Linear System Theory and Design Answer

8/20/2019 Linear System Theory and Design Answer

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Linear System Theory and Design SA01010048 LING QING

1

2.1 Consider the memoryless system with characteristics shown in Fig 2.19, in which u denotes

the input and y the output. Which of them is a linear system? Is it possible to introduce a new

output so that the system in Fig 2.19(b) is linear?

Figure 2.19

Translation: 考虑具有图 2.19中表示的特性的无记忆系统。其中u表示输入， y表示输出。下面哪一个是线性系统？可以找到一个新的输出，使得图 2.19(b)中的系统是线性

的吗？

Answer: The input-output relation in Fig 2.1(a) can be described as:

ua y *=

Here a is a constant. It is a memoryless system. Easy to testify that it is a linear system.

The input-output relation in Fig 2.1(b) can be described as:

bua y += *

Here a and b are all constants. Testify whether it has the property of additivity. Let:

bua y += 11 *

bua y += 22 *

then:

buua y y *2)(*)( 2121 ++=+

So it does not has the property of additivity, therefore, is not a linear system.

But we can introduce a new output so that it is linear. Let:

b y z −=

ua z *= z is the new output introduced. Easy to testify that it is a linear system.

The input-output relation in Fig 2.1(c) can be described as:

uua y *)(=

a(u) is a function of input u. Choose two different input, get the outputs:

111 *ua y =


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2

222 *ua y =

Assure:

21 aa ≠

then:

221121 **)( uaua y y +=+

So it does not has the property of additivity, therefore, is not a linear system.

2.2 The impulse response of an ideal lowpass filter is given by

)(2

)(2sin2)(

0

0

t t

t t t g

−

−=

ω

ω ω

for all t, where w and to are constants. Is the ideal lowpass filter causal? Is is possible to built

the filter in the real world?

Translation: 理想低通滤波器的冲激响应如式所示。对于所有的 t，w 和 to，都是常数。理

想低通滤波器是因果的吗？现实世界中有可能构造这种滤波器吗？

Answer: Consider two different time: ts and tr, ts < tr, the value of g(ts-tr) denotes the output at

time ts, excited by the impulse input at time tr. It indicates that the system output at time

ts is dependent on future input at time tr. In other words, the system is not causal. We

know that all physical system should be causal, so it is impossible to built the filter in

the real world.

2.3 Consider a system whose input u and output y are related by

>

≤==

at for

at for t ut uPt y a

0

)(:))(()(

where a is a fixed constant. The system is called a truncation operator, which chops off the

input after time a. Is the system linear? Is it time-invariant? Is it causal?

Translation: 考虑具有如式所示输入输出关系的系统， a是一个确定的常数。这个系统称作

截断器。它截断时间 a之后的输入。这个系统是线性的吗？它是定常的吗？是因果的吗？

Answer: Consider the input-output relation at any time t, ta:

0= y

Easy to testify that it is linear. So for any time, the system is linear.

Consider whether it is time-invariable. Define the initial time of input to, system input is

u(t), t>=to. Let to=to:


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≤≤

=t other for

at t for t ut y

0

)()(

0

Shift the initial time to to+T . Let to+T>a , then input is u(t-T), t>=to+T . System output:

0)(' =t y

Suppose that u(t) is not equal to 0, y’(t) is not equal to y(t-T). According to the definition,

this system is not time-invariant.

For any time t, system output y(t) is decided by current input u(t) exclusively. So it is a

causal system.

2.4 The input and output of an initially relaxed system can be denoted by y=Hu, where H is some

mathematical operator. Show that if the system is causal, then

u HPP HuP yPaaaa

==

where Pa is the truncation operator defined in Problem 2.3. Is it true PaHu=HPau?

Translation: 一个初始松弛系统的输入输出可以描述为： y=Hu，这里 H 是某种数学运算，

说明假如系统是因果性的，有如式所示的关系。这里 Pa是题 2.3中定义的截断函

数。 PaHu=HPau是正确的吗？

Answer: Notice y=Hu, so:

HuP yP aa =

Define the initial time 0, since the system is causal, output y begins in time 0.

If a0, we can divide u to 2 parts:

≤≤

=t other for

at for t ut p

0

0)()(

>

=t other for

at for t ut q

0

)()(

u(t)=p(t)+q(t). Pay attention that the system is casual, so the output excited by q(t) can’t

affect that of p(t). It is to say, system output from 0 to a is decided only by p(t). Since

PaHu chops off Hu after time a, easy to conclude PaHu=PaHp(t). Notice that p(t)=Pau,

also we have:

u HPP HuP aaa =

It means under any condition, the following equation is correct:

u HPP HuP yPaaaa

==

PaHu=HPau is false. Consider a delay operator H , Hu(t)=u(t-2), and a=1, u(t) is a step

input begins at time 0, then PaHu covers from 1 to 2, but HPau covers from 1 to 3.


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2.5 Consider a system with input u and output y. Three experiments are performed on the system

using the inputs u1(t), u2(t) and u3(t) for t>=0. In each case, the initial state x(0) at time t=0 is

the same. The corresponding outputs are denoted by y1,y2 and y3. Which of the following

statements are correct if x(0)0?

1. If u3=u1+u2, then y3=y1+y2.

2. If u3=0.5(u1+u2), then y3=0.5(y1+y2).

3. If u3=u1-u2, then y3=y1-y2.

Translation: 考虑具有输入 u 输出 y的系统。在此系统上进行三次实验，输入分别为 u1(t),

u2(t) 和 u3(t)， t>=0。每种情况下，零时刻的初态 x(0)都是相同的。相应的输出表

示为 y1,y2 和 y3。在 x(0)不等于零的情况下，下面哪种说法是正确的？

Answer: A linear system has the superposition property:

02211

02211

022011),()(

),()(

)()(t t t yt y

t t t ut u

t xt x≥+→

≥+

+α α

α α

α α

In case 1:

11 =α 12 =α

)0()0(2)()( 022011 x xt xt x ≠=+α α

So y3y1+y2.

In case 2:

5.01 =α 5.02 =α

)0()()( 022011 xt xt x =+α α

So y3=0.5(y1+y2).

In case 3:

11 =α 12 −=α

)0(0)()( 022011 xt xt x ≠=+α α

So y3y1-y2.

2.6 Consider a system whose input and output are related by

=−

≠−−=

0)1(0

0)1()1(/)()(

2

t uif

t uif t ut ut y

for all t.

Show that the system satisfies the homogeneity property but not the additivity property.

Translation: 考虑输入输出关系如式的系统,证明系统满足齐次性,但是不满足可加性.

Answer: Suppose the system is initially relaxed, system input:

)()( t ut p α =

a is any real constant. Then system output q(t):


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=−

≠−−=

0)1(0

0)1()1(/)()(

2

t pif

t pif t pt pt q

=−

≠−−=

0)1(0

0)1()1(/)(2

t uif

t uif t ut uα

So it satisfies the homogeneity property.

If the system satisfies the additivity property, consider system input m(t) and n(t),

m(0)=1, m(1)=2; n(0)=-1, n(1)=3. Then system outputs at time 1 are:

4)0(/)1()1( 2 == mmr

9)0(/)1()1( 2 −== nns

0)]0()0(/[)]1()1([)1( 2 =++= nmnm y

）1()1( sr +≠

So the system does not satisfy the additivity property.

2.7 Show that if the additivity property holds, then the homogeneity property holds for all rational

numbers a . Thus if a system has “continuity” property, then additivity implies homogeneity.

Translation: 说明系统如果具有可加性，那么对所有有理数 a具有齐次性。因而对具有某种

连续性质的系统，可加性导致齐次性。

Answer: Any rational number a can be denoted by:

nma /= Here m and n are both integer. Firstly, prove that if system input-output can be described

as following:

y x →

then:

mymx →

Easy to conclude it from additivity.

Secondly, prove that if a system input-output can be described as following:

y x →

then:

n yn x // →

Suppose:

un x →/ Using additivity:

nu xn xn →=)/(*

So:

nu y =

n yu /=


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It is to say that:

n yn x // →

Then:

nm ynm x /*/* →

ayax →

It is the property of homogeneity.

2.8 Let g(t,T)=g(t+a,T+a) for all t,T and a. Show that g(t,T) depends only on t-T .

Translation: 设对于所有的 t,T 和 a， g(t,T)=g(t+a,T+a)。说明 g(t,T)仅依赖于 t-T 。

Answer: Define:

T t x += T t y −=

So:

2

y xt

+=

2

y xT

−=

Then:

)2

,2

(),( y x y x

gT t g −+

=

)2

,2

( a y x

a y x

g +−

++

=

)

22

,

22

( y x y x y x y x

g +−

+−+−

++

=

)0,( yg=

So:

0)0,(),(=

∂

∂=

∂

∂

x

yg

x

T t g

It proves that g(t,T) depends only on t-T .

2.9 Consider a system with impulse response as shown in Fig2.20(a). What is the zero-state

response excited by the input u(t) shown in Fig2.20(b)?

Fig2.20


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Translation: 考虑冲激响应如图 2.20(a)所示的系统，由如图 2.20(b)所示输入 u(t)激励的零状

态响应是什么？

Answer: Write out the function of g(t) and u(t):

≤≤−

≤≤=

212

10)(

t t

t t t g

≤≤−

≤≤=

211

101)(

t

t t u

then y(t) equals to the convolution integral:

∫ −=t

dr r t ur gt y0

)()()(

If 0=


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2.10 Consider a system described by

uu y y y −=−+ ••••

32

What are the transfer function and the impulse response of the system?

Translation: 考虑如式所描述的系统，它的传递函数和冲激响应是什么？

Answer: Applying the Laplace transform to system input-output equation, supposing that the

System is initial relaxed:

)()()(3)(2)(2 sY ssY sY ssY sY s −=−+

System transfer function:

3

1

32

1

)(

)()(

2 +=

−+

−==

sss

s

sY

sU sG

Impulse response:

t e

s LsG Lt g

311 ]3

1[)]([)( −−− =+

==

2.11 Let y(t) be the unit-step response of a linear time-invariant system. Show that the impulse

response of the system equals dy(t)/dt.

Translation: y(t)是线性定常系统的单位阶跃响应。说明系统的冲激响应等于 dy(t)/dt.

Answer: Let m(t) be the impulse response, and system transfer function is G(s):

ssGsY

1*)()( =

)()( sGs M =

ssY s M *)()( =

So:

dt t dyt m /)()( =

2.12 Consider a two-input and two-output system described by

)()()()()()()()(212111212111

t u p N t u p N t y p Dt y p D +=+

)()()()()()()()( 222121222121 t u p N t u p N t y p Dt y p D +=+

where Nij and Dij are polynomials of p:=d/dt. What is the transfer matrix of the system?

Translation: 考虑如式描述的两输入两输出系统， Nij 和 Dij 是 p:=d/dt的多项式。系统的

传递矩阵是什么？

Answer: For any polynomial of p, N(p), its Laplace transform is N(s).

Applying Laplace transform to the input-output equation:

)()()()()()()()( 212111212111 sU s N sU s N sY s DsY s D +=+

)()()()()()()()( 222121222121 sU s N sU s N sY s DsY s D +=+


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Write to the form of matrix:

)(

)(

)()(

)()(

2

1

2221

1211

sY

sY

s Ds D

s Ds D=

)(

)(

)()(

)()(

2

1

2221

1211

sU

sU

s N s N

s N s N

)(

)(

2

1

sY

sY =

1

2221

1211

)()(

)()( −

s Ds D

s Ds D

)(

)(

)()(

)()(

2

1

2221

1211

sU

sU

s N s N

s N s N

So the transfer function matrix is:

)(sG =

1

2221

1211

)()(

)()( −

s Ds D

s Ds D

)()(

)()(

2221

1211

s N s N

s N s N

By the premising that the matrix inverse:

1

2221

1211

)()(

)()( −

s Ds D

s Ds D

exists.

2.11 Consider the feedback systems shows in Fig2.5. Show that the unit-step responses of the

positive-feedback system are as shown in Fig2.21(a) for a=1 and in Fig2.21(b) for a=0.5.

Show also that the unit-step responses of the negative-feedback system are as shown in Fig

2.21(c) and 2.21(d), respectively, for a=1 and a=0.5.

Fig 2.21

Translation: 考虑图 2.5中所示反馈系统。说明正反馈系统的单位阶跃响应，当 a=1时，如

图 2.21(a)所示。当 a=0.5 时，如图 2.21(b)所示。说明负反馈系统的单位阶跃响应

如图 2.21(c)和 2.21(b)所示，相应地，对 a=1和 a=0.5。


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Answer: Firstly, consider the positive-feedback system. It’s impulse response is:

∑∞

=

−=1

)()(i

iit at g δ

Using convolution integral:

∑∞

=

−=1

)()(i

i it r at y

When input is unit-step signal:

∑=

=n

i

ian y1

)(

)()( n yt y = 1+≤≤ nt n

Easy to draw the response curve, for a=1 and a=0.5, respectively, as Fig 2.21(a) and Fig

2.21(b) shown.

Secondly, consider the negative-feedback system. It’s impulse response is:

∑∞

=

−−−=1

)()()(i

iit at g δ

Using convolution integral:

∑∞

=

−−−=1

)()()(i

i it r at y

When input is unit-step signal:

∑=

−−=n

i

ian y

1

)()(

)()( n yt y = 1+≤≤ nt n

Easy to draw the response curve, for a=1 and a=0.5, respectively, as Fig 2.21(c) and Fig

2.21(d) shown.

2.14 Draw an op-amp circuit diagram for

u x x

−+

−=

•

4

2

50

42

[ ] u x y 2103 −=

2.15 Find state equations to describe the pendulum system in Fig 2.22. The systems are useful to

model one- or two-link robotic manipulators. If θ , 1θ and 2θ are very small, can you

consider the two systems as linear?


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Translation: 试找出图 2.22所示单摆系统的状态方程。这个系统对研究一个或两个连接的机

器人操作臂很有用。假如角度都很小时，能否考虑系统为线性？

Answer: For Fig2.22(a), the application of Newton’s law to the linear movements yields:

)cossin()cos(cos

2

2

2

θ θ θ θ θ θ

•••

−−==− mlldt

d

mmg f

)sincos()sin(sin2

2

2

θ θ θ θ θ θ

•••

−==− mlldt

d m f u

Assuming θ and•

θ to be small, we can use the approximation θ sin =θ , θ cos =1.

By retaining only the linear terms in θ and•

θ , we obtain mg f = and:

umll

g 1+−=

••

θ θ

Select state variables as θ =1 x ,•

=θ 2 x and output θ = y

uml

xlg

x

+

−=

•

/1

1

0/

10

[ ] x y 01=

For Fig2.22(b), the application of Newton’s law to the linear movements yields:

)cos(coscos 1122

112211 θ θ θ ldt

d mgm f f =−−

)cossin( 12

11111 θ θ θ θ

•••

−−= lm

)sin(sinsin 112

2

11122 θ θ θ ldt

d m f f =−

)sincos( 12

11111 θ θ θ θ

•••

−= lm

)coscos(cos 22112

2

2222 θ θ θ lldt

d mgm f +=−

)cossin( 12

11112 θ θ θ θ

•••

−−= lm )cossin( 22

22222 θ θ θ θ

•••

−−+ lm

)sinsin(sin 22112

2

222 θ θ θ lldt

d m f u +=−

)sincos( 12

11112 θ θ θ θ

•••

−= lm )sincos( 22

22222 θ θ θ θ

•••

−+ lm


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Assuming 1θ , 2θ and 1•

θ , 2•

θ to be small, we can use the approximation 1sinθ = 1θ ,

2sinθ = 2θ , 1cosθ =1, 2cosθ =1. By retaining only the linear terms in 1θ , 2θ and

1

•

θ , 2•

θ , we obtain gm f 22 = , gmm f )( 211 += and:

2

11

21

11

211

)(θ θ θ

lm

gm

lm

gmm+

+−=

••

ulmlm

gmm

lm

gmm

22

2

21

211

21

212

1)()(+

+−

+=

••

θ θ θ

Select state variables as 11 θ = x , 12•

=θ x , 23 θ = x , 24•

=θ x and output

=

2

1

2

1

θ

θ

y

y:

+−+

+−=

•

•

•

•

4

3

2

1

22212121

1121121

4

3

2

1

0/)(0/)(

1000

0/0/)(

0010

x

x

x

x

lmgmmlmgmm

lmgmlmgmm

x

x

x

x

+ u

lm

22/1

0

0

0

=

0100

0001

2

1

y

y

4

3

2

1

x

x

x x

2.17 The soft landing phase of a lunar module descending on the moon can be modeled as shown

in Fig2.24. The thrust generated is assumed to be proportional to the derivation of m, where

m is the mass of the module. Then the system can be described by

mgmk ym −−=

•••

Where g is the gravity constant on the lunar surface. Define state variables of the system as:

y x =1 ,•

= y x2 , m x =3 ,•

= m y

Find a state-space equation to describe the system.

Translation: 登月舱降落在月球时，软着陆阶段的模型如图 2.24 所示。产生的冲激力与 m

的微分成正比。系统可以描述如式所示形式。g是月球表面的重力加速度常数。定

义状态变量如式所示，试图找出系统的状态空间方程描述。

Answer: The system is not linear, so we can linearize it.

Suppose:


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−

+−= ygt y 2/2

−••

+−= ygt y

−••••

+−= yg y

−

+= mmm 0

−••

= mm

So:

)( 0

−

+ mm

−••

+− )( yg =

−•

− mk gmm )( 0−

+−

+−− −

gmgm0

−••

= ym0

−•

− mk gmgm−

−− 0

−••

= ym0

−•

− mk

Define state variables as:

−−

= y x1 ,

−•−

= y x 2 ,−−

= m x3 ,

−•−

= m y

Then:

•−

•−

•−

3

2

1

x

x

x

=

000

000

010

−

−

−

3

2

1

x

x

x

+

−

1

/

0

0mk −

u

−

y = [ ]001

−

−

−

3

2

1

x

x

x

2.19 Find a state equation to describe the network shown in Fig2.26.Find also its transfer function.

Translation: 试写出描述图 2.26所示网络的状态方程，以及它的传递函数。

Answer: Select state variables as:

1 x : Voltage of left capacitor

2 x : Voltage of right capacitor

3 x : Current of inductor

Applying Kirchhoff’s current law:


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3 x = R x L x /)( 32

•

−

R x xC u 11+= •

32 x xC u += •

2 x y =

From the upper equations, we get:

C uCR x x //11 +−=•

u x +−= 1

C uC x x //32 +−=•

u x +−= 3

L Rx L x x // 323 −=•

32 x x −=

2 x y =

They can be combined in matrix form as:

•

•

•

3

2

1

x

x

x

=

−

−

−

110

100

001

3

2

1

x

x

x

+ u

0

1

1

[ ]

=

3

2

1

010

x

x

x

y

Use MATLAB to compute transfer function. We type:

A=[-1,0,0;0,0,-1;0,1,-1];

B=[1;1;0];

C=[0,1,0];

D=[0];

[N1,D1]=ss2tf(A,B,C,D,1)Which yields:

N1 =

0 1.0000 2.0000 1.0000

D1 =

1.0000 2.0000 2.0000 1.0000

So the transfer function is:

122

12)(

23

2^

+++

++=

sss

sssG

1

12 ++

+=

ss

s


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2.20 Find a state equation to describe the network shown in Fig2.2. Compute also its transfer

matrix.

Translation: 试写出描述图 2.2所示网络的状态方程，计算它的传递函数矩阵。

Answer: Select state variables as Fig 2.2 shown. Applying Kirchhoff’s current law:

1u = 323121111 x R x L x x xC R ++++ ••

22211

••

=+ xC u xC

223

•

= xC x

3231212311 )( x R x L x xu x Ru ++++−= •

3231 x R x L y += •

From the upper equations, we get:

12131 // C uC x x −=•

232 /C x x =•

12111131112113 ///)(// Lu R Lu L x R R L x L x x +++−−−=•

2113121 u Ru x R x x y ++−−−=

They can be combined in matrix form as:

•

•

•

3

2

1

x

x

x

=

+−−− 12111

2

1

/)(

/100

/100

L R R L L

C

C

3

2

1

x

x

x

+

−

2

1

11

1

1 /

0

/1

/1

0

0

u

u

L R

C

L

[ ]

−−−=

3

2

1

111

x

x

x

R y + [ ]

2

1

21u

u R

Applying Laplace Transform to upper equations:

)(/1/1

)( 1^

21111

21^

susC sC R Rs L

Rs Ls y

++++

+=

)(/1/1

)/1)((2

^

21111

2121 susC sC R Rs L

sC R Rs L

++++

+++

++++

+=

sC sC R Rs L

Rs LsG

21111

21^

/1/1)(

++++

++

sC sC R Rs L

sC R Rs L

21111

2121

/1/1

)/1)((


16/106


16

2.18 Find the transfer functions from u to 1 y and from 1 y to y of the hydraulic tank system

shown in Fig2.25. Does the transfer function from u to y equal the product of the two

transfer functions? Is this also true for the system shown in Fig2.14?

Translation: 试写出图 2.25所示水箱系统从u到 1 y 的传递函数和从 1 y 到 y的传递函数。从

u到 y的传递函数等于两个传递函数的乘积吗？这对图2.14所示系统也是正确的吗？

Answer: Write out the equation about u , 1 y and y :

111 / R x y =

222 / R x y =

dt yudx A /)(111

−=

dt y ydx A /)( 122 −=

Applying Laplace transform:

)1/(1/ 11

^

1

^

s R Au y +=

)1/(1/ 221

^^

s R A y y +=

)1)(1/(1/ 2211

^^

s R As R Au y ++=

So:

=^^

/u y )/(^

1

^

u y )/( 1

^^

y y

But it is not true for Fig2.14, because of the loading problem in the two tanks.


17/106

3.1consider Fig3.1 ,what is the representation of the vector x with respect to the basis ( )21 iq ?

What is the representation of 1q with respect ro ( )22 qi ?图 3.1 中,向量 x 关于 ( )21 iq 的表

示是什么? 1q 关于 ( )22 qi 的表示有是什么?

If we draw from x two lines in parallel with 2i and 1q , they tutersect at 13

1q and 2

3

8i as

shown , thus the representation of x with respect to the basis ( )21 iq is′

3

8

3

1, this can

be verified from [ ]

=

=

=

38

31

11

03

3

83

1

3

122 iq x

To find the representation of 1q with respect to ( )22 qi ,we draw from 1q two lines in

parallel with 2q and 2i , they intersect at -2 2i and 22

3q , thus the representation of 1q

with respect to ( )22 qi , is′

−

2

32 , this can be verified from

[ ]

−

=

−=

=

23

2

21

20

23

2

1

3

221

qiq

3.2 what are the 1-morm ,2-norm , and infinite-norm of the vectors [ ] [ ]′=′−= 111,132 21 x x , 问向量

[ ] [ ]′=′−= 111,132 21 x x 的 1-范数,2-范数和 −∞ 范数是什么?

1max3max

3)111(14)1)3(2(

31116132

2211

21

222

222

1222

1

'

121

12

3

1

11

1

=====++==+−+==

=++==++==

∞∞

=∑

iiii

i

i

x x x x x x

x x x x

x x x

3.3 find two orthonormal vectors that span the same space as the two vectors in problem 3.2 , 求与题 3.2 中的两个向量

张成同一空间的两个标准正交向量.

Schmidt orthonormalization praedure ,


18/106

[ ] [ ]

[ ]′===−=

′−==

′−==

1113

1)(

13214

1132

2

22212

'

122

1

1111

u

uq xq xq xu

u

uq xu

The two orthomormal vectors are

=

−=

1

1

1

3

1

1

3

2

14

121 qq

In fact , the vectors 1 x and 2 x are orthogonal because 01221 =′=′ x x x x so we can only

normalize the two vectors

==

−==

1

1

1

3

1

1

3

2

14

1

2

2

2

1

1

1 x

xq

x

xq

3.4 comsider an mn × matrix A with mn ≥ , if all colums of A are orthonormal , then

m I A A =′ , what can you say abort A A ′ ? 一个 mn× 阶矩阵A( mn ≥ ), 如果 A 的所有列都是

标准正交的,则 m I A A =′ 问 A A ′ 是怎么样?

Letmnijm

aaaa A×

== 21 , if all colums of A are orthomormal , that is

∑=

=≠==

n

i

liliii jiif jiif aaaa

1

'

10

then

[ ]

[ ]

=≠

≠≠=

==

=′

=

=

=

=′

∑

∑∑

=

×==

jiif

jiif aageneralin

aaaa

a

a

a

aaa A A

I

aaaaaa

aaaaaa

aaa

a

a

a

A A

jl

m

i

il

nn jl

m

i

il

m

i

ii

m

mi

m

mmmm

m

mi

m

1

0

)(

100

010

001

1

11

'

'

'2

'

1

''

2

'

'

2

'

1

'

'

12

'

11

'

1

''

2

'

'

'

2

'

1

if A is a symmetric square matrix , that is to say , n=m jlil aa for every nli 2,1, =

then m I A A =′


19/106

3.5 find the ranks and mullities of the following matrices 求下列矩阵的秩和化零度

−−=

−

=

=

1000

2210

4321

011

023

114

100

000

010

321 A A A

134)(033)(123)(

3)(3)(2)(

321

321

=−==−==−=

===

A Nullity A Nullity A Nullity

Arank Arank Arank

3.6 Find bases of the range spaces of the matrices in problem 3.5 求题 3.5 中矩阵值域空间的基

the last two columns of 1 A are linearly independent , so the set

1

0

0

,

0

0

1

can be used as a

basis of the range space of 1 A , all columns of 2 A are linearly independent ,so the set

−

0

0

1

1

2

2

1

3

4

can be used as a basis of the range spare of 2 A

let 43213 aaaa A = ,where ia denotes of 3 A 4321 aand aand aand a are

linearly independent , the third colums can be expressed as 213 2aaa +−= , so the set

321 aaa can be used as a basis of the range space of 3 A

3.7 consider the linear algebraic equation x x −

=

−

−

−

1

0

1

21

33

12

it has three equations and two

unknowns does a solution x exist in the equation ? is the solution unique ? does a solution exist if

[ ]′= 111 y ?

线性代数方程 x x −

=

−

−

−

1

0

1

21

33

12

有三个方程两个未知数, 问方程是否有解?若有解是否

唯一?若 [ ]′= 111 y 方程是否有解?


20/106

Let [ ],21

33

12

21 aa A =

−

−

−

= clearly 1a and 2a are linearly independent , so rank(A)=2 ,

y is the sum of 1a and 2a ,that is rank([A y ])=2, rank(A)= rank([A y ]) so a solution x

exists in A x = y

Nullity(A)=2- rank(A)=0

The solution is unique [ ]′= 11 x if [ ]′= 111 y , then rank([A y ])=3 ≠ rank(A), that is

to say there doesn’t exist a solution in A x = y

3.8 find the general solution of

=

−−

1

2

3

1000

2210

4321

x how many parameters do you have ?

求方程

=

−−

1

2

3

1000

2210

4321

x 的同解,同解中用了几个参数?

Let

=

−−=12

3

10002210

4321

y A we can readily obtain rank(A)= rank([A y ])=3 so

this y lies in the range space of A and [ ]′−= 101 o x p is a solution Nullity(A)=4-3=1 that

means the dimension of the null space of A is 1 , the number of parameters in the general solution

will be 1 , A basis of the null space of A is [ ]′−= on 121 thus the general solution of

A x = y can be expressed an

−

+

−

=+=

0

1

2

1

1

0

0

1

α α n x x p for any real α

α is the only parameter

3.9 find the solution in example 3.3 that has the smallest Euclidean norm 求例 3 中具有最小欧氏

范数的解,


21/106

the general solution in example 3.3 is

−

−

−+=

−

+

−+

−=

2

1

21

1

21

42

1

0

2

0

0

1

1

1

0

0

4

0

α

α

α α

α

α α x for

any real 1α and 2α

the Euclidean norm of x is

16168453

)()()42(

2121

2

2

2

1

2

2

2

1

2

21

2

1

+−−++=

−+−+−++=

α α α α α α

α α α α α x

−

−

−=⇒

=

=⇒

=−+⇒=∂

=−+⇒=∂

11

1611

411

811

4

11

1611

4

082502

042302

2

1

12

2

21

1 x x

x

α

α

α α α

α α α

has the smallest Euclidean

norm ,

3.10 find the solution in problem 3.8 that has the smallest Euclidean norm 求题 3.8 中欧氏范数

最小的解,

−+

−

=

0

1

2

1

1

0

0

1

α x for any real α

the Euclidean norm of x is

−

−

=⇒=⇒=−⇒=∂∂

+−=++−+−=

16

163

6

5

6

102120

2261)2()1( 2222

x x

x

α α α

α α α α α

has the

smallest Euclidean norm

3.11 consider the equation ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnub Aub Aub A x An x nnn

where A is an nn× matrix and b is an 1×n column vector ,under what conditions on A and b will there


22/106

exist ]1[],1[],[ −nuuou to meet the equation for any ]0[],[ xand n x ?

令 A 是 nn × 的矩阵, b 是 1×n 的列向量,问在 A 和b 满足什么条件时,存在 ]1[],1[],[ −nuuou ,对所有的

]0[],[ xand n x ,它们都满足方程 ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnub Aub Aub A x An x nnn ,

write the equation in this form

−

−

== −

]0[

]2[

]1[

],[]0[}{ 1

u

nu

nu

b Ab Ab x An x nn

where ],[ 1b Ab Ab n− is an nn× matrix and ]0[}{ x An x n− is an 1×n column vector , from the equation we

can see , ]1[],1[],[ −nuuou exist to meet the equation for any ]0[],[ xand n x ,if and only if

nb Ab Ab n =− ],[ 1 ρ under this condition , there will exist ]1[],1[],[ −nuuou to meet the equation for any

]0[],[ xand n x .

3.12 given

=

=

=

1

3

2

1

1

0

0

1000

0200

0120

0012

bb A what are the representations of A with respect to

the basis b Ab Ab Ab 32 and the basis b Ab Ab Ab 32 , respectively? 给定

=

=

=

1

3

2

1

1

0

0

1000

0200

0120

0012

bb A 请问 A 关于 b Ab Ab Ab 32

和基 b Ab Ab Ab 32 的表

示分别是什么?

=⋅=

=⋅=

=

1

16

32

24

,

1

4

4

1

,

1

2

1

0

232 b A Ab Ab A Ab Ab A

we have[ ] [ ]

−

−

==∴

+−+−=

7100

18010

20001

8000

718208

3232

324

b Ab Ab Abb Ab Ab Ab A

b Ab Ab Abb A

thus the representation of A


23/106

with respect to the basis b Ab Ab Ab 32 is

−

−

=

7100

18010

20001

8000

A

b Ab Ab Abb A

b Ab Ab Ab A

324

432

718208

1

48

128

152

,

1

24

52

50

,

1

12

20

15

,

1

6

7

4

+−+−=

=

=

=

=

[ ] [ ]

−

−

==∴

710018010

20001

8000

3232 b Ab Ab Abb Ab Ab Ab A thus the representation of A with

respect to the basis b Ab Ab Ab 32 is

−

−

=

7100

18010

20001

8000

A

3.13 find Jordan-form representations of the following matrices

写出下列矩阵的 jordan 型表示:

.

20250

16200

340

.

200

010

101

.

342

100

010

,

300

020

1041

4321

−−

=

−

=

−−−

=

= A A A A

the characteristic polynomial of 1 A is )3)(2)(1()det( 11 −−−=−=∆ λ λ λ λ A I thus the eigenvelues of 1 A are

1 ,2 , 3 , they are all distinct . so the Jordan-form representation of 1 A will be diagonal .

the eigenvectors associated with 3,2,1 === λ λ λ ,respectively can be any nonzero solution of

[ ]

[ ]

[ ]′=⇒=

′=⇒=

′=⇒=

1053

0142

001

3331

2221

1111

qqq A

qqq A

qqq A

thus the jordan-form representation of 1 A with respect to

321 qqq is

=

300

020

001

ˆ1 A

the characteristic polynomial of 2 A is


24/106

223

22 )1)(1)(1(243()det()( Aii A I −++++=+++=−=∆ λ λ λ λ λ λ λ λ has

eigenvalues jand j −−+−− 11,1 the eigenvectors associated with

jand j −−+−− 11,1 are , respectively

[ ] [ ] [ ]′−−′−−′− j jand j j 211211,111 the we have

Q AQ

j

j Aand

j j

j jQ 21

2

100

010

001

ˆ

220

111

111−=

−−

+−

−

=

−

−−+−−=

the characteristic polynomial of 3 A is )2()1()det()( 2

33 −−=−=∆ λ λ λ λ A I theus the

eigenvalues of 3 A are 1 ,1 and 2 , the eigenvalue 1 has multiplicity 2 , and nullity

213)(3)( 33 =−=−−=− I Arank I A the 3 A has two tinearly independent eigenvectors

associated with 1 ,[ ] [ ]

[ ]′−=⇒=−

′=

′=⇒=−

1010)(

0100010)(

333

213

qq I A

qqq I A thus we have

Q AQ Aand Q 31

3

200010

001

ˆ100011

101−

=

=

−−=

the characteristic polynomial of 4 A is3

44 )det()( λ λ λ =−=∆ A I clearly 4 A has lnly one

distinct eigenvalue 0 with multiplicity 3 , Nullity( 4 A -0I)=3-2=1 , thus 4 A has only one

independent eigenvector associated with 0 , we can compute the generalized , eigenvectors

of 4 A from equations below

[ ]

[ ]

[ ]′−=⇒=

′−=⇒=

′=⇒=

430

540

0010

3231

2121

111

vvv A

vvv A

vv A

, then the representation of 4 A

with respect to the basis 321 vvv is

−

−==

= −

450

340

001

000

100

010

ˆ4

1

4 QwhereQ AQ A


25/106

3.14 consider the companion-form matrix

−−−−

=

0100

0010

0001

4321 α α α α

A show that its

characterisic polynomial is given by 433

2

3

1

4)( α λ α λ α λ α λ λ ++++=∆

show also that if iλ is an eigenvalue of A or a solution of 0)( =∆ λ then

133 ′iii λ λ λ is an eigenvector of A associated with iλ

证明矩阵 A的特征项式 433

2

3

1

4)( α λ α λ α λ α λ λ ++++=∆ 并且,如果 iλ 是 iλ 的一

个特征值, 0)( =∆ λ 的一个解, 那么向量

[ ]133 ′

iii

λ λ λ 是 A 关于i

λ 的一个特征向量

proof:

43223

14

432

31

4

432

1

4321

2

1det

1

0det

10

01det

10

01

00

det)(

100

010

001det)det()(

α λ α λ α λ α λ

λ

α α

λ

λ α λ α λ

λ

λ

α α α

λ

λ

λ

α λ

λ

λ

λ

α α α α λ

λ λ

++++=

−+

−++=

−

−+

−

−+=

−

−

−

+

=−=∆ A I

if iλ is an eigenvalue of A , that is to say , 0)( =∆ iλ then we have

=

=

−−−−

=

−−−−

1110100

0010

0001 2

3

2

3

4

2

43

2

2

3

1

2

34321

i

i

i

I

i

i

i

i

i

i

iii

i

i

i

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

α λ α λ α λ α

λ

λ

λ α α α α

that is to say [ ]133 ′iii λ λ λ is an eigenvetor oa A associated with iλ


26/106

3.15 show that the vandermonde determinant

1111

4321

2

4

2

3

2

2

2

1

3

4

3

3

3

2

3

1

λ λ λ λ

λ λ λ λ

λ λ λ λ

equals

)(41 i j ji λ λ −Π ≤


27/106

))(())()()((

1111

0

0

0

det

1111

det222

333

2222

3333

bd cd bcd abacbd cd d cb

d cb

d cb

d cba

d cba

d cba

−−=−−−−=

=

so we can see

−=

1111

0

0

0

det

1111

0

0

0

det222

333

222

333

d cb

d cb

d cb

d cb

d cb

d cb

,

that is to say d and cbabd cd bcd ,,,0))(( ⇒=−− are not distinet

this implies the assumption is not true , that is , the matrix is nonsingular leti

q be the

eigenvectors of A ,44332211

qaq Aqaq Aqaq Aqaq A ====

( )

t independenlinenrlyqqand qso

d d d

d d c

bbb

aaa

qqqq

qd qcqbqa

qd qcqbqa

qd qcqbqa

qqqq

ii

ni

nii⇒=≠

=

=

⇒

=+++

=+++

=+++

=+++

×

×

×

000

0

1

1

1

1

0

0

0

0

1

1

44

32

32

32

32

44332211

443

333

223

113

442

332

222

112

44332211

44332211

α α

α α α α

α α α α

α α α α

α α α α

α α α α

3.16 show that the companion-form matrix in problem 3.14 is nonsingular if and only if

04 ≠α , under this assumption , show that its inverse equals

−−−−

=−

43

42

41

4

1

1

1000

0100

0010

α α α α α α α

A 证明题 3.14 中的友矩阵非奇异当且仅当

04 ≠α , 且矩阵的逆为

−−−−

=−

4

3

4

2

4

1

4

1

1

1000

0100

0010

α α

α α

α α

α

A

proof: as we know , the chacteristic polynomial is

43

2

2

3

1

4

)det()( α λ α λ α λ α λ λ λ ++++=−=∆ A I so let 0=λ , we have


28/106

44 )det()det()1()det( α ==−= A A A

A is nonsingular if and only if

04 ≠α

−−−−

=∴

=

⋅

−−−−

⋅

−−−−

=

⋅

−−−−

⋅

−−−−

−

4

3

4

2

4

1

4

1

4

4

3

4

2

4

1

4

4321

4

4321

4

3

4

2

4

1

4

1

1000

0100

0010

1000

0100

0010

0001

1

1000

0100

0010

0100

0010

0001

1000

0100

0010

0001

0100

0010

0001

1

1000

0100

0010

α α

α α

α α

α

α α α α α α α

α α α α

α α α α

α α

α α

α α

α

A

I

I

3.17 consider

=λ λ λ

λ λ λ

000

2/2

tT

tT T

A with 0≠λ and T>0 show that [ ]′

100 is an

generalized eigenvector of grade 3 and the three columns of

=

000

00

02/222

tT

T T

Q λ

λ λ

constitute a chain of generalized eigenvectors of length 3 , venty

=−

λ

λ

λ

00

10

011

t AQQ

矩阵 A 中 0≠λ ,T>0 , 证明[ ]′100 是 3 级广义特征向量, 并且矩阵 Q 的 3 列组成长度

是 3 的广义特征向量链,验证

=−

λ

λ

λ

00

10

011

t AQQ

Proof : clearly A has only one distinct eigenvalue λ with multiplicity 3


29/106

0

1

00

000

000000

1

00

000

002/0

000

00000

1

00

)(

0

0

0

1

0

0

000

000

00

1

0

0

)(

222

3

2222

2

=

=

=

=−

≠

=

=

=−

T T T T

I A

T T

I A

λ λ λ λ

λ

λ λ

λ

these two equation imply that [ ]′100 is a generalized eigenvctor of grade 3 ,

and

=

=

=−

=

=

=−

0

0

0000

00

2/0

0

)(

01

0

0

000

00

2/0

1

0

0

)(

2222222

222

T

T

T

T

T T

T

T

I A

T

T

T

T T

I A

λ

λ

λ

λ

λ λ

λ

λ

λ

λ

λ

λ

λ λ

λ

that is the three columns of Q consititute a chain of generalized eigenvectors of length 3

=

=

=

=

=

−

λ

λ

λ

λ

λ λ λ λ λ

λ

λ λ

λ λ λ

λ

λ λ

λ

λ λ

λ λ λ

λ

λ λ

λ

λ λ

λ λ λ

00

10

01

00

02/2/3

00

1001

100

0002/

00

1001

00

0

2/2/3

100

00

02/

00

0

2/

1

2

22223222

222232222

AQQ

T T T T T

T T T

Q

T T

T T T

T

T T

T

T T

AQ

3.18 Find the characteristic polynomials and the minimal polynomials of the following matrices求下列矩阵的特征多项式和最小多项式,

)(

000

000

000

000

)(

000

000

000

001

)(

000

000

010

001

)(

000

000

010

001

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

d cba

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ


30/106

)()()()()(

)()()()()(

)()()()()(

)()()()()()()(

14414

213

413

312

412

23

1123

11

λ λ λ λ λ λ

λ λ λ λ λ λ

λ λ λ λ λ λ

λ λ λ λ λ λ λ λ λ λ

−=Ψ−=∆

−=Ψ−=∆

−=Ψ−=∆

−−=Ψ−−=∆

d

c

b

a

3.19 show that if λ is an eigenvalue of A with eigenvector x then )(λ f is an eigenvalue

of )( A f with the same eigenvector x

证明如果λ 是 A 的关于λ 的特征向量,那么 )(λ f 是 )( A f 的特征值, x 是 )( A f 关于

)(λ f 的特征向量,

proof let A be an nn× matrix , use theorem 3.5 for any function )( x f we can define

1

110)( −

−+++= n

n x x xh β β β which equals )( x f on the spectrum of A

if λ is an eigenvalue of A , then we have )()( λ λ h f = and

)()( Ah A f =

x f xh

x x x

x A A I x Ah x A f

x x A x x A x x A x A x x A f

n

n

n

n

k k

)()(

)()()(

,,)(

1

110

1

110

3322

λ λ

λ β λ β β

β β β

λ λ λ λ λ λ

==

+++=+++==∴

====⇒

−−

−−

which implies that )(λ f is an eigenvalue of )( A f with the same eigenvector x

3.20 show that an nn× matrix has the property 0=k A for mk ≥ if and only if A has

eigenvalues 0 with multiplicity n and index m of less , such a matrix is called a nilpotent matrix

证明 nn× 的矩阵在 0=k A 当且仅当 A 的 n 重 0 特征值指数不大于 m ,这样的矩阵被称为

归零矩阵,

proof : if A has eigenvalues 0 with multiplicity n and index M or less then the Jordan-form

representation of A is

=

3

2

1

ˆ

J

J

J

A wheremnma

nn J

ii

l

i

i

nn

i

ii

≤×

=

= ∑

=

×

1

0

10

10

from the nilpotent property , we have ik

i nk for J ≥= 0 so if


31/106

,i

inmamk ×≥≥ liall J k i ,,2,10 == and 0

ˆ

3

2

1

=

=k

k

k

J

J

J

A

then )0)ˆ(0)((0 === A f if onlyand if A f Ak

If ,0 mk for Ak ≥= then ,0ˆ mk for Ak ≥= where

=

3

2

1

ˆ

J

J

J

A , nn

J

l

i

i

nni

i

i

i

ii

=

= ∑=

×

11

1

λ

λ

λ

So we have

li for mnand

li

nnk

k

k

J

ii

k

i

k

i

nk

iii

k

i

k

i

ik

i

,2,1,0

,2,10

)!1)(1(

! 11

=≤=∴

==

−+−=

+−−

λ

λ

λ

λ λ λ

which implies that A has only one distinct eigenvalue o with multiplicity n and index m or less ,

3.21 given

=100100

011

A , find At

eand A A10310

, 求 A 的函数 At

eand A A 10310

, ,

the characteristic polynomial of A is2

2 )1()det()( −=−=∆ λ λ λ λ A I

let2

210)( λ β λ β β λ ++=h on the spectrum of A , we have

21

10

210

10

0

10

2110)1()1(

1)1()1(

0)0()0(

β β

β β β

β

+=⋅′=′

++==

==

h f

h f

h f

the we have 98,0 220 =−== β β β

=

+

−=

+−=

100

100

911

100

100

101

9

100

100

011

8

98 210 A A A

the compute


32/106

102

101

0

21103

1

0

2

1

0

21103

210103

0103103

=

−=

=

⇒

+=⋅

++=

=

β

β

β

β β

β β β

β A

=

+

−=

+−=

100

100

10211

100

100

011

102

100

100

011

101

102101 2103

A A A

to compute At e :

1

22

1

2 2

1

0

21

210

00

+−=

−−=

=

⇒

+=

++=

=

t t

t t

t

t

ete

tee

te

e

e

β

β

β

β β

β β β

β

−

+−−

=

+−+

−−+

=

++=

t

t

t t t t

t t t t

At

e

e

eteee

eeee

A A I Ae

00

110

11

100

100

111

)12(

100

100

011

)22(

100

010

001

2

210 β β β

3.22 use two different methods to compute At e for A1 and A4 in problem 3.13

用两种方法计算题 3.13 中A1和A4 的函数 t A At ee 2,

method 1 : the Jordan-form representation of A1 with respect to the basis

[ ] [ ] [ ]′′′ 105014001 is { }3,2,1ˆ1 diag A =

−−

=

=

=

==∴

−

−

t

t

t t t t t

t

t

t t t

t

t

t

t At A

e

e

eeeee

e

e

eee

Cd

Cc

e

e

e

Cb

QwhereQQeeCa

3

2

33

3

2

32

1

3

2

1ˆ

00

00

)(5)(4

100

010

541

00

00

54

100

010

541

00

00

00

100

010

541

100

010

541

,11


33/106

+−−

+

++

=

−

=

=∴

−

−==

=

−

120250

161200

232/541

100

10

2/1

450

340

001

450

340

001

000

100

010

ˆ

22

2

1

4

44

4

t t

t t

t t t t

t

t t

QQee

QwhereQQe A

t At A

t A

method 2: the characteristic polynomial of1

A is ).3)(2)(1()( −−−=∆ λ λ λ λ let

2

210)( λ β λ β β λ ++=h on the spectrum of 1 A , we have

)2(2

12

34

2

533

93)3()3(

32)2()2(

)1()1(

323

322

330

22103

2102

10

t t t

t t t

t t t

t

t

t

eee

eee

eee

eh f

eh f

eh f

+−=

−++−=

+−=

⇒

++==′

++==

++==

β

β

β

β β β

β β β

β β β

−−

=

+−+

−+−+

+−

+−+−

=

++=

t

t

t t t t t

t t t

t t t

t t t

t t t

t t t

t A

e

e

eeeee

eee

eee

eee

eee

eee

A A I e

3

2

32

32

32

32

32

32

2

12210

00

00

)(5)(4

900

040

40121

)2(2

1

300

020

1041

)2

34

2

5(

3300

0330

0033

1 β β β

the characteristic polynomial of 4 A is3)( λ λ =∆ , let 2210)( λ β λ β β λ ++=h on the

spectrum of 4 A ,we have

2

2

1

0

2:)0()0(

:)0()0(

1:)0()0(

β

β

β

=′′=′′

=′=′

==

t h f

t h f

h f


34/106

thus

+−−

+

++

=

+

−−

+

+=

++=

120250

161200

232/541

000

000

450

2/

20250

16200

2340

22

2

2

241410

4

t t

t t

t t t t

t

t t

t I

A A I e t A β β β


35/106

3.23 Show that functions of the same matrix ; that is )()()()( A f Ag Ag A f = consequently

we have Ae Ae At At = 证明同一矩阵的函数具有可交换性，即 )()()()( A f Ag Ag A f = 因

此有 Ae Ae At At = 成立

proof: let1

110

1

110

)(

)(

−−

−−

+++=

+++=n

n

n

n

A A I A f

A A I Ag

α α α

β β β

( n is the order of A)

k ik

k

nk i

i

n

nk

k ik

k

i

i

n

k

n

nn

n

in

n

i

i

n

in

n

i

i

A A

A A A A I Ag A f

)()(

)()()(

1

22

0

1

0

22

11

1

1

1

1

0

011000

−+−=

−

=−

=

−

=

−−−−

=

−−−

−

=

∑∑∑∑

∑∑

+=

+++++++=

β α β α

β α β α β α β α β α β α

)()()()()()(1

22

0

1

0

Ag A f A A Ag A f k ik

k

nk i

i

n

nk

k

ik

k

i

i

n

k

=+= −+−=

−

=−

=

−

=∑∑∑∑ β α β α

let At e Ag A A f == )(,)( then we have Ae Ae At At =

3.24 let

=3

2

1

0000

00

λ λ

λ

C , find a B such that C e B

= show that of 0=iλ for some

I ,then B does not exist

let

=

3

2

1

00

00

00

λ

λ

λ

C , find a B such that C e B = Is it true that ,for any nonsingular c ,there

exists a matrix B such that C e B = ?

令

=

3

2

1

00

00

00

λ

λ

λ

C 证明若 0111 =⋅⋅ λ λ λ ,则不存在 B 使 C e B =

若

=

3

2

1

00

00

00

λ

λ

λ

C ,是否对任意非奇异 C 都存在 B 使, C e B = ?

Let λ λ λ == e f ln)( so Be B f B == ln)(


36/106

==

3

2

1

ln00

0ln0

00ln

ln

λ

λ

λ

C B where 3,2,1,01 => iλ , if 0=λ for some i ,

1lnλ does not exist

for

=

λ

λ

λ

00

00

01

C we have

=

′

==

λ

λ λ

λ

λ

λ

λ λ

ln00

0ln0

01ln

ln00

0ln0

0nlln

lnC B ,

where 0,0 ≤> λ λ if then λ ln does mot exist , so B does mot exist , we can conclude

that , it is mot true that , for any nonsingular C THERE EXISTS a B such that cek =

3.25 let )()(

1)( AsI Adj

s AsI −

∆=− and let m(s) be the monic greatest common divisor of

all entries of Adj(Si-A) , Verify for the matrix 2 A in problem 3.13 that the minimal polynomial

of A equals )()( sms∆

令, )()(

1)( AsI Adj

s AsI −

∆=− , 并且令 m(s)是 Adj(Si-A)的所有元素的第一最大公因子,

利用题 3.13 中 2 A 验证 A 的最小多项式为 )()( sms∆

verification :

1)(

)1(00

0)2)(1(0

)1(0)2)(1(

)(,

200

010

101

)2(\)1()()2()1()(

200

010

001ˆ

200

010

101

2

2

33

−=

−

−−

−−−−

=−

−

−

−

=−

−−=−−=∆

=

−

=

ssmso

s

ss

sss

AsI Adj

s

S

S

AsI

ssssss

A A

j

we can easily obtain that )()()( smss ∆=j

3.26 Define [ ]1221101)(

1)( −−

−−− ++++∆

=− nnnn Rs Rs Rs R

s AsI where

in

nn Rand sss AsI s α α α ++++=−=∆ −− 221

1

2:)det()( are constant matrices theis


37/106

definition is valid because the degree in s of the adjoint of (sI-A) is at most n-1 , verify

I ARn

ARtr

I A A A I AR Rn

ARtr

I A A I AR R ARtr

I A I AR R ARtr

I R ARtr

nnn

nn

nn

nnnn

α α

α α α α α

α α α α

α α α

α

+=−=

++++=+=−

−=

++=+=−=

+=+=−=

=−=

−

−−−−

−−−−

11

122

11

1211

1

212

2121

3

11011

2

00

1

0)(

1

)(

2

)(2

)(

1

)(

where tr stands for the trase of a matrix and is defined as the sum of all its diagonal entries this

process of computing ii Rand α is called the leverrier algorithm

定义 ][)(

1:)( 12

21

10

1−−

−−− ++++∆

=− nnnn

RS RS RS Rs

ASI 其中 )(s∆ 是 A 的特征多项

式 innnn Rand sss AsI s α α α ++++=−=∆ −− 22

1

1:)det()( 是常数矩阵,这样定义

是有效的, 因为 SI-A 的伴随矩阵中 S 的阶次不超过 N-1 验证

I AR

n

ARtr

I A A A I AR Rn

ARtr

I A A I AR R ARtr

I A I AR R ARtr

I R ARtr

nnn

nn

nn

nnnn

α α

α α α α α

α α α α

α α α

α

+=−=

++++=+=−

−=

++=+=−=

+=+=−=

=−=

−

−−−−

−−−−

11

122

11

1211

1

212

2121

3

11011

2

00

1

0)(

1

)(

2

)(

2

)(1

)(

其中矩阵的迹 tr 定义为其对角元素之和, 这种计算 iα 和 i R 的程式被称为 leverrier 算法.

verification:

implieswhich

S I

S S S S I

ARS AR RS AR RS AR RS R

RS RS RS R ASI

nn

nnn

nnn

nnn

nn

nn

)(

)(

)()()(

])[(

1

2

2

1

1

121

2

12

1

010

12

2

1

1

0

∆=

+++++=

+−+−+−+=

++++−

−−−

−−−−−

−−−−

α α α α


38/106

][)(

1:)( 12

21

10

1−−

−−− ++++∆

=− nnnn

RS RS RS Rs

ASI ’

where nnnn ssss α α α ++++=∆ −− 22

1

1)(

3.27 use problem 3.26 to prove the cayley-hamilton theorem

利用题 3.26 证明 cayley-hamilton 定理

proof:

I AR

I AR R

I AR R

I AR R

I R

nn

nnn

α

α

α

α

=−

=−

=−

=−

=

−

−−−

1

121

212

101

0

multiplying ith equation by1+−in

A yields ( ni 2,1= )

I AR

A R A AR

A R A R A

A R A R A

A R A

nn

nnn

nnn

nnn

nn

α

α

α

α

=−

=−

=−

=−

=

−

−−−

−−−

−−

1

122

1

221

12

2

1101

1

0

then we can see012

2

101

1

0

1

2

2

1

1

=−−++−++=

+++++

−−−−

−−−

nnn

nnn

nn

nnn

AR R A AR R A R A R A

I A A A A

α α α α that is

0)( =∆ A

3.28 use problem 3.26 to show

[ ] I ss Ass As As

AsI

n

nnnnn )()()()(

1

)(

12

113

2122

11

1

−−−−−−

−

++++++++++∆

=

−

α α α α α

利用题 3.26 证明上式,

Proof:


39/106

[ ] I ss Ass As As

I A A AS A A A

S I A AS I AS s

RS RS RS Rs

ASI

n

nnnnn

nnnn

nnnn

nnn

nn

nn

)()()()(

1

])(

)()([)(

1

][)(

1:)(

1

2

1

13

21

22

1

1

122

11

433

12

3

21

22

1

1

12

2

1

1

0

1

−−−−−−

−−−−−−−−

−−−

−−−−−

++++++++++∆

=

+++++++++

++++++∆

=

++++∆

=−

α α α α α

α α α α α α

α α α

another : let 10 =α

[ ] I ss Ass As As

I A A AS A A A

S I A AS I AS

AS AS s

AS

s

S R

s

RS RS RS Rs

ASI

n

nnnnn

nn

nn

nn

nn

nnn

n

i

I n

i

N

i

I n

i

n

ii

n

i

L

li

n

i

inin

i

nn

nn

)()()()(

1

])(

)()(

)(

1)(

1

)(

1

][)(

1:)(

1

2

1

13

21

22

1

1

12

2

1

1

43

3

1

2

3

21

22

1

1

1

0

11

0

01

1 1

0

1

0

11

12

2

1

1

0

1

−−−−−−

−−−−

−−−−

−−−

−

=

−−−

=

−−

−

=

−

=

−

−

=

−−−−

−−−−−

++++++++++∆

=

+++++++++

++++++

+++∆

=

∆

=

∆

=

++++∆

=−

∑∑

∑ ∑∑

α α α α α

α α α α α α

α α α

α α

α

3.29 let all eigenvalues of A be distinct and let iq be a right eigenvector of A associated with

iλ that is i I i qq A λ = define nqqqQ 21= and define

== −

n p

p

p

QP

2

1

1 :; ,

where i p is the ith row of P , show that i p is a left eigenvector of A associated with iλ , that

is iii p A p λ = 如果 A 的所有特征值互不相同 , iq 是关于 iλ 的一个右特征向量 , 即

i I i qq A λ = ,定义

nqqqQ 21= 并且

== −

n p

p

p

QP

2

1

1 :;

其中i p 是 P 的第 I 行, 证明 i p 是 A 的关于 iλ 的一个左特征向量,即 iii p A p λ =


40/106

Proof: all eigenvalues of A are distinct , andi

q is a right eigenvector of A associated withi

λ ,

andn

qqqQ 21= so we know that11

3

2

1

ˆ −− ==

= PAP AQQ A

λ

λ

λ

P APA ˆ=∴ That is

=

⇒

=

nnnnn

n p

p

p

A p

A p

A p

p

p

p

A

p

p

p

λ

λ

λ

λ

λ

λ

22

11

2

1

2

1

2

12

1

: so iii p A p λ = , that is , i p is a

left eigenvector of A associated withi

λ

3.30 show that if all eigenvalues of A are distinct , then1)( −− ASI can be expressed as

∑−

=− − iii

pqs

ASI λ

1)( 1 where iq and i p are right and left eigenvectors of A associated

with iλ 证明若 A 的所有特征值互不相同 , 则1)( −− ASI 可以表示为

∑−

=− − iii

pqs

ASI λ

1)( 1 其中 iq 和 i p 是 A 的关于 iλ 的右特征值和左特征值,

Proof: if all eigenvalues of A are distinct , leti

q be a right eigenvector of A associated withi

λ ,

then nqqqQ 21= is nonsingular , and

=−

n p

p

p

Q

2

1

1: where is a left eigenvector of

A associated withi

λ ,

∑ ∑

∑∑

∑

=−−

=

−−

=−−

=

−−

iiiii

i

iiiii

i

iiii

i

ii

i

pq pqss

pq pqss

A pq pqss

ASI pqs

))((1

)(1

)(1

)(1

λ λ

λ λ λ

λ

That is ∑−

=− − iii

pq

s

ASI

λ

1)( 1


41/106

3.31 find the M to meet the lyapunov equation in (3.59) with

==

−−=

3

33

22

10C B A what are the eigenvalues of the lyapunov equation ? is the

lyapunov equation singular ? is the solution unique ?

已知 A,B,C,求 M 使之满足(3.59) 的 lyapunov 方程的特征值, 该方程是否奇异>解是否唯一?

=⇒=

−∴

=+

3

0

12

13 M C M

C MB AM

3)det()()1)(1()det()( +=−=∆++−+=−=∆ λ λ λ λ λ λ λ B I j j A I B A

The eigenvalues of the Lyapunov equation are

j j j j −=+−−=+=++−= 231)231 21 η η

The lyapunov equation is nonsingular M satisfying the equation

3.32 repeat problem 3.31 for

Documents

Linear System Theory and Design Answer