这一节的内容不长但是证明很难。核心目的是证明 V V V可以成为有限个cyclic space的直和。首先介绍了 T T T-admissible的定义,是一个比invariant更强的定义,其能够保证多项式运算在子空间中有对应的分项(投影),Theorem 3是Cyclic Decomposition Theorem,与之前的Primary Decomposition Theorem相比,其说明 V V V可以成为惟一的有限个 T T T-admissible space的直和,且每个子空间都是一个cyclic space,其generator的 T T T-annihilator是可以递归整除的。在本节的前文中,作者称这个定理是one of the deepest results in linear algebra,证明确实非常繁复。这一定理有一系列重要的推论,例如每一个 T T T-admissible空间都有一个invariant的互补空间。Theorem 4是广义的Cayley-Hamiltion定理,在之前的最小多项式整除特征多项式的结论之上,还可以推出二者有相同的prime factors,且已知最小多项式就可以得出特征多项式。Theorem 5声明每个矩阵都相似于一个唯一的rational form的矩阵。
Exercises
1.Let
T
T
T be the linear operator on
F
2
F^2
F2 which is represented in the standard ordered basis by the matrix
[
0
0
1
0
]
\begin{bmatrix}0&0\\1&0\end{bmatrix}
[0100]. Let
α
1
=
(
0
,
1
)
\alpha_1=(0,1)
α1=(0,1). Show that
F
2
≠
Z
(
α
1
;
T
)
F^2\neq Z(\alpha_1;T)
F2=Z(α1;T), and that there is no non-zero vector
α
2
\alpha_2
α2 in
F
2
F^2
F2 with
Z
(
α
2
;
T
)
Z(\alpha_2;T)
Z(α2;T) disjoint from
Z
(
α
1
;
T
)
Z(\alpha_1;T)
Z(α1;T).
Solution: We have
T
α
1
=
[
0
0
1
0
]
[
0
1
]
=
0
T\alpha_1=\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=0
Tα1=[0100][01]=0
thus
p
α
1
=
x
p_{\alpha_1}=x
pα1=x, which means
dim
Z
(
α
1
;
T
)
=
1
\dim Z(\alpha_1;T)=1
dimZ(α1;T)=1, so
F
2
≠
Z
(
α
1
;
T
)
F^2\neq Z(\alpha_1;T)
F2=Z(α1;T).
Suppose there is some
α
2
=
(
a
,
b
)
≠
0
\alpha_2=(a,b)\neq 0
α2=(a,b)=0 such that
Z
(
α
2
;
T
)
Z(\alpha_2;T)
Z(α2;T) is disjoint from
Z
(
α
1
;
T
)
Z(\alpha_1;T)
Z(α1;T), then
dim
Z
(
α
2
;
T
)
=
1
\dim Z(\alpha_2;T)=1
dimZ(α2;T)=1, which means
p
α
2
=
x
p_{\alpha_2}=x
pα2=x or
T
α
2
=
(
a
,
0
)
=
0
T\alpha_2=(a,0)=0
Tα2=(a,0)=0, so
α
2
=
(
0
,
b
)
≠
0
\alpha_2=(0,b)\neq 0
α2=(0,b)=0, but this means
α
2
=
b
α
1
\alpha_2=b\alpha_1
α2=bα1, which contradicts the hypothesis that
Z
(
α
2
;
T
)
Z(\alpha_2;T)
Z(α2;T) is disjoint from
Z
(
α
1
;
T
)
Z(\alpha_1;T)
Z(α1;T).
2.Let
T
T
T be a linear operator on the finite-dimensional space
V
V
V, and let
R
R
R be the range of
T
T
T.
( a ) Prove that
R
R
R has a complementary
T
T
T-invariant subspace if and only if
R
R
R is independent of the null space
N
N
N of
T
T
T.
( b ) If
R
R
R and
N
N
N are independent, prove that
N
N
N is the unique
T
T
T-invariant subspace complementary to
R
R
R.
Solution:
( a ) If
R
R
R is independent of
N
N
N, then from
dim
R
+
dim
N
=
dim
V
\dim R+\dim N=\dim V
dimR+dimN=dimV we know that
R
⊕
N
=
V
R\oplus N=V
R⊕N=V, and
N
N
N is obviously
T
T
T-invariant. Conversely, if
R
R
R has a complementary
T
T
T-invariant subspace
R
′
R'
R′, let
β
∈
R
′
\beta\in R'
β∈R′, then
T
β
∈
R
′
T\beta\in R'
Tβ∈R′, but also
T
β
∈
R
T\beta\in R
Tβ∈R, thus
T
β
=
0
T\beta=0
Tβ=0 and
β
∈
N
\beta\in N
β∈N, so
R
′
⊆
N
R'\subseteq N
R′⊆N, since
dim
R
′
=
dim
N
=
dim
V
−
dim
R
\dim R'=\dim N=\dim V-\dim R
dimR′=dimN=dimV−dimR, we know
R
′
=
N
R'=N
R′=N and so
R
∩
N
=
{
0
}
R\cap N=\{0\}
R∩N={0}.
( b ) Let
R
′
R'
R′ be any
T
T
T-invariant subspace complementary to
R
R
R, from the prood of (a) we can see that
R
′
=
N
R'=N
R′=N, given
R
R
R and
N
N
N are independent.
3.Let
T
T
T be the linear operator on
R
3
R^3
R3 which is represented in the standard ordered basis by the matrix
[
2
0
0
1
2
0
0
0
3
]
.
\begin{bmatrix}2&0&0\\1&2&0\\0&0&3\end{bmatrix}.
⎣⎡210020003⎦⎤.
Let
W
W
W be the null space of
T
−
2
I
T-2I
T−2I. Prove that
W
W
W has no complementary
T
T
T-invariant subspace.
Solution: Assume there exists a
T
T
T-invariant subspace
W
′
W'
W′ of
R
3
R^3
R3 such that
R
3
=
W
⊕
W
′
R^3=W{\oplus}W'
R3=W⊕W′, then let
β
=
ϵ
1
\beta=\epsilon_1
β=ϵ1, we have
(
T
−
2
I
)
β
=
ϵ
2
(T-2I)\beta=\epsilon_2
(T−2I)β=ϵ2, since
(
T
−
2
I
)
ϵ
2
=
0
(T-2I)\epsilon_2=0
(T−2I)ϵ2=0 we see that
(
T
−
2
I
)
β
∈
W
(T-2I)\beta\in W
(T−2I)β∈W. On the other hand, since
β
∈
R
3
\beta\in R^3
β∈R3, we can find
α
∈
W
,
γ
∈
W
′
\alpha\in W,\gamma\in W'
α∈W,γ∈W′ such that
β
=
α
+
γ
\beta=\alpha+\gamma
β=α+γ, so
(
T
−
2
I
)
β
=
(
T
−
2
I
)
α
+
(
T
−
2
I
)
γ
∈
W
(T-2I)\beta=(T-2I)\alpha+(T-2I)\gamma\in W
(T−2I)β=(T−2I)α+(T−2I)γ∈W
Since
W
′
W'
W′ is
T
T
T-invariant, we see that
(
T
−
2
I
)
γ
=
0
(T-2I)\gamma=0
(T−2I)γ=0 and
(
T
−
2
I
)
β
=
(
T
−
2
I
)
α
(T-2I)\beta=(T-2I)\alpha
(T−2I)β=(T−2I)α, but
α
∈
W
\alpha\in W
α∈W means
(
T
−
2
I
)
α
=
0
(T-2I)\alpha=0
(T−2I)α=0, but
(
T
−
2
I
)
β
=
ϵ
2
(T-2I)\beta=\epsilon_2
(T−2I)β=ϵ2, this is a contradiction.
4.Let
T
T
T be the linear operator on
F
4
F^4
F4 which is represented in the standard ordered basis by the matrix
[
c
0
0
0
1
c
0
0
0
1
c
0
0
0
1
c
]
.
\begin{bmatrix}c&0&0&0\\1&c&0&0\\0&1&c&0\\0&0&1&c\end{bmatrix}.
⎣⎢⎢⎡c1000c1000c1000c⎦⎥⎥⎤.
Let
W
W
W be the null space of
T
−
c
I
T-cI
T−cI.
( a ) Prove that
W
W
W is the subspace spanned by
ϵ
4
\epsilon_4
ϵ4.
( b ) Find the monic generators of the ideals
S
(
ϵ
4
;
W
)
S(\epsilon_4;W)
S(ϵ4;W),
S
(
ϵ
3
;
W
)
S(\epsilon_3;W)
S(ϵ3;W),
S
(
ϵ
2
;
W
)
S(\epsilon_2;W)
S(ϵ2;W),
S
(
ϵ
1
;
W
)
S(\epsilon_1;W)
S(ϵ1;W).
Solution:
( a ) A direct computation shows that the matrix of
T
−
c
I
T-cI
T−cI in the standard ordered basis is the matrix
[
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
]
\begin{bmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}
⎣⎢⎢⎡0100001000010000⎦⎥⎥⎤
and we have
(
T
−
c
I
)
(
∑
i
=
1
4
a
i
ϵ
i
)
=
a
1
ϵ
2
+
a
2
ϵ
3
+
a
3
ϵ
4
(T-cI)(\sum_{i=1}^4a_i\epsilon_i)=a_1\epsilon_2+a_2\epsilon_3+a_3\epsilon_4
(T−cI)(∑i=14aiϵi)=a1ϵ2+a2ϵ3+a3ϵ4, thus
W
W
W consists of all vectors of the form
a
ϵ
4
a\epsilon_4
aϵ4.
( b ) As
ϵ
4
\epsilon_4
ϵ4 is already in
W
W
W, we have
f
(
T
)
ϵ
4
∈
W
f(T)\epsilon_4\in W
f(T)ϵ4∈W for all
f
∈
F
[
x
]
f\in F[x]
f∈F[x], thus the monic generator of
S
(
ϵ
4
;
W
)
S(\epsilon_4;W)
S(ϵ4;W) is
1
1
1.
We have
T
ϵ
3
=
ϵ
4
T\epsilon_3=\epsilon_4
Tϵ3=ϵ4, so the monic generator of
S
(
ϵ
3
;
W
)
S(\epsilon_3;W)
S(ϵ3;W) is
x
x
x. By the same logic, the monic generator of
S
(
ϵ
2
;
W
)
S(\epsilon_2;W)
S(ϵ2;W) is
x
2
x^2
x2 and the monic generator of
S
(
ϵ
1
;
W
)
S(\epsilon_1;W)
S(ϵ1;W) is
x
3
x^3
x3.
5.Let
T
T
T be a linear operator on the vector space
V
V
V over the field
F
F
F. If
f
f
f is a polynomial over
F
F
F and
α
∈
V
\alpha\in V
α∈V, let
f
α
=
f
(
T
)
α
f\alpha=f(T)\alpha
fα=f(T)α. If
V
1
,
…
,
V
k
V_1,\dots,V_k
V1,…,Vk are
T
T
T-invariable subspaces and
V
=
V
1
⊕
⋯
⊕
V
k
V=V_1\oplus\cdots\oplus V_k
V=V1⊕⋯⊕Vk, show that
f
V
=
f
V
1
⊕
⋯
⊕
f
V
k
fV=fV_1\oplus\cdots\oplus fV_k
fV=fV1⊕⋯⊕fVk.
Solution: For
α
∈
V
\alpha\in V
α∈V, we have
α
=
α
1
+
⋯
+
α
k
\alpha=\alpha_1+\cdots+\alpha_k
α=α1+⋯+αk, in which
α
i
∈
V
i
\alpha_i\in V_i
αi∈Vi for
i
=
1
,
…
,
k
i=1,\dots,k
i=1,…,k, so
f
α
=
f
(
T
)
α
=
f
(
T
)
(
α
1
+
⋯
+
α
k
)
=
∑
i
=
1
k
f
(
T
)
α
i
=
∑
i
=
1
k
f
α
i
f\alpha=f(T)\alpha=f(T)(\alpha_1+\cdots+\alpha_k)=\sum_{i=1}^kf(T)\alpha_i=\sum_{i=1}^kf\alpha_i
fα=f(T)α=f(T)(α1+⋯+αk)=i=1∑kf(T)αi=i=1∑kfαi
this shows
f
V
=
f
V
1
+
⋯
+
f
V
k
fV=fV_1+\cdots+ fV_k
fV=fV1+⋯+fVk. To see the sum is a direct sum, let
β
∈
f
V
i
∩
f
V
j
\beta\in fV_i\cap fV_j
β∈fVi∩fVj with
i
≠
j
i\neq j
i=j, then we can find
β
′
∈
V
i
,
γ
′
∈
V
j
\beta'\in V_i,\gamma'\in V_j
β′∈Vi,γ′∈Vj such that
β
=
f
(
T
)
β
′
=
f
(
T
)
γ
′
\beta=f(T)\beta'=f(T)\gamma'
β=f(T)β′=f(T)γ′, since
V
i
V_i
Vi and
V
j
V_j
Vj are
T
T
T-invariant, we have
β
∈
V
i
\beta\in V_i
β∈Vi and
β
∈
V
j
\beta\in V_j
β∈Vj, so
β
∈
V
i
∩
V
j
\beta\in V_i\cap V_j
β∈Vi∩Vj and
β
=
0
\beta=0
β=0, this shows
V
i
V_i
Vi and
V
j
V_j
Vj are independent.
6.Let
T
,
V
,
F
T,V,F
T,V,F be as in Exercise 5. Suppose
α
\alpha
α and
β
\beta
β are vectors in
V
V
V which have the same
T
T
T-annihilator. Prove that, for any polynomial
f
f
f, the vectors
f
α
f\alpha
fα and
f
β
f\beta
fβ have the same
T
T
T-annihilator.
Solution: Let
p
p
p be the
T
T
T-annihilator of both
α
\alpha
α and
β
\beta
β. Suppose the
T
T
T-annihilator of
f
α
f\alpha
fα is
q
q
q, then
q
f
α
=
0
qf\alpha=0
qfα=0, which means
q
f
qf
qf is in the ideal generated by
p
p
p, so we can find polynomial
h
h
h such that
q
f
=
p
h
qf=ph
qf=ph, this means
q
f
β
=
p
h
β
=
h
p
β
=
0
qf\beta=ph\beta=hp\beta=0
qfβ=phβ=hpβ=0, thus the
T
T
T-annihilator of
f
β
f\beta
fβ divides
q
q
q, with the same logic applying to the
T
T
T-annihilator of
f
β
f\beta
fβ, we see
q
q
q divides the
T
T
T-annihilator of
f
β
f\beta
fβ, thus they are the same.
7.Find the minimal polynomials and the rational forms of each of the following real matrices.
[
0
−
1
−
1
1
0
0
−
1
0
0
]
,
[
c
0
−
1
0
c
1
−
1
1
c
]
,
[
cos
θ
sin
θ
−
sin
θ
cos
θ
]
\begin{bmatrix}0&-1&-1\\1&0&0\\-1&0&0\end{bmatrix},\quad \begin{bmatrix}c&0&-1\\0&c&1\\-1&1&c\end{bmatrix},\quad\begin{bmatrix}\cos\theta&\sin{\theta}\\{-\sin\theta}&{\cos\theta}\end{bmatrix}
⎣⎡01−1−100−100⎦⎤,⎣⎡c0−10c1−11c⎦⎤,[cosθ−sinθsinθcosθ]
Solution: For the first matrix, we compute the characteristic polynomial
∣
x
1
1
−
1
x
0
1
0
x
∣
=
x
3
+
x
−
x
=
x
3
\begin{vmatrix}x&1&1\\-1&x&0\\1&0&x\end{vmatrix}=x^3+x-x=x^3
∣∣∣∣∣∣x−111x010x∣∣∣∣∣∣=x3+x−x=x3
and the minimal polynomial is also
x
3
x^3
x3. Thus the rational form of this matrix is
[
0
0
0
1
0
0
0
1
0
]
\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}
⎣⎡010001000⎦⎤
For the second matrix we compute the characteristic polynomial
∣
x
−
c
0
1
0
x
−
c
−
1
1
−
1
x
−
c
∣
=
(
x
−
c
)
[
(
x
−
c
)
2
−
1
]
−
(
x
−
c
)
=
(
x
−
c
)
[
(
x
−
c
)
2
−
2
]
=
x
3
−
3
c
x
2
+
(
3
c
2
−
2
)
x
−
c
3
+
2
c
\begin{aligned}\begin{vmatrix}x-c&0&1\\0&x-c&-1\\1&-1&x-c\end{vmatrix}&=(x-c)[(x-c)^2-1]-(x-c)\\&=(x-c)[(x-c)^2-2]\\&=x^3-3cx^2+(3c^2-2)x-c^3+2c\end{aligned}
∣∣∣∣∣∣x−c010x−c−11−1x−c∣∣∣∣∣∣=(x−c)[(x−c)2−1]−(x−c)=(x−c)[(x−c)2−2]=x3−3cx2+(3c2−2)x−c3+2c
and the minimal polynomial is also
x
3
−
3
c
x
2
+
(
3
c
2
−
2
)
x
−
c
3
+
2
c
x^3-3cx^2+(3c^2-2)x-c^3+2c
x3−3cx2+(3c2−2)x−c3+2c. Thus the rational form of this matrix is
[
0
0
c
3
−
2
c
1
0
−
3
c
2
+
2
0
1
3
c
]
\begin{bmatrix}0&0&c^3-2c\\1&0&-3c^2+2\\0&1&3c\end{bmatrix}
⎣⎡010001c3−2c−3c2+23c⎦⎤
For the third matrix we compute the characteristic polynomial
∣
x
−
cos
θ
−
sin
θ
sin
θ
x
−
cos
θ
∣
=
x
2
−
2
cos
θ
x
+
1
\begin{vmatrix}x-\cos\theta&-\sin\theta\\\sin\theta&x-\cos\theta\end{vmatrix}=x^2-2\cos\theta x+1
∣∣∣∣x−cosθsinθ−sinθx−cosθ∣∣∣∣=x2−2cosθx+1
and the minimal polynomial is also
x
2
−
2
cos
θ
x
+
1
x^2-2\cos\theta x+1
x2−2cosθx+1. Thus the rational form of this matrix is
[
0
−
1
1
2
cos
θ
]
\begin{bmatrix}0&-1\\1&2\cos\theta\end{bmatrix}
[01−12cosθ].
8.Let
T
T
T be the linear operator on
R
3
R^3
R3 which is represented in the standard basis by
[
3
−
4
−
4
−
1
3
2
2
−
4
−
3
]
.
\begin{bmatrix}3&-4&-4\\-1&3&2\\2&-4&-3\end{bmatrix}.
⎣⎡3−12−43−4−42−3⎦⎤.
Find non-zero vectors
α
1
,
…
,
α
r
\alpha_1,\dots,\alpha_r
α1,…,αr satisfying the conditions of Theorem 3.
Solution: We first compute the characteristic polynomial of
T
T
T:
f
=
∣
x
−
3
4
4
1
x
−
3
−
2
−
2
4
x
+
3
∣
=
∣
x
−
3
4
4
1
x
−
3
−
2
0
2
x
−
2
x
−
1
∣
=
(
x
−
1
)
3
f=\begin{vmatrix}x-3&4&4\\1&x-3&-2\\-2&4&x+3\end{vmatrix}=\begin{vmatrix}x-3&4&4\\1&x-3&-2\\0&2x-2&x-1\end{vmatrix}=(x-1)^3
f=∣∣∣∣∣∣x−31−24x−344−2x+3∣∣∣∣∣∣=∣∣∣∣∣∣x−3104x−32x−24−2x−1∣∣∣∣∣∣=(x−1)3
Now the matrix of
T
−
I
T-I
T−I is obviously not zero and the matrix of
(
T
−
I
)
2
(T-I)^2
(T−I)2 is
[
2
−
4
−
4
−
1
2
2
2
−
4
−
4
]
[
2
−
4
−
4
−
1
2
2
2
−
4
−
4
]
=
0
\begin{bmatrix}2&-4&-4\\-1&2&2\\2&-4&-4\end{bmatrix}\begin{bmatrix}2&-4&-4\\-1&2&2\\2&-4&-4\end{bmatrix}=0
⎣⎡2−12−42−4−42−4⎦⎤⎣⎡2−12−42−4−42−4⎦⎤=0
thus the minimal polynomial for
T
T
T is
p
=
(
x
−
1
)
2
p=(x-1)^2
p=(x−1)2. Since
T
ϵ
1
=
(
3
,
−
1
,
2
)
T\epsilon_1=(3,-1,2)
Tϵ1=(3,−1,2) which is not a scalar multiple of
ϵ
1
\epsilon_1
ϵ1,
Z
(
ϵ
1
;
T
)
Z(\epsilon_1;T)
Z(ϵ1;T) has dimension 2 and consists of all vectors
a
ϵ
1
+
b
T
ϵ
1
=
a
(
1
,
0
,
0
)
+
b
(
3
,
−
1
,
2
)
=
(
a
+
3
b
,
−
b
,
2
b
)
a\epsilon_1+bT\epsilon_1=a(1,0,0)+b(3,-1,2)=(a+3b,-b,2b)
aϵ1+bTϵ1=a(1,0,0)+b(3,−1,2)=(a+3b,−b,2b)
So we can let
α
1
=
ϵ
1
\alpha_1=\epsilon_1
α1=ϵ1, the vector
α
2
\alpha_2
α2 must be a characteristic vector of
T
T
T which is not in
Z
(
ϵ
1
;
T
)
Z(\epsilon_1;T)
Z(ϵ1;T), As we can see if
α
=
(
x
1
,
x
2
,
x
3
)
\alpha=(x_1,x_2,x_3)
α=(x1,x2,x3), then
T
α
=
α
T\alpha=\alpha
Tα=α means
α
\alpha
α is in the form
(
2
a
+
2
b
,
a
,
b
)
(2a+2b,a,b)
(2a+2b,a,b), let
a
=
1
,
b
=
1
a=1,b=1
a=1,b=1 we see that we can make
α
2
=
(
4
,
1
,
1
)
\alpha_2=(4,1,1)
α2=(4,1,1).
9.Let
A
A
A be the real matrix
A
=
[
1
3
3
3
1
3
−
3
−
3
−
5
]
.
A=\begin{bmatrix}1&3&3\\3&1&3\\-3&-3&-5\end{bmatrix}.
A=⎣⎡13−331−333−5⎦⎤.
Find an invertible
3
×
3
3\times 3
3×3 real matrix
P
P
P such that
P
−
1
A
P
P^{-1}AP
P−1AP is in rational form.
Solution: First compute the characteristic polynomial for
A
A
A
det
(
x
I
−
A
)
=
∣
x
−
1
−
3
−
3
−
3
x
−
1
−
3
3
3
x
+
5
∣
=
∣
x
−
1
−
3
−
3
−
3
x
−
1
−
3
0
x
+
2
x
+
2
∣
=
(
x
+
2
)
(
x
2
−
2
x
+
1
−
9
+
3
x
−
3
+
9
)
=
(
x
+
2
)
2
(
x
−
1
)
\begin{aligned}\det (xI-A)&=\begin{vmatrix}x-1&-3&-3\\-3&x-1&-3\\3&3&x+5\end{vmatrix}=\begin{vmatrix}x-1&-3&-3\\-3&x-1&-3\\0&x+2&x+2\end{vmatrix}\\&=(x+2)(x^2-2x+1-9+3x-3+9)\\&=(x+2)^2(x-1)\end{aligned}
det(xI−A)=∣∣∣∣∣∣x−1−33−3x−13−3−3x+5∣∣∣∣∣∣=∣∣∣∣∣∣x−1−30−3x−1x+2−3−3x+2∣∣∣∣∣∣=(x+2)(x2−2x+1−9+3x−3+9)=(x+2)2(x−1)
and since
(
A
+
2
I
)
(
A
−
I
)
=
[
3
3
3
3
3
3
−
3
−
3
−
3
]
[
0
3
3
3
0
3
−
3
−
3
−
6
]
=
0
(A+2I)(A-I)=\begin{bmatrix}3&3&3\\3&3&3\\-3&-3&-3\end{bmatrix}\begin{bmatrix}0&3&3\\3&0&3\\-3&-3&-6\end{bmatrix}=0
(A+2I)(A−I)=⎣⎡33−333−333−3⎦⎤⎣⎡03−330−333−6⎦⎤=0
the minimal polynomial for
A
A
A is
(
x
+
2
)
(
x
−
1
)
=
x
2
+
x
−
2
(x+2)(x-1)=x^2+x-2
(x+2)(x−1)=x2+x−2.
Since
A
ϵ
1
=
(
1
,
3
,
−
3
)
A\epsilon_1=(1,3,-3)
Aϵ1=(1,3,−3) is not a scalar multiple of
ϵ
1
\epsilon_1
ϵ1, one subspace can be
(
ϵ
1
,
A
ϵ
1
)
(\epsilon_1,A\epsilon_1)
(ϵ1,Aϵ1), which consists of vectors like
(
a
+
b
,
3
b
,
−
3
b
)
(a+b,3b,-3b)
(a+b,3b,−3b), choose a characteristic vector associated with the characteristic value
−
2
-2
−2, which may be
(
1
,
1
,
−
2
)
(1,1,-2)
(1,1,−2), then let
P
=
[
1
1
1
0
3
1
0
−
3
−
2
]
P=\begin{bmatrix}1&1&1\\0&3&1\\0&-3&-2\end{bmatrix}
P=⎣⎡10013−311−2⎦⎤
we have
det
P
=
−
3
≠
0
\det P=-3\neq 0
detP=−3=0, thus
P
P
P is invertible, and
A
P
=
[
1
3
3
3
1
3
−
3
−
3
−
5
]
[
1
1
1
0
3
1
0
−
3
−
2
]
=
[
1
1
−
2
3
−
3
−
2
−
3
3
4
]
AP=\begin{bmatrix}1&3&3\\3&1&3\\-3&-3&-5\end{bmatrix}\begin{bmatrix}1&1&1\\0&3&1\\0&-3&-2\end{bmatrix}=\begin{bmatrix}1&1&-2\\3&-3&-2\\-3&3&4\end{bmatrix}
AP=⎣⎡13−331−333−5⎦⎤⎣⎡10013−311−2⎦⎤=⎣⎡13−31−33−2−24⎦⎤
the rational form of
A
A
A is clearly
[
0
2
0
1
−
1
0
0
0
−
2
]
\begin{bmatrix}0&2&0\\1&-1&0\\0&0&-2\end{bmatrix}
⎣⎡0102−1000−2⎦⎤, and we have
P
[
0
2
0
1
−
1
0
0
0
−
2
]
=
[
1
1
1
0
3
1
0
−
3
−
2
]
[
0
2
0
1
−
1
0
0
0
−
2
]
=
[
1
1
−
2
3
−
3
−
2
−
3
3
4
]
P\begin{bmatrix}0&2&0\\1&-1&0\\0&0&-2\end{bmatrix}=\begin{bmatrix}1&1&1\\0&3&1\\0&-3&-2\end{bmatrix}\begin{bmatrix}0&2&0\\1&-1&0\\0&0&-2\end{bmatrix}=\begin{bmatrix}1&1&-2\\3&-3&-2\\-3&3&4\end{bmatrix}
P⎣⎡0102−1000−2⎦⎤=⎣⎡10013−311−2⎦⎤⎣⎡0102−1000−2⎦⎤=⎣⎡13−31−33−2−24⎦⎤
thus
P
P
P is the matrix we need.
10.Let
F
F
F be a subfield of the complex numbers and let
T
T
T be the linear operator on
F
4
F^4
F4 which is represented in the standard ordered basis by the matrix
[
2
0
0
0
1
2
0
0
0
a
2
0
0
0
b
2
]
.
\begin{bmatrix}2&0&0&0\\1&2&0&0\\0&a&2&0\\0&0&b&2\end{bmatrix}.
⎣⎢⎢⎡210002a0002b0002⎦⎥⎥⎤.
Find the characteristic polynomial for
T
T
T. Consider the cases
a
=
b
=
1
a=b=1
a=b=1;
a
=
b
=
0
a=b=0
a=b=0;
a
=
0
,
b
=
1
a=0,b=1
a=0,b=1. In each of these cases, find the minimal polynomial for
T
T
T and non-zero vectors
α
1
,
…
,
α
r
\alpha_1,\dots,\alpha_r
α1,…,αr which satisfy the conditions of Theorem 3.
Solution: The characteristic polynomial for
T
T
T is
(
x
−
2
)
4
(x-2)^4
(x−2)4.
In the case
a
=
b
=
1
a=b=1
a=b=1, the minimal polynomial for
T
T
T is
(
x
−
2
)
4
(x-2)^4
(x−2)4, since for
ϵ
1
=
(
1
,
0
,
0
,
0
)
\epsilon_1=(1,0,0,0)
ϵ1=(1,0,0,0), we have
T
ϵ
1
=
(
1
,
1
,
0
,
0
)
T\epsilon_1=(1,1,0,0)
Tϵ1=(1,1,0,0),
T
2
ϵ
1
=
(
1
,
2
,
1
,
0
)
T^2\epsilon_1=(1,2,1,0)
T2ϵ1=(1,2,1,0),
T
3
ϵ
1
=
(
1
,
3
,
3
,
1
)
T^3\epsilon_1=(1,3,3,1)
T3ϵ1=(1,3,3,1), we have
α
1
=
ϵ
1
\alpha_1=\epsilon_1
α1=ϵ1.
In the case
a
=
b
=
0
a=b=0
a=b=0 and
a
=
0
,
b
=
1
a=0,b=1
a=0,b=1, the minimal polynomial for
T
T
T is
(
x
−
2
)
2
(x-2)^2
(x−2)2. The non-zero vectors
α
i
\alpha_i
αi satisfing Theorem 3 can be found using techniques in the proof of Theorem 3.
11.Prove that if
A
A
A and
B
B
B are
3
×
3
3\times 3
3×3 matrices over a field
F
F
F, a necessary and sufficient condition that
A
A
A and
B
B
B be similar over
F
F
F is that they have the same characteristic polynomial and the same minimal polynomial. Give an example which shows that this is false for
4
×
4
4\times 4
4×4 matrices.
Solution: If
A
A
A and
B
B
B are similar, then both are similar to the same matrix
R
R
R which is in rational form, thus the minimal polynomial for
A
A
A and
B
B
B are the same. Let it be
p
p
p.
If
deg
p
=
3
\deg p=3
degp=3, then it is the characteristic polynomial for both
A
A
A and
B
B
B.
If
deg
p
=
2
\deg p=2
degp=2, then
R
R
R must have the form
[
A
1
0
0
a
]
\begin{bmatrix}A_1&0\\0&a\end{bmatrix}
[A100a], where
A
1
A_1
A1 is a
2
×
2
2\times 2
2×2 matrix in the rational form, then the characteristic polynomial for
A
A
A and
B
B
B are
p
(
x
−
a
)
p(x-a)
p(x−a).
If
deg
p
=
1
\deg p=1
degp=1, then
R
R
R must be diagonal, then it is apparent the characteristic polynomial for
A
A
A and
B
B
B are equal.
Conversely, if
A
A
A and
B
B
B have the same characteristic polynomial
f
f
f and the same minimal polynomial
p
p
p. We find the unique matrix in the rational form that
A
A
A and
B
B
B are similar to, namely
R
A
R_A
RA and
R
B
R_B
RB, then
If
deg
p
=
3
\deg p=3
degp=3, then we must have
R
A
=
R
B
R_A=R_B
RA=RB.
If
deg
p
=
2
\deg p=2
degp=2, then
R
A
=
[
A
1
0
0
a
]
R_A=\begin{bmatrix}A_1&0\\0&a\end{bmatrix}
RA=[A100a],
R
B
=
[
A
1
0
0
b
]
R_B=\begin{bmatrix}A_1&0\\0&b\end{bmatrix}
RB=[A100b], where
A
1
A_1
A1 is a
2
×
2
2\times 2
2×2 matrix in the rational form, and
f
/
p
=
x
−
a
=
x
−
b
f/p=x-a=x-b
f/p=x−a=x−b means
a
=
b
a=b
a=b, so
R
A
=
R
B
R_A=R_B
RA=RB.
If
deg
p
=
1
\deg p=1
degp=1, then
R
A
R_A
RA and
R
B
R_B
RB must be diagonal, since the characteristic polynomial for
A
A
A and
B
B
B are equal ,we have
R
A
=
R
B
R_A=R_B
RA=RB.
Since
A
A
A is similar to
R
A
R_A
RA and
B
B
B is similar to
R
B
R_B
RB, we have
A
A
A similar to
B
B
B.
For a counterexample of
4
×
4
4\times 4
4×4 matrix, we let
A
=
[
0
−
1
1
2
1
1
]
,
B
=
[
0
−
1
1
2
0
−
1
1
2
]
A=\begin{bmatrix}0&-1\\1&2\\&&1\\&&&1\end{bmatrix},\quad B=\begin{bmatrix}0&-1\\1&2\\&&0&-1\\&&1&2\end{bmatrix}
A=⎣⎢⎢⎡01−1211⎦⎥⎥⎤,B=⎣⎢⎢⎡01−1201−12⎦⎥⎥⎤
The characteristic polynomial of
A
A
A and
B
B
B is
(
x
−
1
)
4
(x-1)^4
(x−1)4, the minimal polynomial of
A
A
A and
B
B
B is
(
x
−
1
)
2
(x-1)^2
(x−1)2, but
A
A
A and
B
B
B are not similar.
12.Let
F
F
F be a subfield of the field of complex numbers, and let
A
A
A and
B
B
B be
n
×
n
n\times n
n×n matrices over
F
F
F. Prove that if
A
A
A and
B
B
B are similar over the field of complex numbers, then they are similar over
F
F
F.
Solution: The rational form of
A
A
A is a matrix over
F
F
F, thus a matrix over
C
C
C, likewise for
B
B
B. Thus if
A
A
A and
B
B
B are similar over the field of complex numbers, due to Theorem 5, the rational form of
A
A
A is the same as
B
B
B over
C
C
C, which means
A
A
A and
B
B
B are similar to the same rational form over
F
F
F, and the conclusion follows.
13.Let
A
A
A be an
n
×
n
n\times n
n×n matrix with complex entires. Prove that if every characteristic value of
A
A
A is real, then
A
A
A is similar to a matrix with real entries.
Solution: The characteristic polynomial for
A
A
A contains only linear factors, and so is the minimal polynomial
p
p
p for
A
A
A, since every characteristic value of
A
A
A is real, we see
p
p
p consists only real coefficients.
Let
T
T
T be the linear operator on
F
n
F^n
Fn which is represented by
A
A
A in the standard basis, then there is an ordered basis
B
\mathfrak B
B for
V
V
V such that
[
T
]
B
=
[
A
1
A
2
⋱
A
r
]
[T]_{\mathfrak B}=\begin{bmatrix}A_1\\&A_2\\&&{\ddots}\\&&&&A_r\end{bmatrix}
[T]B=⎣⎢⎢⎡A1A2⋱Ar⎦⎥⎥⎤
where each
A
i
A_i
Ai is the companion matrix of some polynomial
p
i
p_i
pi, and
p
1
=
p
p_1=p
p1=p, and
p
i
∣
p
p_i|p
pi∣p for
i
=
2
,
…
,
r
i=2,\dots,r
i=2,…,r, since
p
p
p consists only real coefficients, so are all
p
i
p_i
pi, which means all
A
i
A_i
Ai have real entries, and so is
[
T
]
B
[T]_{\mathfrak B}
[T]B, apparently,
A
A
A is similar to
[
T
]
B
[T]_{\mathfrak B}
[T]B.
14.Let
T
T
T be a linear operator on the finite-dimensional space
V
V
V. Prove that there exists a vector
α
∈
V
\alpha\in V
α∈V with this property. If
f
f
f is a polynomial and
f
(
T
)
α
=
0
f(T)\alpha=0
f(T)α=0, then
f
(
T
)
=
0
f(T)=0
f(T)=0. (Such a vector
α
\alpha
α is called a separating vector for the algebra of polynomials in
T
T
T.) When
T
T
T has a cyclic vector, give a direct proof that any cyclic vector is a separating vector for the algebra of polynomials in
T
T
T.
Solution: We first prove if
α
\alpha
α is a cyclic vector of
T
T
T, then
α
\alpha
α is a separating vector for the algebra of polynomials in
T
T
T. Suppose
dim
V
=
n
\dim V=n
dimV=n, then
α
,
…
,
T
n
−
1
α
\alpha,\dots,T^{n-1}\alpha
α,…,Tn−1α span
V
V
V, so for any
β
∈
V
\beta\in V
β∈V, we have
β
=
g
(
T
)
α
\beta=g(T)\alpha
β=g(T)α for some polynomial
g
g
g. Now if
f
f
f is a polynomial and
f
(
T
)
α
=
0
f(T)\alpha=0
f(T)α=0, we have
f
(
T
)
β
=
f
(
T
)
g
(
T
)
α
=
g
(
T
)
f
(
T
)
α
=
g
(
T
)
0
=
0
f(T)\beta=f(T)g(T)\alpha=g(T)f(T)\alpha=g(T)0=0
f(T)β=f(T)g(T)α=g(T)f(T)α=g(T)0=0
thus
f
(
T
)
=
0
f(T)=0
f(T)=0.
Now for any linear operator
T
T
T on
V
V
V, use the Cyclic Decomposition Theorem, we can write
V
=
Z
(
α
1
;
T
)
⊕
⋯
⊕
Z
(
α
r
;
T
)
V=Z(\alpha_1;T)\oplus\cdots\oplus Z(\alpha_r;T)
V=Z(α1;T)⊕⋯⊕Z(αr;T), let
α
=
α
1
+
⋯
+
α
r
\alpha=\alpha_1+\cdots+\alpha_r
α=α1+⋯+αr, then
f
(
T
)
α
=
0
f(T)\alpha=0
f(T)α=0 means
f
(
T
)
α
i
=
0
f(T)\alpha_i=0
f(T)αi=0 since each
Z
(
α
i
;
T
)
Z(\alpha_i;T)
Z(αi;T) is invariant under
T
T
T, and
V
V
V is the direct sum of all
Z
(
α
i
;
T
)
Z(\alpha_i;T)
Z(αi;T). It follows that
f
(
T
)
=
0
f(T)=0
f(T)=0 on
Z
(
α
i
;
T
)
Z(\alpha_i;T)
Z(αi;T) for
i
=
1
,
…
,
r
i=1,\dots,r
i=1,…,r, which means
f
(
T
)
=
0
f(T)=0
f(T)=0 on
V
V
V.
15.Let
F
F
F be a subfield of the field of complex numbers, and let
A
A
A be an
n
×
n
n\times n
n×n matrix over
F
F
F. Let
p
p
p be the minimal polynomial for
A
A
A. If we regard
A
A
A as a matrix over
C
C
C, then
A
A
A has a minimal polynomial
f
f
f as an
n
×
n
n\times n
n×n matrix over
C
C
C. Use a theorem on linear equations to prove
p
=
f
p=f
p=f. Can you also see how this follows from the cyclic decomposition theorem?
Solution: If we write
p
=
c
0
+
c
1
x
+
⋯
+
x
k
p=c_0+c_1x+\cdots+x^k
p=c0+c1x+⋯+xk, then
p
(
A
)
=
0
p(A)=0
p(A)=0 means
(
c
0
,
c
1
,
…
,
1
)
(c_0,c_1,\dots,1)
(c0,c1,…,1) is a solution for the system
x
1
I
+
⋯
+
x
k
+
1
A
k
=
0
x_1I+\cdots+x_{k+1}A^k=0
x1I+⋯+xk+1Ak=0
in the field
F
F
F, and likewise,
f
=
d
0
+
d
1
x
+
⋯
+
x
k
f=d_0+d_1x+\cdots+x^k
f=d0+d1x+⋯+xk is the polynomial which has coefficients
(
d
0
,
d
1
,
…
,
1
)
(d_0,d_1,\dots,1)
(d0,d1,…,1) as a solution for the same system in the field
C
C
C, then
(
d
0
,
d
1
,
…
,
1
)
(d_0,d_1,\dots,1)
(d0,d1,…,1) is a solution in
F
F
F due to the final remark in Sec 1.4. Thus both
(
c
0
,
c
1
,
…
,
1
)
(c_0,c_1,\dots,1)
(c0,c1,…,1) and
(
d
0
,
d
1
,
…
,
1
)
(d_0,d_1,\dots,1)
(d0,d1,…,1) are in the solution space for
x
1
I
+
⋯
+
x
k
+
1
A
k
=
0
x_1I+\cdots+x_{k+1}A^k=0
x1I+⋯+xk+1Ak=0. To prove
p
=
f
p=f
p=f, assume there is some
c
i
≠
d
i
c_i\neq d_i
ci=di, then
(
c
0
,
c
1
,
…
,
c
k
−
1
)
−
(
d
0
,
d
1
,
…
,
d
k
−
1
)
(c_0,c_1,\dots,c_{k-1})-(d_0,d_1,\dots,d_{k-1})
(c0,c1,…,ck−1)−(d0,d1,…,dk−1) is a non-trivial solution for the system
x
1
I
+
⋯
+
x
k
A
k
−
1
=
0
x_1I+\cdots+x_{k}A^{k-1}=0
x1I+⋯+xkAk−1=0
Let
h
i
=
c
i
−
d
i
h_i=c_i-d_i
hi=ci−di and
h
=
h
0
+
h
1
x
+
⋯
+
h
k
−
1
x
k
−
1
h=h_0+h_1x+\cdots+h_{k-1}x^{k-1}
h=h0+h1x+⋯+hk−1xk−1, we see
h
(
A
)
=
0
h(A)=0
h(A)=0, but
deg
h
<
deg
p
\deg h<\deg p
degh<degp, a contradiction.
To get this result from the cyclic decomposition theorem, notice that by Exercise 12,
A
A
A has the same rational form in
F
F
F and
C
C
C, and the first block matrix of the rational form of
A
A
A is the companion matrix of
p
p
p in
F
F
F and
f
f
f in
C
C
C, we have
p
=
f
p=f
p=f.
16.Let
A
A
A be an
n
×
n
n\times n
n×n matrix with real entries such that
A
2
+
I
=
0
A^2+I=0
A2+I=0. Prove that
n
n
n is even, and if
n
=
2
k
n=2k
n=2k, then
A
A
A is similar over the field of real numbers to a matrix of the block form
[
0
−
I
I
0
]
\begin{bmatrix}0&-I\\I&0\end{bmatrix}
[0I−I0] where
I
I
I is the
k
×
k
k\times k
k×k identity matrix.
Solution: The minimal polynomial for
A
A
A is
x
2
+
1
x^2+1
x2+1, by the generalized Cayley-Hamilton Theorem, the characteristic polynomial for
A
A
A must be of the form
f
=
(
x
2
+
1
)
k
f=(x^2+1)^k
f=(x2+1)k, so
n
=
deg
f
n=\deg f
n=degf is even.
If
n
=
2
k
n=2k
n=2k, we know
A
A
A is similar to one and only one matrix
B
B
B in the rational form. If we write
B
=
[
A
1
⋱
A
r
]
B=\begin{bmatrix}A_1\\&\ddots\\&&A_r\end{bmatrix}
B=⎣⎡A1⋱Ar⎦⎤
where each
A
i
A_i
Ai is the companion matrix of
p
i
p_i
pi, and
p
i
+
1
p_{i+1}
pi+1 divides
p
i
p_i
pi, from the proof of Theorem 3 we know
p
1
=
x
2
+
1
p_1=x^2+1
p1=x2+1, and the only possible polynomial which divides
x
2
+
1
x^2+1
x2+1 is
x
2
+
1
x^2+1
x2+1 and
1
1
1. Since
1
1
1 can only be the annihilator of zero vectors, we see that
B
=
[
A
1
⋱
A
k
]
,
A
i
=
[
−
1
1
]
,
i
=
1
,
…
,
k
B=\begin{bmatrix}A_1\\&\ddots\\&&A_k\end{bmatrix}, \quad A_i=\begin{bmatrix}&-1\\1\end{bmatrix},i=1,\dots,k
B=⎣⎡A1⋱Ak⎦⎤,Ai=[1−1],i=1,…,k
Let
B
=
{
ϵ
1
,
…
,
ϵ
n
}
\mathscr B=\{\epsilon_1,\dots,\epsilon_n\}
B={ϵ1,…,ϵn} be a basis for
R
n
R^n
Rn and
T
T
T is the linear operator with
[
T
]
B
=
B
[T]_{\mathscr B}=B
[T]B=B, then
T
ϵ
2
i
−
1
=
ϵ
2
i
,
T
ϵ
2
i
=
−
ϵ
2
i
−
1
,
i
=
1
,
…
,
k
T\epsilon_{2i-1}=\epsilon_{2i},\quad T\epsilon_{2i}=-\epsilon_{2i-1},\quad i=1,\dots,k
Tϵ2i−1=ϵ2i,Tϵ2i=−ϵ2i−1,i=1,…,k
If we let
α
i
=
ϵ
2
i
−
1
\alpha_i=\epsilon_{2i-1}
αi=ϵ2i−1 for
i
=
1
,
…
,
k
i=1,\dots,k
i=1,…,k and
α
i
=
ϵ
2
i
−
2
k
\alpha_i=\epsilon_{2i-2k}
αi=ϵ2i−2k for
i
=
k
+
1
,
…
,
n
i=k+1,\dots,n
i=k+1,…,n, then
B
′
=
{
α
1
,
…
,
α
n
}
\mathscr B'=\{\alpha_1,\dots,\alpha_n\}
B′={α1,…,αn} is a basis for
V
V
V, and we can verify
[
T
]
B
′
=
[
0
−
I
I
0
]
[T]_{\mathscr B'}=\begin{bmatrix}0&-I\\I&0\end{bmatrix}
[T]B′=[0I−I0], which means
B
B
B is similar to
[
0
−
I
I
0
]
\begin{bmatrix}0&-I\\I&0\end{bmatrix}
[0I−I0] and so is
A
A
A.
17.Let
T
T
T be a linear operator on a finite-dimensional vector space
V
V
V. Suppose that
( a ) the minimal polynomial for
T
T
T is a power of an irreducible polynomial;
( b ) the minimal polynomial is equal to the characteristic polynomial.
Show that no non-trivial
T
T
T-invariant subspace has a complementary
T
T
T-invariant subspace.
Solution: Let
W
W
W be a non-trivial
T
T
T-invariant subspace of
V
V
V, assume there is
W
′
W'
W′ which is
T
T
T-invariant such that
W
⊕
W
′
=
V
W\oplus W'=V
W⊕W′=V, let
T
W
=
U
T_W=U
TW=U and
T
W
′
=
U
′
T_{W'}=U'
TW′=U′, then the minimal polynomial
p
p
p of
T
W
T_W
TW and
p
′
p'
p′ of
T
W
′
T_{W'}
TW′ divide the minimal polynomial for
T
T
T. Since the minimal polynomial for
T
T
T is of the form
q
n
q^n
qn where
q
q
q is irreducible, we have
p
=
q
r
p=q^r
p=qr and
p
′
=
q
s
p'=q^s
p′=qs, where
r
+
s
≤
n
r+s\leq n
r+s≤n. As
W
W
W is non-trivial, we have
r
≥
1
r\geq 1
r≥1.
Now if
s
≥
1
s\geq 1
s≥1, we can get a contradiction by the following procedure: from (b) we know that
T
T
T has a cyclic vector
α
\alpha
α such that the
T
T
T-annihilator of
α
\alpha
α is
q
n
q^n
qn, and there is
α
1
∈
W
,
α
2
∈
W
′
\alpha_1\in W,\alpha_2\in W'
α1∈W,α2∈W′ such that
α
=
α
1
+
α
2
\alpha=\alpha_1+\alpha_2
α=α1+α2, we let
k
=
max
(
r
,
s
)
k=\max(r,s)
k=max(r,s), then
1
≤
k
<
n
1\leq k<n
1≤k<n, and
q
k
(
T
)
α
1
=
q
k
(
T
)
α
2
=
0
q^k(T)\alpha_1=q^k(T)\alpha_2=0
qk(T)α1=qk(T)α2=0, which means
q
k
(
T
)
α
=
0
q^k(T)\alpha=0
qk(T)α=0, this is a contradiction.
Thus
s
=
0
s=0
s=0, or the minimal polynomial for
T
W
′
T_{W'}
TW′ is
1
1
1, which means
18.If
T
T
T is a diagonalizable linear operator, then every
T
T
T-invariant subspace has a complementary
T
T
T-invariant subspace.
Solution:
T
T
T is diagonalizable means if we let
c
1
,
…
,
c
k
c_1,\dots,c_k
c1,…,ck be distinct characteristic values of
T
T
T and let
V
i
=
null
(
T
−
c
i
I
)
V_i=\text{null }(T-c_iI)
Vi=null (T−ciI), then
V
=
V
1
⊕
⋯
⊕
V
k
V=V_1\oplus\cdots\oplus V_k
V=V1⊕⋯⊕Vk
Let
W
W
W be a
T
T
T-invariant subspace of
V
V
V, then by Exercise 10 of Section 6.8, we have
W
=
(
W
∩
V
1
)
⊕
⋯
⊕
(
W
∩
V
k
)
W=(W\cap V_1)\oplus\cdots\oplus(W\cap V_k)
W=(W∩V1)⊕⋯⊕(W∩Vk)
Consider
W
∩
V
i
W\cap V_i
W∩Vi, for any
β
∈
W
∩
V
i
\beta\in W\cap V_i
β∈W∩Vi, we have
β
∈
V
i
\beta\in V_i
β∈Vi, so
T
β
=
c
i
β
T\beta=c_i\beta
Tβ=ciβ. Since
W
∩
V
i
W\cap V_i
W∩Vi is a subspace, we can find
{
α
1
,
…
,
α
r
i
}
\{\alpha_1,\dots,\alpha_{r_i}\}
{α1,…,αri} to be a basis for it, then it can be extended to a basis for
V
i
V_i
Vi, namely
{
α
1
,
…
,
α
s
i
}
\{\alpha_1,\dots,\alpha_{s_i}\}
{α1,…,αsi}, all of which are characteristic vectors associated with
c
i
c_i
ci. Let
U
i
U_i
Ui be the space spanned by
{
α
r
i
+
1
,
…
,
α
s
i
}
\{\alpha_{{r_i}+1},\dots,\alpha_{s_i}\}
{αri+1,…,αsi}, then
(
W
∩
V
i
)
⊕
U
i
=
V
i
(W\cap V_i)\oplus U_i=V_i
(W∩Vi)⊕Ui=Vi, let
U
=
U
1
⊕
⋯
⊕
U
k
U=U_1\oplus\cdots\oplus U_k
U=U1⊕⋯⊕Uk, we see that
V
=
W
⊕
U
V=W\oplus U
V=W⊕U, and as each
U
i
U_i
Ui is invariant under
T
T
T, so is
U
U
U.
19.Let
T
T
T be a linear operator on the finite-dimensional space
V
V
V. Prove that
T
T
T has a cyclic vector if and only if the following is true: Every linear operator
U
U
U which commutes with
T
T
T is a polynomial in
T
T
T.
Solution: First suppose
α
\alpha
α is a cyclic vector of
T
T
T, then if
dim
V
=
n
\dim V=n
dimV=n, we have
{
α
,
T
α
,
…
,
T
n
−
1
α
}
\{\alpha,T\alpha,\dots,T^{n-1}\alpha\}
{α,Tα,…,Tn−1α} being a basis for
V
V
V. Given an operator
U
U
U which commutes with
T
T
T, we have
U
α
=
a
0
α
+
⋯
+
a
n
−
1
T
n
−
1
α
=
f
(
T
)
α
U\alpha=a_0\alpha+\cdots+a_{n-1}T^{n-1}\alpha=f(T)\alpha
Uα=a0α+⋯+an−1Tn−1α=f(T)α
where
f
(
x
)
=
a
0
+
a
1
x
+
⋯
+
a
n
−
1
x
n
−
1
f(x)=a_0+a_1x+\cdots+a_{n-1}x^{n-1}
f(x)=a0+a1x+⋯+an−1xn−1, notice that
U
T
k
α
=
T
k
U
α
=
T
k
f
(
T
)
α
=
f
(
T
)
T
k
α
,
k
=
2
,
…
n
−
1
UT^k\alpha=T^kU\alpha=T^kf(T)\alpha=f(T)T^k\alpha,\quad k=2,\dots n-1
UTkα=TkUα=Tkf(T)α=f(T)Tkα,k=2,…n−1
We can see that
U
=
f
(
T
)
U=f(T)
U=f(T) on a basis for
V
V
V, thus on
V
V
V.
Conversely, if every linear operator
U
U
U which commutes with
T
T
T is a polynomial in
T
T
T, let the cyclic decomposition of
V
V
V by
T
T
T be
V
=
Z
(
α
1
;
T
)
⊕
⋯
⊕
Z
(
α
r
;
T
)
V=Z(\alpha_1;T)\oplus\cdots\oplus Z(\alpha_r;T)
V=Z(α1;T)⊕⋯⊕Z(αr;T)
and
p
i
p_i
pi is the
T
T
T-annihilator for
α
i
\alpha_i
αi with
p
i
+
1
∣
p
i
p_{i+1}|p_i
pi+1∣pi. Define
U
U
U as follows:
U
α
=
0
U\alpha=0
Uα=0 if
α
∈
Z
(
α
1
;
T
)
\alpha\in Z(\alpha_1;T)
α∈Z(α1;T) and
U
α
=
α
U\alpha=\alpha
Uα=α if
α
∈
Z
(
α
i
;
T
)
,
i
=
2
,
…
,
r
\alpha\in Z(\alpha_i;T),i=2,\dots,r
α∈Z(αi;T),i=2,…,r. For any
β
∈
V
\beta\in V
β∈V, we have
β
=
β
1
+
⋯
+
β
r
\beta=\beta_1+\cdots+\beta_r
β=β1+⋯+βr where each
β
i
∈
Z
(
α
i
;
T
)
\beta_i\in Z(\alpha_i;T)
βi∈Z(αi;T), so
U
T
β
=
U
(
T
β
1
+
T
β
2
+
⋯
+
T
β
r
)
=
U
(
T
β
2
+
⋯
+
T
β
r
)
=
T
(
β
2
+
⋯
+
β
r
)
=
T
(
U
β
2
+
⋯
+
U
β
r
)
=
T
U
β
\begin{aligned}UT\beta&=U(T\beta_1+T\beta_2+\cdots+T\beta_r)=U(T\beta_2+\cdots+T\beta_r)\\&=T(\beta_2+\cdots+\beta_r)=T(U\beta_2+\cdots+U\beta_r)=TU\beta\end{aligned}
UTβ=U(Tβ1+Tβ2+⋯+Tβr)=U(Tβ2+⋯+Tβr)=T(β2+⋯+βr)=T(Uβ2+⋯+Uβr)=TUβ
Then
U
U
U commutes with
T
T
T, thus is a polynomial for
T
T
T. Let
U
=
q
(
T
)
U=q(T)
U=q(T), since
q
(
T
)
α
1
=
0
q(T)\alpha_1=0
q(T)α1=0, we know
p
1
∣
q
p_1|q
p1∣q, which means
p
i
∣
q
p_i|q
pi∣q for
i
≥
2
i\geq 2
i≥2, so
α
i
=
U
α
i
=
q
(
T
)
α
i
=
0
\alpha_i=U\alpha_i=q(T)\alpha_i=0
αi=Uαi=q(T)αi=0 for
i
≥
2
i\geq 2
i≥2, which means
Z
(
α
i
;
T
)
=
{
0
}
Z(\alpha_i;T)=\{0\}
Z(αi;T)={0} for
i
≥
2
i\geq 2
i≥2, so
V
=
Z
(
α
1
;
T
)
V=Z(\alpha_1;T)
V=Z(α1;T) and
T
T
T has a cyclic vector.
20.Let
V
V
V be a finite-dimensional vector space over the field
F
F
F, and let
T
T
T be a linear operator on
V
V
V. We ask when it is true that every non-zero vector in
V
V
V is a cyclic vector for
T
T
T. Prove that this is the case if and only if the characteristic polynomial for
T
T
T is irreducible over
F
F
F.
Solution: Let
dim
V
=
n
\dim V=n
dimV=n. First suppose the characteristic polynomial
f
f
f for
T
T
T is irreducible over
F
F
F, then by the Generalized Cayley-Hamiltion Theorem, the minimal polynomial
p
p
p for
T
T
T is equal to
f
f
f and irreducible over
F
F
F. For any nonzero vector
α
∈
V
\alpha\in V
α∈V, if
α
,
T
α
,
…
,
T
n
−
1
α
\alpha,T\alpha,\dots,T^{n-1}\alpha
α,Tα,…,Tn−1α is linearly dependent, then there is
g
∈
F
[
x
]
g\in F[x]
g∈F[x] with
deg
g
<
n
\deg g<n
degg<n such that
g
(
T
)
α
=
0
g(T)\alpha=0
g(T)α=0, let
p
α
p_{\alpha}
pα be the
T
T
T-annihilator for
α
\alpha
α, then
p
α
∣
g
p_{\alpha}|g
pα∣g and
deg
p
α
<
n
\deg p_{\alpha}<n
degpα<n, further from
α
≠
0
\alpha\neq 0
α=0 we know
deg
p
α
>
1
\deg p_{\alpha}>1
degpα>1, since
p
(
T
)
α
=
0
p(T)\alpha=0
p(T)α=0, we see
p
α
∣
p
p_{\alpha}|p
pα∣p, a contradiction to
p
p
p being irreducible. Then it means
α
,
T
α
,
…
,
T
n
−
1
α
\alpha,T\alpha,\dots,T^{n-1}\alpha
α,Tα,…,Tn−1α is linearly independent, or
Z
(
α
;
T
)
=
V
Z(\alpha;T)=V
Z(α;T)=V.
Conversely, if every non-zero vector in
V
V
V is a cyclic vector for
T
T
T, and assume the characteristic polynomial
f
f
f for
T
T
T is not irreducible over
F
F
F, then if the minimal polynomial
p
p
p for
T
T
T is not the same as
f
f
f, we have
deg
p
<
deg
f
=
n
\deg p<\deg f=n
degp<degf=n, there is a vector
α
∈
V
\alpha\in V
α∈V such that the
T
T
T-annihilator for
α
\alpha
α is
p
p
p, so
Z
(
α
;
T
)
Z(\alpha;T)
Z(α;T) has dimension
deg
p
\deg p
degp, which means
α
\alpha
α is not a cyclic vector for
T
T
T.
If
p
=
f
p=f
p=f and
f
=
g
h
f=gh
f=gh where
deg
g
≥
1
,
deg
h
≥
1
\deg g\geq1,\deg h\geq1
degg≥1,degh≥1, then it is apparent
deg
h
<
n
\deg h<n
degh<n, let
h
=
h
0
+
h
1
x
+
⋯
+
x
k
h=h_0+h_1x+\cdots+x^k
h=h0+h1x+⋯+xk, there is a vector
α
∈
V
\alpha\in V
α∈V such that the
T
T
T-annihilator for
α
\alpha
α is
p
p
p, thus
g
(
T
)
h
(
T
)
α
=
0
g(T)h(T)\alpha=0
g(T)h(T)α=0, and
β
=
g
(
T
)
α
≠
0
\beta=g(T)\alpha\neq 0
β=g(T)α=0. Notice that
h
0
β
+
h
1
T
β
+
⋯
+
T
k
β
=
h
(
T
)
β
=
h
(
T
)
g
(
T
)
α
=
0
h_0\beta+h_1T\beta+\cdots+T^k\beta=h(T)\beta=h(T)g(T)\alpha=0
h0β+h1Tβ+⋯+Tkβ=h(T)β=h(T)g(T)α=0
this shows
β
,
T
β
,
…
,
T
k
β
\beta,T\beta,\dots,T^k\beta
β,Tβ,…,Tkβ is linearly dependent, thus by Theorem 1,
dim
Z
(
β
;
T
)
≤
k
=
deg
h
<
n
\dim Z(\beta;T)\leq k=\deg h<n
dimZ(β;T)≤k=degh<n, so
β
\beta
β is not a cyclic vector for
T
T
T.
21.Let
A
A
A be an
n
×
n
n\times n
n×n matrix with real entries. Let
T
T
T be the linear operator on
R
n
R^n
Rn which is represented by
A
A
A in the standard ordered basis, and let
U
U
U be the linear operator on
C
n
C^n
Cn which is represented by
A
A
A in the standard ordered basis. Use the result of Exercise 20 to prove the following: If the only subspace invariant under
T
T
T are
R
n
R^n
Rn and the zero subspace, then
U
U
U is diagonalizable.
Solution: Since
A
A
A is real, the characteristic polynomial for
T
T
T and
U
U
U are equal, both are
f
=
det
(
x
I
−
A
)
f=\det(xI-A)
f=det(xI−A). Now given any nonzero vector
α
∈
R
n
\alpha\in R^n
α∈Rn, the cyclic space
Z
(
α
;
T
)
Z(\alpha;T)
Z(α;T) must be
R
n
R^n
Rn since it is invariant under
T
T
T and contains
α
\alpha
α, so by Exercise 20,
f
f
f is irreducible over
R
R
R, which means
f
f
f must be of the form
x
−
c
x-c
x−c or
x
2
+
d
x^2+d
x2+d where
d
>
0
d>0
d>0. Then in the field
C
C
C,
f
f
f can be factored into prime factors and thus
U
U
U is diagonalizable.
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7.2 Cyclic Decompositions and the Rational Form
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