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# Lecture5-6.pdf

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University of Toronto Scarborough

Mathematics

MATB24H3

Sophie Chrysostomou

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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