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Lecture

Lecture 23.pdf

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Department
Mathematics
Course Code
MATH 136
Professor
Robert Sproule

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Wednesday, March 5 − Lecture 23 : Coordinates of a vector with respect to a basis. Concepts: 1. Recall: When expressing a vector v in V as a linear combination of the vectors in a basis, the coefficients are unique. 2. Define and find the coordinates of v with respect to a given basis B of a vector space V. 23.1 Recall − The unique representation theorem. If V is a vector space and B = { v ,1v ,2...., vk}forms a basis for V then for every v in V there is a unique set of coefficients α , 1 ..2, α in k such that v = α v + α v1 1.... 2 2 v . k k 23.2 Definition − If v is a vector in the vector space V which has an ordered basis B = {v ,1v ,2...., v k Then the unique coefficients α , α .1., 2,used tokexpress v as a linear combination of B = {v ,1v ,2...., vk} are called the coordinates of v with respect to the basis B. We will represent thesek unique scalars α , α ...., α by [v] , i.e., 1 2, k B [v] B (α , 1 ..2,, α ).k Since is expressed as a k-tuple, the order of α’s musi respect the order in which the elements of the basis B={v 1, v2,..., vk}are given. We refer to B as an ordered basis. 23.2.1 Examples a) Suppose S = {e , e 1 is2the standard basis of V = ℝ . Consider v = (−1, 7). Since (−1, 7) = −1e + 71 , the2 the unique coordinates [v] of v with Sespect to the basis S are [v]S= [(−1,7)] = S−1, 7). The coordinates of v with respect to the standard basis S of areℝ are called the standard coordinates. Notice that [v] S = v. Equality [v] =Sv holds true only when the vector space V = ℝ . n 2 b) Suppose we have another basis B = {u , u } for 1 , 2here u = (1, 2) and 1 u 2 = (3, 5). Find the unique coordinates [v] of v B (−1,7) with respect to the basis B. Solution: We solve for since [v]B= (α, β). Solving the matrix equation is equivalent to solving for or to obtain as unique solution α= 26 and β= −9. Notation – We will denote the matrix [u , u1] 2s P .S B (We will formally define tS Bmatrix P below.) The matrix P = [u , u ] is seen as mapping theB-coordinates [(−1, 7)] = (26, −9) S B 1 2 B with respect to the basis B to [(−1, 7)]S= (−1, 7), the coordinates of (−1, 7) with respect to the standard basisS. That is, PS[(B1, 7)] = B(−1, 7)] S. We found [(−1, 7)] =B(26, −9). 23.3 We now show how coordinates with respect to a basis are computed in three types of vectors spaces: m 1) ℝ 2) the vector space M m×n of all m by n matrices. 3) the vector space P on all polynomials of degree at most n. 23.3.1 Computation of coordinates of a vector with respect to a basis in ℝ m . Suppose B = {v , v , ...., v } is an ordered basis of ℝ . Suppose that v is a vector of m 1 2 m ℝ for which we would like to determine its coordinates [v] = (α B α ..1., 2,) withm respect to the ordered basis B. To do this we must solve for the α’s in the system α 1 1 α 2 2 .... + α m m v Since the v’i are m-tuples solving for the α’siis the same thing as find a solution set for this system [ v1v 2... v |mv]. In this case finding coordinates of v with respect to the ordered basis B is done by applying the Gauss-Jordan elimination algorithm. m 23.3.1.1 Definition − Suppose B = {v , v1, .2., v }mis an ordered basis of ℝ . The matrix whose columns are the vectors of the basisB = {v , v ,1...2, v } ms called coordinate matrix with respect to the ordered basis B. We will denote it by SP B [v v1...2 v ] m The columns of the matrix are the coordinates [v] = vi Sth iespect to the standard basis S. The matrix P Ss Bee as mapping [v] to [vB = v. TSat is, P [v] = Sv]B. B S 23.3.1.2 Example − Consider the basis B = {(1, 0, 0), (3, 1, 0), (1, 1, 1)} of ℝ . 3 a) Find the coordinates of u =1(3, 4, 5) and of u = 22, 9, 5) with respect to the basis B. b) Find the coordinates of e ,1e ,2e 3ith respect to the basis B. Solution: a) We first construct the coordinates matrix P .S Be matrix P is theBmatrix whose columns are
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