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MATB24H3 (13)
Lecture

# Lecture7-8.pdf

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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 7-8 Theorem : Bases and Linear Transformations Let V and V ′ be vector spaces over R. Let T : V −→ V ′ be a linear transformation. Let B be a basis for V . Suppose that T(b) is given ∀ b ∈ B. Then for any v ∈ V , T(v) is uniquely determined by the vectors T(b) where b ∈ B. proof. To say that T(v) is uniquely determined by the vectors T(b) for all v ∈ V it is equivalent to saying that if T is a linear transformation and T(b) = T(b) ∀ b ∈ B then, T = T ¯ Let v ∈ V . Since B is a basis of V . Then there are b ,b 1▯▯▯2,b ∈ B k and r ,r ,▯▯▯ ,r ∈ F such that 1 2 k v = r 1 1 r b2+ 2▯▯ + r b k k Therefore: T(v) = T(r b + 1 b1+ ▯▯2 2 r b ) k k = r T1b ) 1 r T(2 ) +2▯▯▯ + r T(bk) k = r T1b ) 1 r T(2 ) +2▯▯▯ + r T(bk)¯ k = T(r b1+ 1 b +2▯2▯ + r b ) k k = T(v) ¯ Thus T = T The above theorem says that we only need to know the images of all vectors in a basis to determine what the linear transformation is. 1 Example : Let {2,(x − 2),(x − 2) ,(x − 2) } be a basis of P and suppose 3 that T : P 3→ P is 3 linear transformation with: 2 1) T(2) = 0, 2)T(x − 2) = 1, 3)T((x − 2) ) = 2x − 2 3 2 4)T((x − 2) ) = 3x − 6x Find T(p(x)) for all polynomials p(x) ∈ P . 3 Solution: T(1) = .5T(2) = 0 T(x) = T(x − 2) + T(2) = 1 + 0 = 1 T(x ) = T((x − 2) + 4(x − 2) + 2(2)) = 2x − 2 + 4(1) + 0 = 2x + 2 3 3 2 T(x ) = T((x − 2) + 6(x − 2) + 12(x − 2) + 4(2)) 2 2 = 3x − 6x + 6(2x − 2) + 12(1) + 4(0) = 3x + 6x Thus: T(ax + bx + cx + d) = a(3x + 6x) + b(2x + 2) + c 2 Theorem : Suppose: ′ a) V and V are ﬁnite dimensional vector spaces. ′ b) T : V −→ V is a linear transformation. c) B = (b ,b ,▯▯▯ ,b ) and B = (b ,b ,▯▯▯ ,b ) are ordered bases for 1 2 n 1 2 m V and V respectively. n m d) T : R −→ R is the linear transformation T(v ) = (T(B)) B ′ ∀ v ∈ V. Then ¯ i) The standard matrix representation of T is the mxn matrix [T] B,B′with jth column the column (T(b )) ′. j B ii) Furthermore, (T(v)) B′ = [T] B,B′(v B for all v ∈ V . iii) The matrix [T] B,B′ in this theorem is called the matrix representation ′ of T relative to the bases B, B . iv) If the dim(ker(T)) = nullity of T and dim(range(T)) = rank of T then the rank equation is nullity of T + rank of T = dim(V ) Example : T : P −→ P 3y T(p) 2 p + p . ′ ′′ Let B = (2,(x − 2),(x − 2) ,(x − 2) ) be an ordered basis of P . 3 B = (1,x,x ) be an ordered basis of P . 2 Find [T] B,B′, its kernel and its range. Then: 3 T(2) = 0 = 0(1) + 0(x) + 0(x ) thus (T(2))B′ = [0,0,0] T(x − 2) = 1 − 0 2 = 1(1) + 0(x) + 0(x )) thus (T(x − 2))B′= [1.0,0] 2 T((x − 2) ) = 2(x − 2) + 2 = 2x − 2 = −2(1) + 2(x) + 0(x ) thus (T(x − 2) )B′= [−2,2,0] T((x − 2) ) = 3(x − 2) + 6(x − 2) 2 = 3(x − 4x + 4) + 6x − 12 = 3x − 6x 2 3 = 0(1) − 6(x) + 3(x ) thus (T(x − 2) )B′= [0,−6,3]   0 1 −2 0   so [TB,B′= 0 0 2 −6 0 0 0 3 We are familiar with ﬁnding the range and kernel of T using its matrix representation [T] . We simply ﬁnd the column space ,and the nullspace B,B of this matrix.     0 1 −2 0 1 0 1 −2 0  0 0 2 −6  −−−→ 0 0 1 −3  1R 0 0 0 3 3 3 0 0 0 1     0 1 −2 0 0 1 0 0 −−−−−→1 0 0 1 0 −−−−R→ 0 0 1 0  0 0 0 1 0 0 0 1 ¯ The column space of [T]B,B= sp([1,0,0],[−2,2,0],[0,−6,3]) = range(T). Thus, range(T) = sp(1,−2 + 2x,−6x + 3x ) = sp(1,x,x ) since rank T = 3 = dim(P2). Similarly: nullspace of [B,B = {[a,0,0,0]| a ∈ R} thus ker(T) = sp([1,0,0,0]) and therefore ker(T) = sp(2) the constant polynomials. The 4 nullity of T is 1. Thus we have rank of T + nullity of T = 4 = dim3P ) Remember that if the range is the codomain then the transformation is onto. Similarly, T is 1-1 if and only if the kerT = {0} . Thus, the above transformation is onto but not one to one. Example : Let T : P4−→ M 2x2R) by ▯ ▯ 2 3 4 a + c b + d T(a + bx + cx + dx + ex ) = a + c a + b + d 2 3 4 We use the standard basis α = {1,x,x ,x ,x }f4r P and ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ 1 0 0 1
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