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Let
T and
U be the linear operators on
R2 defined by
T(x1,x2)=(x2,x1)andU(x1,x2)=(x1,0)
( a ) How would you describe
T and
U geometrically?
( b ) Give rules like the ones definning
T and
U for each of the transformations
(U+T),UT,TU,T2,U2.
Solution:
( a )
T is a symmetric transformation with respect to the line
y=x,
U is the projection on
x-axis.
( b ) We have
(U+T)(x1,x2)UT(x1,x2)TU(x1,x2)T2(x1,x2)U2(x1,x2)=(x1+x2,x1)=(x2,0)=(0,x1)=(x1,x2)=(x1,0)
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Let
T be the (unique) linear operator on
C3 for which
Tϵ1=(1,0,i),Tϵ2=(0,1,1),Tϵ3=(i,1,0).
Is
T invertible?
Solution: No, since
T(ϵ3−iϵ1)=(i,1,0)−(i,0,−1)=(0,1,1)=Tϵ2, thus
T(ϵ3−iϵ1−ϵ2)=0
but
ϵ3−iϵ1−ϵ2=(−i,−1,1)=0.
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Let
T be the linear operator on
R3 defined by
T(x1,x2,x3)=(3x1,x1−x2,2x1+x2+x3)
Is
T invertible? If so, find a rule for
T−1 like the one which defines
T.
Solution:
T is invertible, the rule for
T−1 can be described as
T−1(x1,x2,x3)=(31x1,31x1−x2,−x1+x2+x3)
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For the liear operator
T of Exercise 3, prove that
(T2−I)(T−3I)=0.
Solution: Since
(T−3I)(x1,x2,x3)=T(x1,x2,x3)−3(x1,x2,x3)=(0,x1−4x2,2x1+x2−2x3)
and
(T2−I)(x1,x2,x3)=T2(x1,x2,x3)−(x1,x2,x3)=T(3x1,x1−x2,2x1+x2+x3)−(x1,x2,x3)=(9x1,3x1−(x1−x2),6x1+(x1−x2)+(2x1+x2+x3))−(x1,x2,x3)=(9x1,2x1+x2,9x1+x3)−(x1,x2,x3)=(8x1,2x1,9x1)
So
(T2−I)(T−3I)(x1,x2,x3)=(T2−I)(0,x1−4x2,2x1+x2−2x3)=(0,0,0)
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Let
C2×2 be the complex vector space of
2×2 matrices with complex entries. Let
B=[1−4−14]
and let
T be the linear operator on
C2×2 defined by
T(A)=BA. What is the rank of
T? Can you describe
T2?
Solution: If
A=[acbd], then
T(A)=BA=[1−4−14][acbd]=[a−c4c−4ab−d4d−4b]
the rank of
T is 2. Also we have
T2(A)=T(T(A))=[1−4−14][a−c4c−4ab−d4d−4b]=[5a−5c20c−20a5b−5d20d−20b].
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Let
T be a linear transformation from
R3 into
R2, and let
U be a linear transformation from
R2 into
R3. Prove that the transformation
UT is not invertible. Generalize the theorem.
Solution: Let
a=Tϵ1,b=Tϵ2,c=Tϵ3, then
a,b,c∈R2, thus are linearly dependent, without loss of generality, we suppose
c=k1a+k2b,k1,k2∈R, then
k1a+k2b−c=0, or
k1Tϵ1+k2Tϵ2−Tϵ3=T(k1ϵ1+k2ϵ2−ϵ3)=0
Notice
k1ϵ1+k2ϵ2−ϵ3=0, we can have
UT(k1ϵ1+k2ϵ2−ϵ3)=U(0)=0
thus
UT is not invertible.
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Find two linear operators
T and
U on
R2 such that
TU=0 but
UT=0.
Solution:
Let
U(x1,x2)=(x1,0),T(x1,x2)=(x2,0), then
TU(x1,x2)=T(x1,0)=(0,0),∀x1,x2∈RUT(0,1)=U(1,0)=(1,0)
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Let
V be a vector space over the field
F and
T a linear operator on
V. If
T2=0, what can you say about the relation of the range of
T to the null space of
T? Give an example of a linear operator
T on
R2 such that
T2=0 but
T=0.
Solution: If
T2=0, then for any
α∈range T, we have
α=Tβ,β∈V, thus
Tα=T2β=0, so
α also belongs to the null space of
T, it follows that
range T is a subspace of the null space of
T.
T(x1,x2)=(x2,0) is an example of
T2=0, while
T=0.
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Let
T be a linear operator on the finite-dimensional space
V. Suppose there is a linear operator
U on
V such that
TU=I. Prove that
T is invertible and
U=T−1. Give an example which shows that this is false when
V is not finite dimensional.
Solution: For any
α∈V, we have
α=(TU)(α)=T(Uα), this means
α∈range T, so the range of
T is
V, by Theorem 9,
T is invertible. To see
U=T−1, notice that
T−1T=I and
TU=I, thus
U=IU=(T−1T)U=T−1(TU)=T−1I=T−1
If
T=D on the space of polynomial functions, then let
U be the integrable operator on the space of polynomial functions, we have
DU=I, but
D is not invertible.
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Let
A be an
m×n with entries in
F and let
T be the linear transformation from
Fn×1 into
Fm×1 defined by
T(X)=AX. Show that if
m<n it may happen that
T is onto without being non-singular. Similarly, show that if
m>n we may have
T non-singular but not onto.
Solution: If
m<n, then the system
AX=b can have a solution if rank
A=m, so
T can be onto, but
AX=0 must have an nonzero solution, since the solution space has dimension
n−m≥1.
If
m>n, then let
AX=x1A1+⋯+xnAn, each
Ai a
m×1 vector, as long as those
Ai are linearly independent,
AX=0 will mean
xi=0 and
X=0, so
T is non-singular, but
dimFm×1=m and
A1,…,An cannot span
Fm×1, thus
T is not onto.
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Let
V be a finite-dimensional vector space and let
T be a linear operator on
V. Suppose that
rank (T2)=rank (T). Prove that the range and null space of
T are disjoint, i.e., have only the zero vector in common.
Solution: Assume
α=0,α∈range T∩null T, then
Tα=0, and
α is a linear independent set in range
T, thus can be expanded to a basis of range
T, namely
(α,β1,…,βn), so
rank (T)=n+1.
Notice that
range T2 is the set of vectors
T2β,β∈V, and
T2β=T(Tβ), as
Tβ∈range T, we have
Tβ=yα+∑i=1nyiβi, in which
y,y1,⋯,yn are scalars. So
T2β=T(yα+∑i=1nyiβi)=∑i=1nyiTβi, which means
Tβ1,…,Tβn spans range
T2, so
rank (T2)≤n, a contradiction.
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Let
p,m and
n be positive integers and
F a field. Let
V be the space of
m×n matrices over
F and
W the space of
p×n matrices over
F. Let
B be a fixed
p×m matrix and let
T be the linear transformation from
V into
W defined by
T(A)=BA. Prove that
T is invertible if and only if
p=m and
B is an invertible
m×m matrix.
Solution: If
p=m and
B is an invertible
m×m matrix, then define
U(A)=B−1A, we have
TU(A)=T(B−1A)=B(B−1A)=(BB−1)A=A and
UT(A)=U(BA)=B−1(BA)=(BB−1)A=A. Thus
T is invertible and
T−1(A)=B−1A.
Conversely, if
T is invertible, then
T is non-singular and
range T=W, use Exercise 3.2.10 we know if
p>m, then
T cannot be onto, if
p<m, then
T cannot be non-singular. Thus
p=m and
B is a
m×m matrix, if B is not invertible then the column vectors of
B are linearly dependent, thus there is
x1,…,xm not all 0 such that
x1b1+⋯+xmbm=0, then let
A be a matrix with
x1,…,xm be in the first column and 0 otherwise, we have
T(A)=0, this contradicts
T being non-singular. Thus
B must be invertible.