1 ; : : : ; wm : (8.4.3) Lemma 8.4.11. In the situation described above, fw1 ; : : : ; wn g is a basis of F if, and only if, C is invertible in Matn .R/. Proof. If fw1 ; : : : ; wn g is a basis, we can also write each vj as an R–linear combinations of the wi ’s, X di;j wi : vj D i Let D denote the matrix D D .di;j /. We can write the n previous equations as the single matrix equation: Œw1 ; : : : ; wm  D D Œv1 ; : : : ; vn : (8.4.4) Combining (8.4.4) and (8.4.3), we obtain Œv1 ; : : : ; vn  D Œv1 ; : : : ; vn  CD: Using the linear independence of fv1 ; : : : ; vn g, as in the proof of Lemma 8.4.1, we conclude that CD D En , the n–by–n identity matrix. Thus C has a right inverse in Matn .R/. It follows from this that det.C / is a unit in R, and, therefore, C is invertible in Matn .R/, by Corollary 8.3.16. Conversely, suppose that C is invertible in Matn .R/ with inverse C 1 . Then Œv1 ; : : : ; vn  D Œw1 ; : : : ; wn C 1 : This shows that fv1 ; : : : ; vn g is contained in the R–span of fw1 ; : : : ; wn g, so the latter set spansP . F Finally, suppose j ˛j wj D 0. Then 23 23 ˛1 ˛1 6:7 : 5 D Œv1 ; : : : ; vn  C 6 : 7 : 0 D Œw1 ; : : : ; wn  4 : 4:5 : ˛n ˛n ...
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