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Lecture

Lecture 24.pdf

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Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
Friday, March 7 − Lecture 24 : Change of coordinates matrix . Concepts: 1. Computing the matrix which transforms the coordinates of a vector with respect to a basis to coordinates of the same vector with respect to another basis. 24.1 Theorem – Let B = {b , b ,1…, 2 } and Cm= {c , c , …, c1} b2 two bames of an abstract vector space Vof dimension m. Let x ∈ V. Then there exists a unique m by m matrices PCanB P sucB Chat CPB[x] B [x] C BPC[x] C [x] B The matrices CP Bnd P aBe Cs follows Furthermore, P P = I. C BB C Proof outline: Let B = {b 1, b2, …, b }mand C = {c , c 1 …2 c } be mwo bases of an abstract vector space Vof dimension m. Suppose x belongs to V. Recall that T(x) = [x] is B linear mapping and so [ax + by] = a[xB + b[y] B B. (See 23.4 in the previous lecture) So P = [ [b ] [b ] [b ] [b ] ] maps [x] to [x] . C B 1 C 2 C 3 C ... m C B C Similarly, PB= C [c ] [1 B [c 2B 3 B ...cm]B] maps [x] to Cx] . B Remark – Note that P = C [B ] [b ] 1C ] 2 C 3 C ...[bm]C] is a generalization of the matrix SP Biscussed in the previous lecture. 24.2 Example – Suppose we are given two bases B = {b , b } and C = {c , c } for an 1 2 1 2 abstract vector space V. Suppose that Suppose x is some vector in V such that x = 3b + b . Th1t is,2suppose Find [x] C . Solution: Method 1: Method 2: We modify the question : Suppose, on the other hand, that [x] = (6, 4) iC known. That is, suppose we know that x = 6c + 4c .1Find [2] . B Solution: Alternatively: By solving for the systems [ [b ] [b ]1|C[c ]2]Cand 1 Cb ] [b ] | [1 C ] )2 C 2 C we deduce [c ] 1 C1/10, –1/10) and [c ] = (325B 2/5) or c1= (1/10) b + 1–1/10) b 2 c = (3/5) b + (2/5) b . 2 1 2 We then compute 24.2.1 Example – Consider the basis B = {(1, 0, 0), (3, 1, 0), (1, 1, 1)} of ℝ 3, and B denote the standard basis S = { e , e 1 e2}. 3 a) Find the matrix P wSicB converts B-coordinates to S-coordinates. 3 b) Suppose u an1 u are 2wo vectors in ℝ for which it is known that [u ] = (1, −1, 1)B and [u ]2=B(−15, 4, 5). That is, u1= 1(1, 0, 0) + –1(3, 1, 0) + 5(1, 1, 1) u = –15(1, 0, 0) + 4(3, 1, 0) + 5(1, 1, 1) 2 Use the matrix P to determine what the vectorsu and u are (with respect to the S B
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