-
Let
T be the linear operator on
C2 defined by
T(x1,x2)=(x1,0). Let
B be the stantard ordered basis for
C2 and let
B′={α1,α2} be the ordered basis defined by
α1=(1,i),α2=(−i,2).
( a ) What is the matrix of
T relative to the pair
B,B′?
( b ) What is the matrix of
T relative to the pair
B′,B?
( c ) What is the matrix of
T in the ordered basis
B′?
( d ) What is the matrix of
T in the ordered basis
{α2,α1}?
Solution:
( a )
Tϵ1=2α1−iα2,Tϵ2=0α1+0α2, thus the matrix of
T relative to the pair
B,B′ is
[2−i00].
( b )
Tα1=ϵ1,Tα2=−iϵ1, thus the matrix of
T relative to the pair
B′,B is
[10−i0].
( c )
Tα1=2α1−iα2,Tα2=−2iα1−α2, thus
[T]B′=[2−i−2i−1].
( d )
Tα2=−α2−2iα1,Tα1=−iα2+2α1, thus
[T]{α2,α1}=[−1−2i−i2].
-
Let
T be the linear transformation from
R3 into
R2 defined by
T(x1,x2,x3)=(x1+x2,2x3−x1)
( a ) If
B is the standard ordered basis for
R3 and
B′ is the standard ordered basis for
R2, what is the matrix of
T relative to the pair
B,B′?
( b ) If
B={α1,α2,α3} and
B′={β1,β2}, where
α1=(1,0,−1),α2=(1,1,1),α3=(1,0,0),β1=(0,1),β2=(1,0)
what is the matrix of
T relative to the pair
B,B′?
Solution:
( a ) We let
ϵ1,ϵ2,ϵ3 be the standard ordered basis for
R3 and
ϵ1′,ϵ2′ be the standard ordered basis for
R2, then
Tϵ1=(1,−1),Tϵ2=(1,0),Tϵ3=(0,2)
thus the matrix of
T relative to the pair
B′,B is
A=[1−11002]
( b ) A direct calculation shows
Tα1=(1,−3)=−3β1+β2Tα2=(2,1)=β1+2β2,Tα3=(1,−1)=−β1+β2
thus the matrix of
T relative to the pair
B′,B is
A=[−3112−11]
-
Let
T be a linear operator on
Fn, let
A be the matrix of
T in the standard ordered basis for
Fn, and let
W be the subspace of
Fn spanned by the column vectors of
A. What does
W have to do with
T?
Solution: By the condition given, if we write
A=⎣⎢⎡A11⋮An1⋯⋱⋯A1n⋮Ann⎦⎥⎤, then
Tϵj=∑i=1nAijϵi=Aj, the
j-th column vector of
A, thus if
W is spanned by
A1,…,An, it is easy to see
W=range T.
-
Let
V be a two-dimensional vector space over the field
F, and let
B be an ordered basis for
V. If
T is a linear operator on
V and
[T]B=[acbd]
prove that
T2−(a+d)T+(ad−bc)I=0.
Solution: We write
B={α1,α2}, then for any
α∈V, we have
α=x1α1+x2α2,x1,x2∈F, notice that
[Tα1]B=[T]B[α1]B=[acbd][10]=[ac],
[Tα2]B=[T]B[α2]B=[acbd][01]=[bd],
[T2α1]B=[T]B[Tα1]B=[acbd][ac]=[a2+bcac+cd],
[T2α2]B=[T]B[Tα2]B=[acbd][bd]=[ab+bdbc+d2]
thus
(T2−(a+d)T+(ad−bc)I)α1=T2α1−(a+d)Tα1+(ad−bc)α1=(a2+bc)α1+(ac+cd)α2−(a+d)(aα1+cα2)+(ad−bc)α1=(a2+bc−a2−ad+ad−bc)α1+(ac+cd−ac−cd)α2=0
(T2−(a+d)T+(ad−bc)I)α2=T2α2−(a+d)Tα2+(ad−bc)α2=(ab+bd)α1+(bc+d2)α2−(a+d)(bα1+dα2)+(ad−bc)α2=(ab+bd−ab−bd)α1+(bc+d2−ad−d2+ad−bc)α2=0
Now we have
(T2−(a+d)T+(ad−bc)I)α1=(T2−(a+d)T+(ad−bc)I)α2=0, and so
(T2−(a+d)T+(ad−bc)I)α=(T2−(a+d)T+(ad−bc)I)(x1α1+x2α2)=x1(T2−(a+d)T+(ad−bc)I)α1+x2(T2−(a+d)T+(ad−bc)I)α2=0
-
Let
T be the linear operator on
R3, the matrix of which in the standard ordered basis is
A=⎣⎡10−1213114⎦⎤
Find a basis for the range of
T and a basis for the null space of
T.
Solution: Using Exercise 3 we know the range of
T is spanned by the column vectors of
A, using elementary column operations we have
A=⎣⎡10−1213114⎦⎤→⎣⎡10−1015015⎦⎤→⎣⎡10−1015000⎦⎤
so a basis for the range of
T is
(1,0,−1),(0,1,5), notice that
Tϵ3−Tϵ1=Tϵ2−2Tϵ1, thus
T(ϵ3+ϵ1−ϵ2)=T(1,−1,1)=0
and the dimension of the null space of
T is
1, so a basis for the null space of
T is
(1,−1,1).
-
Let
T be the linear operator on
R2 defined by
T(x1,x2)=(−x2,x1)
( a ) What is the matrix of
T in the standard ordered basis for
R2?
( b ) What is the matrix of
T in the ordered basis
B={α1,α2}, where
α1=(1,2) and
α2=(1,−1)?
( c ) Prove that for every real number
c the operator
(T−cI) is invertible.
( d ) Prove that if
B is any ordered basis for
R2 and
[T]B=A, then
A12A21=0.
Solution:
( a )
Tϵ1=(0,1)=ϵ2,Tϵ2=(−1,0)=−ϵ1, so the matrix of
T in the standard ordered basis for
R2 is
[01−10].
( b ) We have
Tα1=(−2,1)=−31α1−35α2,Tα2=(1,1)=32α1+31α2
thus
[T]B=[−1/3−5/32/31/3]
( c ) The matrix of
T−cI in the standard ordered basis for
R2 is
[−c1−1−c], thus
(T−cI)ϵ1=(−c,1),(T−cI)ϵ2=(−1,−c)
since
(−c,1),(−1,−c) are linearly independent for any
c∈R, thus a basis of
R2, this means
T−cI is invertible.
( d ) Let
B′={ϵ1,ϵ2}, then
[T]B′=[01−10], given any
B, we can find an invertible
P s.t.
[T]B=A=P[T]B′P−1
we can assume
P=[acbd], then
P−1=ad−bc1[d−c−ba], obviously
a,b,c,d cannot be all zero.
A=ad−bc1[acbd][01−10][d−c−ba]=ad−bc1[bd−a−c][d−c−ba]=ad−bc1[bd+acd2+c2−b2−a2−bd−ac]
thus
A12A21=−(a2+b2)(c2+d2)=0.
-
Let
T be the linear operator on
R3 defined by
T(x1,x2,x3)=(3x1+x3,−2x1+x2,−x1+2x2+4x3).
( a ) What is the matrix of
T in the standard ordered basis for
R3?
( b ) What is the matrix of
T in the ordered basis
{α1,α2,α3}, where
α1=(1,0,1),α2=(−1,2,1),α3=(2,1,1)?
( c ) Prove that
T is invertible and give a rule for
T−1 like the one which defines
T.
Solution:
( a ) The matrix of
T in the standard ordered basis for
R3 is
⎣⎡3−2−1012104⎦⎤.
(b) We form
P where
Pj=[αj]{ϵ1,ϵ2,ϵ3}, it is easy to see
P=⎣⎡101−121211⎦⎤, and we perform
⎣⎡101−121211100010001⎦⎤→⎣⎡100−12221−110−1010001⎦⎤→⎣⎡100−12021−210−101−1001⎦⎤→⎣⎡100−1200010−1/21/2−11/21/211/2−1/2⎦⎤→⎣⎡100010001−1/4−1/41/2−3/41/41/25/41/4−1/2⎦⎤
thus
P−1=1/4⎣⎡−1−12−31251−2⎦⎤
and then
[T]{α1,α2,α3}=P−1[T]{ϵ1,ϵ2,ϵ3}P=41⎣⎡−1−12−31251−2⎦⎤⎣⎡3−2−1012104⎦⎤⎣⎡101−121211⎦⎤=41⎣⎡−2−6473−2193−6⎦⎤⎣⎡101−121211⎦⎤=41⎣⎡17−3−23515−1422−60⎦⎤
( c ) It is enough to prove
[T]{ϵ1,ϵ2,ϵ3} is invertible, this is true since
⎣⎡3−2−1012104⎦⎤−1=⎣⎡4/98/9−1/32/913/9−2/3−1/9−2/91/3⎦⎤
since we know
[T−1]{ϵ1,ϵ2,ϵ3}=([T]{ϵ1,ϵ2,ϵ3})−1, it’s able to describe
T−1(x1,x2,x3)=(94x1+92x2−91x3,98x1+913x2−92x3,−31x1−32x2+31x3)
-
Let
θ be a real number. Prove that the following two matrices are similar over the field of complex numbers:
[cosθsinθsinθcosθ],[eiθ00e−iθ]
Solution: Let
T be the linear operator on
C2 which is represented by the first matrix in the standard ordered basis, then
T(1,0)=(cosθ,sinθ),T(0,1)=(−sinθ,cosθ), let
α1=(i,1),α2=(1,i), then
Tα1=i(cosθ,sinθ)+(−sinθ,cosθ)=(icosθ−sinθ,isinθ+cosθ)=eiθ(i,1)=eiθα1
and similarly we can see
Tα2=e−iθα2, it is easy to see
{α1,α2} are linearly independent, thus a basis of
C2, since
[T]{α1,α2}=[eiθ00e−iθ], the two matrices are similar, with
P=[i11i] and
[T]{α1,α2}=P−1[cosθsinθ−sinθcosθ]P
-
Let
V be a finite-dimensional vector space over the field
F and let
S and
T be linear operators on
V. We ask: When do there exist ordered bases
B and
B′ for
V such that
[S]B=[T]B′? Prove that such bases exist if and only if there is an invertible linear operator
U on
V such that
T=USU−1.
Solution: If
[S]B=[T]B′, then let
U be the linear operator that carries
B onto
B′, i.e., if we let
B={a1,…,an},B′={b1,…,bn}
and define
U by
Uai=bi,i=1,…,n. Then
U is invertible since it carries a basis onto another basis, and
U−1bi=ai,i=1,…,n. If we denote
[S]B=[T]B′=A=(Aij), then by definition we have
Saj=∑i=1nAijai,Tbj=∑i=1nAijbi, so
USU−1(bj)=USaj=U(i=1∑nAijai)=i=1∑nAijUai=i=1∑nAijbi=Tbj,j=1,…,n
Since
USU−1 and
T are equal on a basis of
V, we have
T=USU−1.
Conversely, if
T=USU−1 for some invertible
U, we let
B={a1,…,an} be an ordered basis for
V, and
B′={Ua1,…,Uan}, since
U is invertible,
B′ is a basis for
V. Notice that if
α∈V, then
α=k1a1+⋯+knan,k1,…,kn∈F
thus
[α]B=⎣⎢⎡k1⋮kn⎦⎥⎤, and
Uα=k1Ua1+⋯+knUan, so
[Uα]B′=⎣⎢⎡k1⋮kn⎦⎥⎤, so we have
[α]B=[Uα]B′, from this and the fact that
TU=US we can have
[S]B[α]B=[Sα]B=[USα]B′=[TUα]B′=[T]B′[Uα]B′=[T]B′[α]B
and it must follow that
[S]B=[T]B′.
[Alternatively, one easier proof is using Theorem 14 and the consequence of Theorem 13, since in this case we have
[T]B′=[U−1]B[T]B[U]B=[U−1TU]B=[U−1(USU−1)U]B=[S]B
].
-
We have seen that the linear operator
T on
R2 defined by
T(x1,x2)=(x1,0) is represented in the standard ordered basis by the matrix
A=[1000].
This operator satisfies
T2=T. Prove that if
S is a linear operator on
R2 such that
S2=S, then
S=0,or
S=I, or there is an ordered basis
B for
R2 such that
[S]B=A (above).
Solution: If
S=0 or
S=I, we obviously have
S2=S, now suppose
S=0,S=I, but
S2=S, then it is able to find
α,β∈R2, s.t.
Sα=0,Sβ=β, since
Sα∈range S, we have
dimrange S≥1, also since
S(Sβ−β)=S2β−Sβ=Sβ−Sβ=0, we know
Sβ−β∈null S, and
dimnull S≥1, as we discuss in
R2,
dimrange S+dimnull S=2, thus
dimrange S=dimnull S=1. Now if we let
a=Sα,b=Sβ−β, and
B={a,b}, then
B is an ordered basis for
R2 and
[S]B=A.
-
Let
W be the space of all
n×1 column matrices over a field
F. If $A4 is an
n×n matrix over
F, then
A defines a linear operator
LA on
W through left multiplicaition:
LA(X)=AX. Prove that every linear operator on
W is left multiplication by some
n×n matrix, i.e., is
LA for some
A.
Solution: If
T is a linear operator on
W, let
B′={ϵ1,…,ϵn} be the standard basis on
W, for each
X=⎣⎢⎡x1⋮xn⎦⎥⎤, we have
X=∑j=1nxjϵj, and if we define
A:=[T]B′, then
T(X)=T(∑j=1nxjϵj)=∑j=1nxjTϵj=∑j=1nxj∑i=1nAijϵi=∑i=1n(∑j=1nAijxj)ϵi=AX, thus
T=LA.
For the second question, let
B={a1,…,an}, if
[α]B=[β]B=⎣⎢⎡x1⋮xn⎦⎥⎤, then
α=β=∑j=1nxjaj , this shows
(Uα=Uβ)⇒(α=β), so
U is injective. Also for any
X=⎣⎢⎡x1⋮xn⎦⎥⎤∈W, define
α=∑j=1nxjaj, then
Uα=X, thus
U is surjective, combined we show
U is an isomorphism.
If
T is a linear operator on
V, then let
X=⎣⎢⎡x1⋮xn⎦⎥⎤,Y=⎣⎢⎡y1⋮yn⎦⎥⎤∈W, by definition we have
UTU−1(cX+Y)=UT(U−1(cX+Y))=UT(j=1∑n(cxj+yj)aj)=U(T(j=1∑n(cxj+yj)aj))=U(j=1∑n(cxj+yj)Taj)=j=1∑n(cxj+yj)UTaj=cj=1∑nxjUTaj+j=1∑nyjUTaj=c(UTj=1∑nxjaj)+UT(j=1∑nyjaj)=c(UT(U−1(X)))+(UT(U−1(Y)))=c(UTU−1)(X)+(UTU−1)(Y)
thus
UTU−1 is a linear operator on
W.
If
T is a linear operator on
V, then let
C=[T]B, i.e.
Taj=∑j=1nCijai, in the first part we proved
A=[UTU−1]B′, to compute
A, we see that
UTU−1(ϵj)=UTaj=U(∑j=1nCijai)=∑j=1nCijUai=∑j=1nCijϵj , thus
[UTU−1]B′=C=[T]B, or
A=[T]B.
-
Let
V be an
n-dimensional vector space over the field
F, and let
B={α1,…,αn} be an ordered basis for
V.
( a ) According th Theorem 1, there is a unique linear operator
T on
V such that
Tαj=αj+1,j=1,…,n−1,Tαn=0.
What is the matrix
A of
T in the ordered basis
B?
( b ) Prove that
Tn=0 but
Tn−1=0.
( c ) Let
S be any linear operator on
V such that
Sn=0 but
Sn−1=0. Prove that there is an ordered basis
B′ for
V such that the matrix of
S in the ordered basis
B′ is the matrix
A of part (a).
( d ) Prove that if
M and
N are
n×n matrices over
F such that
Mn=Nn=0 but
Mn−1=0=Nn−1, then
M and
N are similar.
Solution:
( a ) A direct computation shows
A=⎣⎢⎢⎡01000⋱…⋱100⎦⎥⎥⎤.
( b ) We have
Tkαn+1−k=0,k=1,…,n, thus
Tn=0, but
Tn−1α1=αn=0.
( c ) It is able to choose
α s.t.
Sn−1α=0 but
Snα=0, notice
α=0, and
{α,Sα,…,Sn−1α} is linearly independent, for if we have
k1α+k2Sα+⋯+knSn−1α=0
then
Sn−1(k1α+k2Sα+⋯+knSn−1α)=k1Sn−1α=0, thus
k1=0, the above becomes
k2Sα+⋯+knSn−1α=0
then
Sn−2(k2Sα+⋯+knSn−1α)=k2Sn−1α=0, thus
k2=0, continue this step we eventually have
k1=⋯=kn=0. Thus we can define
B′={α,Sα,…,Sn−1α}, and
[S]B′=A.
( d ) Let
T and
S be linear operators which satisfies
[T]B=M,[S]B=N, in which
B={ϵ1,…,ϵn}. From ( c ) we can find two ordered basis
B1 and
B2 s.t.
[T]B1=[S]B2=A, by Theorem 14, let
P be the
n×n matrix with columns
Pj=[ϵj]B1, and
Q be the
n×n matrix with columns
Qj=[ϵj]B2, then
M=[T]B=P−1AP,N=[S]B=Q−1AQ
thus
M=P−1QAQ−1P=(Q−1P)−1A(Q−1P).
-
Let
V and
W be finite-dimensional vector spaces over the field
F and let
T be a linear transformation from
V into
W. If
B={α1,…,αn} and B′={β1,…,βm}
are ordered base for
V and
W, respectively, define the linear transformations
Ep,q as in the proof of Theorem 5:
Ep,q(αi)=δiqβp. Then the
Ep,q,1≤p≤m,1≤q≤n, form a basis for
L(V,W), and so
T=p=1∑mq=1∑nApqEp,q
for certain scalars
Apq (the coordinates of
T in this basis for
L(V,W)). Show that the matrix
A with entries
A(p,q)=Apq is precisely the matrix of
T relative to the pair
B,B′.
Solution:
Tαj=(∑p=1m∑q=1nApqEp,q)(αj)=∑p=1m∑q=1nApqEp,q(αj)=∑p=1m∑q=1nApqδjqβp=∑p=1mApjβp, and the conclusion follows.