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College Algebra Exam Review 159

College Algebra Exam Review 159 - S into a vector space W...

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3.3. VECTOR SPACES 169 V over K , we have a linear isomorphism S B W V ! K n given by S B W X i ˛ i v i 7! X i ˛ i O e i D 2 6 6 6 4 ˛ 1 ˛ 2 : : : ˛ n 3 7 7 7 5 ; where . O e 1 ; : : : ; O e n / is the standard ordered basis of K n . S B .v/ is called the coordinate vector of v with respect to B . Proposition 3.3.31. Any two n –dimensional vector spaces over K are lin- early isomorphic. Proof. The case n D 0 is left to the reader. For n 1 , any two n dimensional vector spaces over K are each isomorphic to K n , and hence isomorphic to each other. n This proposition reveals that (finite–dimensional) vector spaces are not very interesting, as they are completely classified by their dimension. That is why the actual subject of finite–dimensional linear algebra is not vector spaces but rather linear maps, which have more interesting structure than vector spaces themselves. Proposition 3.3.32. (The universal property of bases.) Let V be a vector space over K and let S be a basis of V . Then any function f W S ! W from S into a vector space
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Unformatted text preview: S into a vector space W extends uniquely to a linear map T W V ! W . Proof. We will assume that S D f v 1 ;:::;v n g is ﬁnite, in order to simplify the notation, although the result is equally valid if S is inﬁnite. Let f W S ! W be a function. Any element v 2 V has a unique expression as a linear combination of elements of S , v D P i ˛ i v i . There is only one possible way to deﬁne T.v/ , namely T.v/ D P i ˛ i f.v i / . It is then straightforward to check that T is linear. n Direct sums and complements The (external) direct sum of several vectors spaces V 1 , V 2 , . . . , V n over a ﬁeld K is the Cartesian product V 1 ² V 2 ²³³³² V n with component–by– component operations: .a 1 ;a 2 ;:::;a n / C .b 1 ;b 2 ;:::;b n / D .a 1 C b 1 ;a 2 C b 2 ;:::;a n C b n //...
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