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Unformatted text preview: Lecture Note 43: Feb 7, 2007 Dr. Jeff ChakFu WONG Department of Mathematics Chinese University of Hong Kong [email protected] MAT 2310C Linear Algebra and Its Applications Fall, 2007 Produced by Jeff ChakFu WONG 1 C OORDINATES AND C HANGE OF B ASIS COORDINATES AND CHANGE OF BASIS 2 C OORDINATES COORDINATES 3 If V is an ndimensional vector space, we know that V has a basis S with n vectors in it; so far we have not paid much attention to the order of the vectors in S . However, in the discussion of this lecture we speak of an ordered basis S = { v 1 , v 2 ,..., v n } for V ; thus S = { v 2 , v 1 ,..., v n } is a different ordered basis for V . Ordered basis: A set of vectors S = { v 1 , v 2 , ··· , v n } in a vector space V is called an ordered basis for V provided S is a basis for V and if we reorder the vectors in S , the new ordering of the vectors in S is considered a different basis for V . COORDINATES 4 If S = { v 1 , v 2 ,..., v n } is an ordered basis for the ndimensional vector space V , then every vector v in V can be uniquely expressed in the form v = c 1 v 1 + c 2 v 2 + ··· + c n v n , where c 1 ,c 2 ,...,c n are real numbers. We shall refer to [ v ] S = c 1 c 2 . . . c n as the coordinate vector of v with respect to the ordered basis S . The entries of [ v ] S are called the coordinate of v with respect to S . COORDINATES 5 Observe that the coordinate vector [ v ] S depends on the order in which the vectors in S are listed; a change in the order of this listing may change the coordinates of v with respect to S . All bases considered in this section are assumed to be ordered bases. COORDINATES 6 Example 1 Let S = { v 1 , v 2 , v 3 , v 4 } be a basis for R 4 , where v 1 = (1 , 1 , , 0) , v 2 = (2 , , 1 , 0) , v 3 = (0 , 1 , 2 , 1) , v 4 = (0 , 1 , 1 , 0) . If v = (1 , 2 , 6 , 2) , compute [ v ] S . Solution To find [ v ] S we need to compute constants c 1 ,c 2 ,c 3 and c 4 such that c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = v , which is just a linear combination problem. The previous equation leads to the linear system whose augmented matrix is (verify) 1 2 1 1 1 1 2 1 2 1 6 1 2 , (1) or equivalently, [ v T 1 v T 2 v T 3 v T 4  v T ] . COORDINATES 7 Transforming the matrix in (1) to reduced row echelon form, we obtain the solution (verify) c 1 = 3 , c 2 = 1 , c 3 = 2 , c 4 = 1 , so the coordinate vector of v with respect to the basis S is [ v ] S = 3 1 2 1 . COORDINATES 8 Example 2 Let S = { e 1 , e 2 , e 3 } be the natural basis for R 3 and let v = (2 , 1 , 3) ....
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 Spring '11
 JeffWong
 Linear Algebra, Algebra, Vector Space, transition matrices, PS ←T

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