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# Lecture7-8.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 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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