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MATB24H3 (13)
Lecture

# Lecture5-6.pdf

7 Pages
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School
University of Toronto Scarborough
Department
Mathematics
Course
MATB24H3
Professor
Sophie Chrysostomou
Semester
Winter

Description
University of Toronto at Scarborough Department of Computer & Mathematical Sciences MAT B24S Fall 2011 Lectures 5-6 1 Coordinatization of Vectors Deﬁnition: Let V be a vector space over the ﬁeld F. B = (b ,b ,▯▯▯ ,1 ) 2 n denotes an ordered basis of n vectors in V, if : • B is a basis of V. • the order in which the vectors appear in B is ﬁxed. Now if v ∈ V, and v = r 1 +1r b 2 ▯2▯ + r b n n is the unique expression of v as a linear combination of vectors from B, then the vector v = [r ,r ,▯▯▯ ,r ] is the coordinate vector of v relative to B 1 2 n B. Example : Let B = (2,(x − 2),(x − 2) ,(x − 2) ) be an ordered basis of 3 2 the vector space P o3er the ﬁeld R. Let p(x) = x + 5x − x + 4. Find the coordinate vector of p relative to B. solution. Since B is a basis for P , th3n there are r ,r ,r 1r 2 R 3uc4 that: p(x) = r (2) + r (x − 2) + r (x − 2) + r (x − 2) 3 1 2 3 4 = 2r 1 r (x2− 2) + r (x 3 4x + 4) + r (x − 4x + 12x − 8) or: 3 2 3 2 x +5x −x+4 = r x +(4 −6r )x3+(r 44r +12r 2x+(2r3−2r +44 −8r ) 1 2 3 4 Thus we have: 1 2r −2r +4r −8r = 4 1 2 3 4 2 −4r 3 +12r 4 = −1 r3 −6r 4 = 5 r4 = 1     2 −2 4 −8 ▯ 4 1 −1 2 −4 ▯ 2 0 1 −4 12 ▯−1  R1() 0 1 −4 12 ▯ −1  ▯  =⇒ 2  ▯  0 0 1 −6 ▯ 5  0 0 1 −6 ▯ 5  0 0 0 1 1 0 0 0 1 1     1 −1 2 0 ▯ 6 1 −1 0 0 ▯−16  0 1 −4 0 ▯−13  0 1 −4 0 ▯−13  −−−−−→4  ▯  −−−−→3  ▯  R2−124  0 0 1 0 ▯ 11  0 0 1 0 ▯ 11  R3+6R4 0 0 0 1 1 0 0 0 1 1     1 −1 0 0 ▯ −16 1 0 0 0 ▯15 R +4R 0 1 0 0 ▯ 31 R +R  0 1 0 0 ▯31 −−−−−→3  ▯  −−−−→2  ▯  0 0 1 0 ▯ 11  0 0 1 0 ▯11 0 0 0 1 1 0 0 0 1 1 The above implies: p(x) = 15(2) + 31(x − 2) + 11(x − 2) + 1(x − 2) and (p(x))B= [15,31,11,1] Alternatively we may use the Taylor polynomial of p(x) about a = 2 to ﬁnd these coeﬃcients. Note: 3 2 3 2 p(x) = x + 5x − x + 4 therefore p(2) = 2 + 5(2 ) − 2 + 4 = 30 p (x) = 3(x ) + 10(x) − 1 therefore p (2) = 3(2 ) + 10(2) − 1 = 31 ′′ ′′ p (x) = 6(x) + 10 therefore p (2) = 6(2) + 10 = 22 p (x) = 6 therefore p (2) = 6. Thus, using Taylor’s formula, since p is a polynomial of degree 3: p (2) p (2) p(x) = p(2) + p (2)(x − 2) + (x − 2) + (x − 2)3 2! 3! therefore 22 2 6 3 2 3 p(x) = 30+31(x−2)+ (x−2) + (x−2) = 15▯2+31(x−2)+11(x−2) +(x−2) . 2! 3! 2 and (p(x)) B [15,31,11,1] Note: that the polynomial has now been represented by a vector ∈ R . 4 In fact every polynomial from P can be3represented by a vector ∈ R and 4 4 every vector ∈ R can be represented by a polynomial from P . Thus we ma3 say that P a3d R are isomorphic. ie. there a one to one and onto mapping between the two vector spaces. 2 Linear Transformations Deﬁnition: A function T that maps a vector space V over F into a vector ′ space V over the ﬁeld F is a linear transformation if ∀ v,u ∈ V and ∀ r ∈ F, the following are satisﬁed: 1. T(u + v) = T(u) + T(v) 2. T(rv) = rT(u) Example : Let F be the set of all functions of R. Let D be the set of all functions twice diﬀerentiable on R. Let T : D → F be deﬁned by T(f) = f + f . ′′ Is T a linear transformation? solution. Let f,g ∈ D and r ∈ R. Then: 1. ′ ′′ T(f + g) = (f + g) + (f + g) = f + g + f + g ′′ ′′ ′ ′′ ′ ′′ = f + f + g + g = T(f) + T(g) 2. ′ ′′ T(rf) = (rf) + (rf) = rf + rf = r(f + f ) ′ ′′ = rT(f) 3 T satisﬁes the properties of a linear transformation. Therefore it is a linear transformation. Example : Let T : F −→ R be deﬁned by T(f) = 1 ∀ f ∈ F. Is T a linear transformation? solution. : Let f,g ∈ F. Then T(f + g) = 1 ▯= 2 = T(f) + T(g). Thus T is not a linear transformation. Deﬁnition: Let T : V −→ V be a linear transformation.Then: 1. V is the domain of T. ′ 2. V is the codomain of T. 3. If W ⊂ V then: ▯ • the image of W under T is T[W] = {T(w) w ∈ W}. ▯
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