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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: uay *= 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: buay += * Here a and b are all constants. Testify whether it has the property of additivity. Let: buay += 11 * buay += 22 * then: buuayy *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: byz −= uaz *= 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: uuay *)(= a(u) is a function of input u. Choose two different input, get the outputs: 111 *uay = Linear System Theory and Design SA01010048 LING QING 2 222 *uay = Assure: 21 aa ≠ then: 221121 **)( uauayy +=+ 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 )(2sin 2)( 0 0 tt tttg − − = 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 > ≤ == atfor atfortu tuPty 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, t<=a: uy = Easy to testify that it is linear. Consider the input-output relation at any time t, t>a: 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<a, so It decides the system output y(t), t>=to: Linear System Theory and Design SA01010048 LING QING 3 ≤≤ = totherfor attfortu ty 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)(' =ty 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 uHPPHuPyP aaaa == 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: HuPyP aa = Define the initial time 0, since the system is causal, output y begins in time 0. If a<=0,then u=Hu. Add operation PaH in both left and right of the equation: uHPPHuP aaa = If a>0, we can divide u to 2 parts: ≤≤ = totherfor atfortu tp 0 0)( )( > = totherfor atfortu tq 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: uHPPHuP aaa = It means under any condition, the following equation is correct: uHPPHuPyP aaaa == 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. Linear System Theory and Design SA01010048 LING QING 4 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 ),()( ),()( )()( tttyty tttutu txtx ≥+→ ≥+ + αα αα αα In case 1: 11 =α 12 =α )0()0(2)()( 022011 xxtxtx ≠=+αα So y3<>y1+y2. In case 2: 5.01 =α 5.02 =α )0()()( 022011 xtxtx =+αα So y3=0.5(y1+y2). In case 3: 11 =α 12 −=α )0(0)()( 022011 xtxtx ≠=+αα So y3<>y1-y2. 2.6 Consider a system whose input and output are related by =− ≠−− = 0)1(0 0)1()1(/)( )( 2 tuif tuiftutu ty 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: )()( tutp α= a is any real constant. Then system output q(t): Linear System Theory and Design SA01010048 LING QING 5 =− ≠−− = 0)1(0 0)1()1(/)( )( 2 tpif tpiftptp tq =− ≠−− = 0)1(0 0)1()1(/)(2 tuif tuiftutuα 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 =++= nmnmy )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: yx → then: mymx→ Easy to conclude it from additivity. Secondly, prove that if a system input-output can be described as following: yx → then: nynx // → Suppose: unx →/ Using additivity: nuxnxn →=)/(* So: nuy = nyu /= Linear System Theory and Design SA01010048 LING QING 6 It is to say that: nynx // → Then: nmynmx /*/* → 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: Ttx += Tty −= So: 2 yxt += 2 yxT −= Then: ) 2 , 2 (),( yxyxgTtg −+= ) 2 , 2 ( ayxayxg +−++= ) 22 , 22 ( yxyxyxyxg +−+−+−++= )0,(yg= So: 0)0,(),( = ∂ ∂ = ∂ ∂ x yg x Ttg 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 Linear System Theory and Design SA01010048 LING QING 7 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 )( tt tt tg ≤≤− ≤≤ = 211 101 )( t t tu then y(t) equals to the convolution integral: ∫ −= t drrturgty 0 )()()( If 0=<t=<1, 0=<r=<1, 0<=t-r<=1: ∫= t rdrty 0 )( 2 2t = If 1<=t<=2: )(ty ∫ − −= 1 0 )()( t drrturg ∫ − −+ 1 1 )()( t drrturg ∫ −+ t drrturg 1 )()( )(1 ty= )(2 ty+ )(3 ty+ Calculate integral separately: )(1 ty ∫ − −= 1 0 )()( t drrturg 10 ≤≤ r 21 ≤−≤ rt ∫ − −= 1 0 t rdr 2 )1( 2−− = t )(2 ty ∫ − −= 1 1 )()( t drrturg 10 ≤≤ r 10 ≤−≤ rt ∫ − = 1 1t rdr 2 )1( 2 1 2− −= t )(3 ty ∫ −= t drrturg 1 )()( 21 ≤≤ r 10 ≤−≤ rt ∫ −= t drr 1 )2( 2 1)1(2 2 − −−= tt )(ty )(1 ty= )(2 ty+ )(3 ty+ 242 3 2 −+−= tt Linear System Theory and Design SA01010048 LING QING 8 2.10 Consider a system described by uuyyy −=−+ •••• 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 sYssYsYssYsYs −=−+ System transfer function: 3 1 32 1 )( )()( 2 + = −+ − == sss s sY sUsG Impulse response: te s LsGLtg 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): s sGsY 1*)()( = )()( sGsM = ssYsM *)()( = So: dttdytm /)()( = 2.12 Consider a two-input and two-output system described by )()()()()()()()( 212111212111 tupNtupNtypDtypD +=+ )()()()()()()()( 222121222121 tupNtupNtypDtypD +=+ 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 sUsNsUsNsYsDsYsD +=+ )()()()()()()()( 222121222121 sUsNsUsNsYsDsYsD +=+ Linear System Theory and Design SA01010048 LING QING 9 Write to the form of matrix: )( )( )()( )()( 2 1 2221 1211 sY sY sDsD sDsD = )( )( )()( )()( 2 1 2221 1211 sU sU sNsN sNsN )( )( 2 1 sY sY = 1 2221 1211 )()( )()( − sDsD sDsD )( )( )()( )()( 2 1 2221 1211 sU sU sNsN sNsN So the transfer function matrix is: )(sG = 1 2221 1211 )()( )()( − sDsD sDsD )()( )()( 2221 1211 sNsN sNsN By the premising that the matrix inverse: 1 2221 1211 )()( )()( − sDsD sDsD 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。 Linear System Theory and Design SA01010048 LING QING10 Answer: Firstly, consider the positive-feedback system. It’s impulse response is: ∑ ∞ = −= 1 )()( i i itatg δ Using convolution integral: ∑ ∞ = −= 1 )()( i i itraty When input is unit-step signal: ∑ = = n i iany 1 )( )()( nyty = 1+≤≤ ntn 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 i itatg δ Using convolution integral: ∑ ∞ = −−−= 1 )()()( i i itraty When input is unit-step signal: ∑ = −−= n i iany 1 )()( )()( nyty = 1+≤≤ ntn 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 uxx − + − = • 4 2 50 42 [ ] uxy 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? Linear System Theory and Design SA01010048 LING QING 11 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 θθθθθθ ••• −−==− mll dt dmmgf )sincos()sin(sin 2 2 2 θθθθθθ ••• −==− mll dt dmfu Assuming θ and • θ to be small, we can use the approximation θsin =θ , θcos =1. By retaining only the linear terms in θ and • θ , we obtain mgf = and: u mll g 1 +−= •• θθ Select state variables as θ=1x , • = θ2x and output θ=y u ml x lg x + − = • /1 1 0/ 10 [ ]xy 01= For Fig2.22(b), the application of Newton’s law to the linear movements yields: )cos(coscos 112 2 112211 θθθ ldt dmgmff =−− )cossin( 1 2 11111 θθθθ ••• −−= lm )sin(sinsin 112 2 11122 θθθ ldt dmff =− )sincos( 1 2 11111 θθθθ ••• −= lm )coscos(cos 22112 2 2222 θθθ lldt dmgmf +=− )cossin( 1 2 11112 θθθθ ••• −−= lm )cossin( 2 2 22222 θθθθ ••• −−+ lm )sinsin(sin 22112 2 222 θθθ lldt dmfu +=− )sincos( 1 2 11112 θθθθ ••• −= lm )sincos( 2 2 22222 θθθθ ••• −+ lm Linear System Theory and Design SA01010048 LING QING 12 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 gmf 22 = , gmmf )( 211 += and: 2 11 2 1 11 21 1 )( θθθ lm gm lm gmm + + −= •• u lmlm gmm lm gmm 22 2 21 21 1 21 21 2 1)()( + + − + = •• θθθ Select state variables as 11 θ=x , 12 • = θx , 23 θ=x , 24 • = θx 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 mgmkym −−= ••• Where g is the gravity constant on the lunar surface. Define state variables of the system as: yx =1 , • = yx2 , mx =3 , • = my 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: Linear System Theory and Design SA01010048 LING QING 13 − +−= ygty 2/2 − •• +−= ygty − •••• +−= ygy − += mmm 0 − •• = mm So: )( 0 − +mm − •• +− )( yg = − • − mk gmm )( 0 − +− +−− − gmgm0 − •• =ym0 − • − mk gmgm − −− 0 − •• =ym0 − • − mk Define state variables as: −− = yx1 , − •− = yx 2 , −− = mx3 , − •− = my 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: 1x : Voltage of left capacitor 2x : Voltage of right capacitor 3x : Current of inductor Applying Kirchhoff’s current law: Linear System Theory and Design SA01010048 LING QING 14 3x = RxLx /)( 32 • − RxxCu 11+= • 32 xxCu += • 2xy = From the upper equations, we get: CuCRxx //11 +−= • ux +−= 1 CuCxx //32 +−= • ux +−= 3 LRxLxx // 323 −= • 32 xx −= 2xy = 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 1 2 ++ + = ss s Linear System Theory and Design SA01010048 LING QING 15 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 xRxLxxxCR ++++ •• 22211 •• =+ xCuxC 223 • = xCx 3231212311 )( xRxLxxuxRu ++++−= • 3231 xRxLy += • From the upper equations, we get: 12131 // CuCxx −= • 232 /Cxx = • 12111131112113 ///)(// LuRLuLxRRLxLxx +++−−−= • 2113121 uRuxRxxy ++−−−= They can be combined in matrix form as: • • • 3 2 1 x x x = +−−− 12111 2 1 /)( /100 /100 LRRLL C C 3 2 1 x x x + − 2 1 11 1 1 / 0 /1 /1 0 0 u u LR C L [ ] −−−= 3 2 1 111 x x x Ry + [ ] 2 1 21 u u R Applying Laplace Transform to upper equations: )( /1/1 )( 1 ^ 21111 21 ^ su sCsCRRsL RsLsy ++++ + = )( /1/1 )/1)(( 2 ^ 21111 2121 su sCsCRRsL sCRRsL ++++ ++ + ++++ + = sCsCRRsL RsLsG 21111 21 ^ /1/1 )( ++++ ++ sCsCRRsL sCRRsL 21111 2121 /1/1 )/1)(( Linear System Theory and Design SA01010048 LING QING 16 2.18 Find the transfer functions from u to 1y and from 1y 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 1y 的传递函数和从 1y 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 , 1y and y : 111 / Rxy = 222 / Rxy = dtyudxA /)( 111 −= dtyydxA /)( 122 −= Applying Laplace transform: )1/(1/ 11 ^ 1 ^ sRAuy += )1/(1/ 221 ^^ sRAyy += )1)(1/(1/ 2211 ^^ sRAsRAuy ++= So: = ^^ /uy )/( ^ 1 ^ uy )/( 1 ^^ yy But it is not true for Fig2.14, because of the loading problem in the two tanks. 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 1 q and 23 8 i as shown , thus the representation of x with respect to the basis ( )21 iq is ′ 3 8 3 1 , this can be verified from [ ] = = = 3 8 3 1 11 03 3 8 3 1 3 1 22 iqx 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 3 q , thus the representation of 1q with respect to ( )22 qi , is ′ − 2 32 , this can be verified from [ ] − = − = = 2 3 2 21 20 2 3 2 1 3 221 qiq 3.2 what are the 1-morm ,2-norm , and infinite-norm of the vectors [ ] [ ]′=′−= 111,132 21 xx , 问向量 [ ] [ ]′=′−= 111,132 21 xx 的 1-范数,2-范数和 −∞ 范数是什么? 1max3max 3)111(14)1)3(2( 31116132 2211 2 1222 22 2 1222 1 ' 121 12 3 1 111 ==== =++==+−+== =++==++== ∞∞ = ∑ iiii i i xxxxxx xxxx xxx 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 , [ ] [ ] [ ]′===−= ′−==′−== 111 3 1)( 132 14 1132 2 2 2212 ' 122 1 1 111 u u qxqxqxu u u qxu The two orthomormal vectors are = −= 1 1 1 3 1 1 3 2 14 1 21 qq In fact , the vectors 1x and 2x are orthogonal because 01221 =′=′ xxxx 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 x q x x q 3.4 comsider an mn× matrix A with mn ≥ , if all colums of A are orthonormal , then mIAA =′ , what can you say abort AA ′ ? 一个 mn× 阶矩阵A( mn ≥ ), 如果A的所有列都是 标准正交的,则 mIAA =′ 问 AA ′是怎么样? Let [ ] [ ] mnijm aaaaA × == 21 , if all colums of A are orthomormal , that is ∑ = = ≠ == n i liliii jiif jiif aaaa 1 ' 1 0 then [ ] [ ] =≠ ≠≠ = == =′ = = = =′ ∑ ∑∑ = × == jiif jiif aageneralin aaaa a a a aaaAA I aaaaaa aaaaaa aaa a a a AA jl m i il nnjl 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 mIAA =′ 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 AAA 134)(033)(123)( 3)(3)(2)( 321 321 =−==−==−= === ANullityANullityANullity ArankArankArank 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 1A are linearly independent , so the set 1 0 0 , 0 0 1 can be used as a basis of the range space of 1A , all columns of 2A 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 2A let [ ]43213 aaaaA = ,where ia denotes of 3A 4321 aandaandaandaare 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 3A 3.7 consider the linear algebraic equation xx − = − − − 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 [ ]′= 111y ? 线性代数方程 xx − = − − − 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 [ ]′= 111y 方程是否有解? Let [ ], 21 33 12 21 aaA = − − − = 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 [ ]′= 11x if [ ]′= 111y , 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 = −−= 1 2 3 1000 2210 4321 yA we can readily obtain rank(A)= rank([A y ])=3 so this y lies in the range space of A and [ ]′−= 101 oxp 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 ααnxx 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, 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 αα α ααx 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 +−−++= −+−+−++= αααααα αααααx − − − =⇒ = = ⇒ =−+⇒= ∂ =−+⇒= ∂ 11 16 11 4 11 8 11 4 11 16 11 4 08250 2 04230 2 2 1 12 2 21 1 x x x α α αα α αα 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 αx for any real α the Euclidean norm of x is − − =⇒=⇒=−⇒= ∂ ∂ +−=++−+−= 1 6 1 6 3 6 5 6 102120 2261)2()1( 2222 x x x αα α ααααα has the smallest Euclidean norm 3.11 consider the equation ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnubAubAubAxAnx nnn where A is an nn× matrix and b is an 1×n column vector ,under what conditions on A and b will there exist ]1[],1[],[ −nuuou to meet the equation for any ]0[],[ xandnx ? 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[],[ xandnx ,它们都满足方程 ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnubAubAubAxAnx nnn , write the equation in this form − − == − ]0[ ]2[ ]1[ ],[]0[}{ 1 u nu nu bAbAbxAnx nn where ],[ 1bAbAb n− is an nn× matrix and ]0[}{ xAnx n− is an 1×n column vector , from the equation we can see , ]1[],1[],[ −nuuou exist to meet the equation for any ]0[],[ xandnx ,if and only if nbAbAb n =− ],[ 1ρ under this condition , there will exist ]1[],1[],[ −nuuou to meet the equation for any ]0[],[ xandnx . 3.12 given = = = 1 3 2 1 1 0 0 1000 0200 0120 0012 bbA what are the representations of A with respect to the basis { }bAbAbAb 32 and the basis { }bAbAbAb 32 , respectively? 给定 = = = 1 3 2 1 1 0 0 1000 0200 0120 0012 bbA 请问 A 关于{ }bAbAbAb 32 和基{ }bAbAbAb 32 的表 示分别是什么? =⋅= =⋅= = 1 16 32 24 , 1 4 4 1 , 1 2 1 0 232 bAAbAbAAbAbA we have [ ] [ ] − − ==∴ +−+−= 7100 18010 20001 8000 718208 3232 324 bAbAbAbbAbAbAbA bAbAbAbbA thus the representation of A with respect to the basis { }bAbAbAb 32 is − − = 7100 18010 20001 8000 A bAbAbAbbA bAbAbAbA 324 432 718208 1 48 128 152 , 1 24 52 50 , 1 12 20 15 , 1 6 7 4 +−+−= = = = = [ ] [ ] − − ==∴ 7100 18010 20001 8000 3232 bAbAbAbbAbAbAbA thus the representation of A with respect to the basis { }bAbAbAb 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 −− = − = −−− = = AAAA the characteristic polynomial of 1A is )3)(2)(1()det( 11 −−−=−=∆ λλλλ AI thus the eigenvelues of 1A are 1 ,2 , 3 , they are all distinct . so the Jordan-form representation of 1A will be diagonal . the eigenvectors associated with 3,2,1 === λλλ ,respectively can be any nonzero solution of [ ] [ ] [ ]′=⇒= ′=⇒= ′=⇒= 1053 0142 001 3331 2221 1111 qqqA qqqA qqqA thus the jordan-form representation of 1A with respect to { }321 qqq is = 300 020 001 ˆ 1A the characteristic polynomial of 2A is 2 23 22 )1)(1)(1(243()det()( AiiAI −++++=+++=−=∆ λλλλλλλλ has eigenvalues jandj −−+−− 11,1 the eigenvectors associated with jandj −−+−− 11,1 are , respectively [ ] [ ] [ ]′−−′−−′− jjandjj 211211,111 the we have QAQ j jAand jj jjQ 2 1 2 100 010 001 ˆ 220 111 111 −= −− +− − = − −−+−−=the characteristic polynomial of 3A is )2()1()det()( 2 33 −−=−=∆ λλλλ AI theus the eigenvalues of 3A are 1 ,1 and 2 , the eigenvalue 1 has multiplicity 2 , and nullity 213)(3)( 33 =−=−−=− IArankIA the 3A has two tinearly independent eigenvectors associated with 1 , [ ] [ ] [ ]′−=⇒=− ′=′=⇒=− 1010)( 0100010)( 333 213 qqIA qqqIA thus we have QAQAandQ 3 1 3 200 010 001 ˆ 100 011 101 −= = − −= the characteristic polynomial of 4A is 3 44 )det()( λλλ =−=∆ AI clearly 4A has lnly one distinct eigenvalue 0 with multiplicity 3 , Nullity( 4A -0I)=3-2=1 , thus 4A has only one independent eigenvector associated with 0 , we can compute the generalized , eigenvectors of 4A from equations below [ ] [ ] [ ]′−=⇒= ′−=⇒= ′=⇒= 430 540 0010 3231 2121 111 vvvA vvvA vvA , then the representation of 4A with respect to the basis { }321 vvv is − −== = − 450 340 001 000 100 010 ˆ 4 1 4 QwhereQAQA 3.14 consider the companion-form matrix −−−− = 0100 0010 0001 4321 αααα A show that its characterisic polynomial is given by 43 3 2 3 1 4)( αλαλαλαλλ ++++=∆ show also that if iλ is an eigenvalue of A or a solution of 0)( =∆ λ then [ ]133 ′iii λλλ is an eigenvector of A associated with iλ 证明友矩阵 A的特征多项式 43 3 2 3 1 4)( αλαλαλαλλ ++++=∆ 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 λλλ 是 A 关于 iλ 的一个特征向量 proof: 4322 3 1 4 43 2 3 1 4 432 1 4321 2 1 det 1 0 det 10 01det 10 01 00 det)( 100 010 001 det)det()( αλαλαλαλ λ αα λ λ αλαλ λ λ ααα λ λ λ αλ λ λ λ ααααλ λλ ++++= − + − ++= − −+ − −+= − − − + =−=∆ AI 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 3 4321 i i i I i i i i i i iii i i i λ λ λ λ λ λ λ λ λ λ αλαλαλα λ λ λαααα that is to say [ ]133 ′iii λλλ is an eigenvetor oa A associated with iλ 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 λλλλ λλλλ λλλλ equals )(41 ijji λλ −Π ≤<≤ , thus we conclude that the matrix is nonsingular or equivalently , the eigenvectors are linearly independent if all eigenvalues are distinct ,证明 vandermonde 行列式 1111 4321 2 4 2 3 2 2 2 1 3 4 3 3 3 2 3 1 λλλλ λλλλ λλλλ 为 )(41 ijji λλ −Π ≤<≤ , 因此如果所有的特征值都互不 相同则该矩阵非奇异, 或者等价地说, 所有特征向量线性无关, proof: 41 241423132412 23132414424142313313 12 24142313 2414423133 141312 24142313 2414423133 141312 144133122 14 2 413 2 312 2 2 141312 144133122 14 2 413 2 312 2 2 4321 2 4 2 3 2 2 2 1 3 4 3 3 3 2 3 1 ))()()()()(( )])()()(())()()(()[( )( ))(())(( ))(())(( det ))(())((0 ))(())((0 det )()()( )()()( det 1111 0 )()()(0 )()()(0 det 1111 det ≤<≤Π= −−−−−−= −−−−−−−−−−−= −⋅ −−−− −−−− −= −−− −−−− −−−− −= −−− −−− −−− = −−− −−− −−− = ji λλλλλλλλλλλλ λλλλλλλλλλλλλλλλλλλλ λλ λλλλλλλλ λλλλλλλλλλ λλλλλλ λλλλλλλλ λλλλλλλλλλ λλλλλλ λλλλλλλλλ λλλλλλλλλ λλλλλλ λλλλλλλλλ λλλλλλλλλ λλλλ λλλλ λλλλ let a, b, c and d be the eigenvalues of a matrix A , and they are distinct Assuming the matrix is singular , that is abcd=0 , let a=0 , then we have ))()(( 111 detdet 1111 0 0 0 det 222 222 333 222 333 bcbdcdbcddcb dcb bcd dcb dcb dcb dcb dcb dcb −−−−= ⋅−= = and from vandermonde determinant ))(())()()(( 1111 0 0 0 det 1111 det 222 333 2222 3333 bdcdbcdabacbdcd dcb dcb dcb dcba dcba dcba −−=−−−−= = so we can see −= 1111 0 0 0 det 1111 0 0 0 det 222 333 222 333 dcb dcb dcb dcb dcb dcb , that is to say dandcbabdcdbcd ,,,0))(( ⇒=−− are not distinet this implies the assumption is not true , that is , the matrix is nonsingular let i q be the eigenvectors of A , 44332211 qaqAqaqAqaqAqaqA ==== ( ) tindependenlinenrlyqqand q so ddd ddc bbb aaa qqqq qdqcqbqa qdqcqbqa qdqcqbqa qqqq ii ni nii ⇒= ≠ = = ⇒ =+++ =+++ =+++ =+++ × × × 00 0 0 1 1 1 1 0 0 0 0 1 1 44 32 32 32 32 44332211 44 3 33 3 22 3 11 3 44 2 33 2 22 2 11 2 44332211 44332211 α α αααα αααα αααα αααα αααα 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 −−−− =− 4 3 4 2 4 1 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()( αλαλαλαλλλ ++++=−=∆ AI so let 0=λ , we have 4 4 )det()det()1()det( α==−= AAA 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 αααα αααα α α α α α α α A I I 3.17 consider = λ λλ λλλ 00 0 2/2 tT tTT 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 TT Q λ λλ constitute a chain of generalized eigenvectors of length 3 , venty =− λ λ λ 00 10 01 1 tAQQ 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 01 1 tAQQ Proof : clearly A has only one distinct eigenvalue λ with multiplicity 3 0 1 0 0 000 000 000 1 0 0 000 00 2/0 000 000 00 1 0 0 )( 0 0 0 1 0 0 000 000 00 1 0 0 )( 222 3 2222 2 = = = =− ≠ = = =− T TTT IA TT IA λ λλλ λ λλ 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 TT T T IA T T T TT IA λ λ λ λ λλ λ λ λ λ λ λ λλ λ that is the three columns of Q consititute a chain of generalized eigenvectors of length 3 = = = = = − λ λ λ λ λλ λλλ λ λ λ λ λλ λ λ λ λ λλ λλλ λ λλ λ λλ λλλ 00 10 01 00 0 2/2/3 00 10 01 100 00 02/ 00 10 01 00 0 2/2/3 100 00 02/ 00 0 2/ 1 2 22223222 222232222 AQQ TT TTT T TT Q TT TTT T TT T TT 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 dcba λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ )()()()()( )()()()()( )()()()()( )()()()()()()( 14 4 14 2 13 4 13 3 12 4 12 2 3 112 3 11 λλλλλλ λλλλλλ λλλλλλ λλλλλλλλλλ −=Ψ−=∆ −=Ψ−=∆ −=Ψ−=∆ −−=Ψ−−=∆ d c b a 3.19 show that if λ is an eigenvalue of A with eigenvector x then )(λf is an eigenvalue of )(Af with the same eigenvector x 证明如果λ是 A 的关于λ的特征向量,那么 )(λf 是 )(Af 的特征值, x是 )(Af 关于 )(λf 的特征向量, proof let A be an nn× matrix , use theorem 3.5 for any function )(xf we can define 1 110)( − −+++= n n xxxh βββ which equals )(xf on the spectrum of A if λ is an eigenvalue of A , then we have )()( λλ hf = and )()( AhAf = xfxh xxx xAAIxAhxAf xxAxxAxxAxAxxAf n n n n kk )()( )()()( ,,)( 1 110 1 110 3322 λλ λβλββ βββ λλλλλλ == +++= +++==∴ ====⇒ − − − − which implies that )(λf is an eigenvalue of )(Af with the same eigenvector x 3.20 show that an nn× matrix has the property 0=kA 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=kA 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 where mnma nn J ii l i i nn i ii ≤× = = ∑= × 1 0 10 10 from the nilpotent property , we have i k i nkforJ ≥= 0 so if ,ii nmamk ×≥≥ liallJ k i ,,2,10 == and 0ˆ 3 2 1 = = k k k J J J A then )0)ˆ(0)((0 === AfifonlyandifAfAk If ,0 mkforAk ≥= then ,0ˆ mkforAk ≥= where = 3 2 1 ˆ J J J A , nnJ l i i nni i i i ii = = ∑ = × 11 1 λ λ λ So we have liformnand li nnk kk J ii k i k i nk i ii k i k i i k i ,2,1,0 ,2,10 )!1)(1( ! 11 =≤=∴ == −+− = +−− λ λ λ λλλ which implies that A has only one distinct eigenvalue o with multiplicity n and index m or less , 3.21 given = 100 100 011 A , find AteandAA 10310 , 求A的函数 AteandAA 10310 , , the characteristic polynomial of A is 22 )1()det()( −=−=∆ λλλλ AI let 2210)( λβλββλ ++=h on the spectrum of A , we have 21 10 210 10 0 10 2110)1()1( 1)1()1( 0)0()0( ββ βββ β +=⋅′=′ ++== == hf hf hf the we have 98,0 220 =−== βββ = + −= +−= 100 100 911 100 100 101 9 100 100 011 8 98 210 AAA the compute 102 101 0 21103 1 0 2 1 0 21 103 210 103 0 103103 = −= = ⇒ +=⋅ ++= = β β β ββ βββ βA = + −= +−= 100 100 10211 100 100 011 102 100 100 011 101 102101 2103 AAA to compute Ate : 1 22 1 2 2 1 0 21 210 0 0 +−= −−= = ⇒ += ++= = tt tt t t ete tee te e e β β β ββ βββ β − +−− = +−+ −−+ = ++= t t tttt tttt At e e eteee eeee AAIAe 00 110 11 100 100 111 )12( 100 100 011 )22( 100 010 001 2 210 βββ 3.22 use two different methods to compute Ate 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 tAAt ee 2, method 1 : the Jordan-form representation of A1 with respect to the basis [ ] [ ] [ ]{ }′′′ 105014001 is { }3,2,1ˆ1 diagA = −− = = = ==∴ − − t t ttttt t t ttt t t t tAtA 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 +−− + ++ = − = =∴ − −== = − 120250 161200 232/541 100 10 2/1 450 340 001 450 340 001 000 100 010 ˆ 22 2 1 4 44 4 tt tt tttt t tt QQee QwhereQQeA tAtA tA method 2: the characteristic polynomial of 1A is ).3)(2)(1()( −−−=∆ λλλλ let 2 210)( λβλββλ ++=h on the spectrum of 1A , we have )2( 2 1 2 34 2 5 33 93)3()3( 32)2()2( )1()1( 32 3 32 2 33 0 2210 3 210 2 10 ttt ttt ttt t t t eee eee eee ehf ehf ehf +−= −++−= +−= ⇒ ++==′ ++== ++== β β β βββ βββ βββ −− = +−+ −+−+ +− +− +− = ++= t t ttttt ttt ttt ttt ttt ttt tA e e eeeee eee eee eee eee eee AAIe 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 βββ the characteristic polynomial of 4A is 3)( λλ =∆ , let 2210)( λβλββλ ++=h on the spectrum of 4A ,we have 2 2 1 0 2:)0()0( :)0()0( 1:)0()0( β β β =′′=′′ =′=′ == thf thf hf thus +−− + ++ = + −− + += ++= 120250 161200 232/541 000 000 450 2/ 20250 16200 2340 22 2 2 2 41410 4 tt tt tttt t tt tI AAIe tA βββ 3.23 Show that functions of the same matrix ; that is )()()()( AfAgAgAf = consequently we have AeAe AtAt = 证明同一矩阵的函数具有可交换性,即 )()()()( AfAgAgAf = 因 此有 AeAe AtAt = 成立 proof: let 1 110 1 110 )( )( − − − − +++= +++= n n n n AAIAf AAIAg ααα βββ ( n is the order of A) k ik k nki i n nk k ik k i i n k n nn n in n i i n in n i i AA AAAAIAgAf )()( )()()( 1 22 0 1 0 22 11 1 1 1 1 0 011000 − +−= − = − = − = − −−− = − −− − = ∑∑∑∑ ∑∑ += +++++++= βαβα βαβαβαβαβαβα )()()()()()( 1 22 0 1 0 AgAfAAAgAf kik k nki i n nk k ik k i i n k =+= − +−= − = − = − = ∑∑∑∑ βαβα let AteAgAAf == )(,)( then we have AeAe AtAt = 3.24 let = 3 2 1 00 00 00 λ λ λ C , find a B such that CeB = 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 CeB = Is it true that ,for any nonsingular c ,there exists a matrix B such that CeB = ? 令 = 3 2 1 00 00 00 λ λ λ C 证明若 0111 =⋅⋅ λλλ ,则不存在 B 使 Ce 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, CeB = ? Let λλ λ == ef ln)( so BeBf B == ln)( == 3 2 1 ln00 0ln0 00ln ln λ λ λ CB 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 λλ λ λ λλ ln00 0ln0 01ln ln00 0ln0 0nlln lnCB , where 0,0 ≤> λλ if then λln 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)( AsIAdj s AsI − ∆ =− and let m(s) be the monic greatest common divisor of all entries of Adj(Si-A) , Verify for the matrix 2A in problem 3.13 that the minimal polynomial of A equals )()( sms∆ 令, )( )( 1)( AsIAdj 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 2A 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()( 200 010 001 ˆ 200 010 101 2 2 33 −= − −− −−−− =− − − − =− −−=−−=∆ = − = ssmso s ss sss AsIAdj s S S AsI ssssss AA j we can easily obtain that )()()( smss ∆=j 3.26 Define [ ]1221101 )( 1)( −− −−− ++++ ∆ =− nn nn RsRsRsR s AsI where in nn RandsssAsIs ααα ++++=−=∆ −− 22 1 1 2:)det()( are constant matrices theis definition is valid because the degree in s of the adjoint of (sI-A) is at most n-1 , verify IAR n ARtr IAAAIARR n ARtr IAAIARRARtr IAIARRARtr IRARtr nnn nn nn nnnn αα ααααα αααα ααα α +=−= ++++=+= − −= ++=+=−= +=+=−= =−= − −− −− −−−− 1 1 12 2 1 1 121 1 1 21 2 212 1 3 1101 1 2 0 0 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 Randα is called the leverrier algorithm 定义 ][ )( 1:)( 12 2 1 1 0 1 −− −−− ++++ ∆ =− nn nn RSRSRSR s ASI 其中 )(s∆ 是 A 的特征多项 式 in nnn RandsssAsIs ααα ++++=−=∆ −− 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 IAR n ARtr IAAAIARR n ARtr IAAIARRARtr IAIARRARtr IRARtr nnn nn nn nnnn αα ααααα αααα ααα α +=−= ++++=+= − −= ++=+=−= +=+=−= =−= − −− −− −−−− 1 1 12 2 1 1 121 1 1 21 2 212 1 3 1101 1 2 0 0 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 iR 的程式被称为 leverrier 算法. verification: implieswhich SI SSSSI ARSARRSARRSARRSR RSRSRSRASI nn nnn nnn nnn nn nn )( )( )()()( ])[( 1 2 2 1 1 121 2 12 1 010 12 2 1 1 0 ∆= +++++= +−+−+−+= ++++− − −− −−− −− −− −− αααα ][ )( 1:)( 12 2 1 1 0 1 −− −−− ++++ ∆ =− nn nn RSRSRSR s ASI ’ where n nnn ssss ααα ++++=∆ −− 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: IAR IARR IARR IARR IR nn nnn α α α α =− =− =− =− = − −−− 1 121 212 101 0 multiplying ith equation by 1+−inA yields ( ni 2,1= ) IAR ARAAR ARARA ARARA ARA nn nnn nnn nnn nn α α α α =− =− =− =− = − −−− −−− −− 1 12 2 1 2 21 12 2 1 101 1 0 then we can see 012 2 101 1 0 1 2 2 1 1 =−−++−++= +++++ −−− − − −− nnn nnn nn nnn ARRAARRARARA IAAAA αααα that is 0)( =∆ A 3.28 use problem 3.26 to show [ ]IssAssAsA s AsI n nnnnn )()()( )( 1 )( 1 2 1 13 21 22 1 1 1 − −−−−− − ++++++++++ ∆ = − ααααα 利用题 3.26 证明上式, Proof: [ ]IssAssAsA s IAAASAAA SIAASIAS s RSRSRSR s ASI n nnnnn nn nn nn nn nnn nn nn )()()( )( 1 ])( )()([ )( 1 ][ )( 1:)( 1 2 1 13 21 22 1 1 12 2 1 1 43 3 1 2 3 21 22 1 1 12 2 1 1 0 1 − −−−−− −− −− −− −− −−− −− −−− ++++++++++ ∆ = +++++++++ ++++++ ∆ = ++++ ∆ =− ααααα αααααα ααα another : let 10 =α [ ]IssAssAsA s IAAASAAA SIAASIAS ASAS s AS s SR s RSRSRSR s ASI n nnnnn nn nn nn nn nnn n i In i N i In 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 1 1 0 01 1 1 0 1 0 11 12 2 1 1 0 1 − −−−−− −− −− −− −− −−− − = −− − = −− − = − = − − = −−−− −− −−− ++++++++++ ∆ = +++++++++ ++++++ +++ ∆ = ∆ = ∆ = ++++ ∆ =− ∑∑ ∑ ∑∑ ααααα αααααα ααα αα α 3.29 let all eigenvalues of A be distinct and let iq be a right eigenvector of A associated with iλ that is iIi qqA λ= define [ ]nqqqQ 21= and define == − np p p QP 2 1 1 :; , where ip is the ith row of P , show that ip is a left eigenvector of A associated with iλ , that is iii pAp 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 iIi qqA λ= ,定义 [ ]nqqqQ 21= 并且 == − np p p QP 2 1 1 :; 其中 ip 是 P 的第 I 行, 证明 ip 是 A的关于 iλ 的一个左特征向量,即 iii pAp λ= Proof: all eigenvalues of A are distinct , and iq is a right eigenvector of A associated with iλ , and [ ]nqqqQ 21= so we know that 11 3 2 1 ˆ −− == = PAPAQQA λ λ λ PAPA ˆ=∴ That is = ⇒ = nnnn n n p p p Ap Ap Ap p p p A p p p λ λ λ λ λ λ 22 11 2 1 2 1 2 1 2 1 : so iii pAp λ= , that is , ip is a left eigenvector of A associated with iλ 3.30 show that if all eigenvalues of A are distinct , then 1)( −− ASI can be expressed as ∑ −=− − ii i pq s ASI λ 1)( 1 where iq and ip 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 ∑ −=− − ii i pq s ASI λ 1)( 1 其中 iq 和 ip 是 A的关于 iλ 的右特征值和左特征值, Proof: if all eigenvalues of A are distinct , let iq be a right eigenvector of A associated with iλ , then [ ]nqqqQ 21= is nonsingular , and =− np p p Q 2 1 1: where is a left eigenvector of A associated with iλ , ∑ ∑ ∑∑ ∑ =− − = − − =− − = − − iiiii i iiiii i iiii i ii i pqpqs s pqpqs s Apqpqs s ASIpq s ))((1 )(1)(1 )(1 λ λ λ λλ λ That is ∑ −=− − ii i pq s ASI λ 1)( 1 3.31 find the M to meet the lyapunov equation in (3.59) with == −− = 3 3 3 22 10 CBA 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 MCM CMBAM 3)det()()1)(1()det()( +=−=∆++−+=−=∆ λλλλλλλ BIjjAI BA The eigenvalues of the Lyapunov equation are jjjj −=+−−=+=++−= 231)231 21 ηη The lyapunov equation is nonsingular M satisfying the equation 3.32 repeat problem 3.31 for − = == −− = 3 3 3 3 1 21 10 21 CCBA with two different C ,用本题给出的 A, B, C, 重复题 3.31 的问题, α α α anyforMCM solutionNoCM CMBAM − =⇒= −− ⇒= −− ∴ =+ 311 11 11 11 2 1 1)()1()( 2 −=∆+=∆ λλλλ BA the eigenvalues of the lyapunov equation are 01121 =+−==ηη the lyapunov equation is singular because it has zero eigenvalue if C lies in the range space of the lyapunov equation , then solution exist and are not unique , 3.33 check to see if the following matrices are positive definite or senidefinite 确定下列矩阵 是否正定或者正半定, − − 332313 322212 312111 201 000 100 202 013 232 aaaaaa aaaaaa aaaaaa 1. ∴<−=−= 202 013 232 0792 13 32 det is not positive definite , nor is pesitive semidefinite 2. )21)(21( 201 00 10 det −−+−= − − λλλ λ λ λ it has a negative eigenvalue 21− , so the second matrix is not positive definite , neither is positive demidefinte ,, 3 the third matrix ‘s prindipal minors ,0,0,0 332211 ≥≥≥ aaaaaa ` 0det0det0det0det 332313 322212 312111 3323 3222 3313 3111 2212 2111 = = = = aaaaaa aaaaaa aaaaaa aaaa aaaa aaaa aaaa aaaa aaaa that is all the principal minors of the thire matrix are zero or positive , so the matrix is positive semidefinite , 3.34 compute the singular values of the following matrices 计算下列矩阵的奇值, − − = − − − − − − − 52 22 01 10 21 012 101 42 21 012 101 )6)(1(674)5)(2()( 2 −−=+−=−−−=∆ λλλλλλλ the eigenvalues of − − − − 01 10 21 012 101 are 6 and 1 , thus the singular values of − − 012 101 are 6 and 1 , = − − 206 65 42 21 42 21 )41 2 3 2 25)(41 2 3 2 25(36)20)(5()( −−+−=−−−=∆ λλλλλ the eigenvalues of − − 42 21 42 21 are 41 2 3 2 2541 2 3 2 25 −+ and , thus the singular values of − 42 21 are 70.1)41 2 3 2 25(70.4)41 2 3 2 25( 2 1 2 1 =−=+ and 3.35 if A is symmetric , what is the ralatimship between its eigenvalues and singular values ? 对 称矩阵 A 的特征值与奇异值之间有什么关系? If A is symmetric , then 2AAAAA =′=′ LET iq be an eigenvector of A associated with eigenvaue iλ that is , ),3,1(, niqqA iii == λ thus we have ),3,1(22niqqAqAqA iiiiiii ==== λλλ Which implies 2iλ is the eigenvalue of niforA ,2,1 2 = , (n is the order of A ) So the singular values of A are || iλ nifor ,2,1= where iλ is the eigenvalue of A 3.36 show [ ] m n m mn n n babbb a a a I ∑ = += + 1 21 2 1 1det 证明上式成立 let [ ]n n bbbB a a a A 21 2 1 = = A is 1×n and B is n×1 use (3.64) we can readily obtain [ ] [ ] ∑ ≤ += += + n m mm n nn n n ba a a a bbbIbbb a a a I 1 2 1 21121 2 1 1 detdet 3.37 show (3.65) 证明(3.65) proof: let − − = = = n m n m n m IsB AIsP IsB IsQ Is AIsN 0 0 then we have − − = − = BAsI AssIQP sIBs ABsI NP n m n m 0 0 because PsPQQP PsPNNP detdetdet)det( detdetdet)det( == == we have det(NP)=det(QP) And )det()det()det()det( )det()det()det()det( BAsISBAsIsIQP ABsISsIABsINP n m nm m n nm −=−= −=−= 3.38 Consider yxA = , where A is nm× and has rank m is yAAA ′′ −1)( a solution ? if not , under what condition will it be a solution? Is yAAA 1)( −′′ a solution ? nm× 阶矩阵 A 秩为 m , yAAA ′′ −1)( 是不是方程 yxA = 的解? 如果不是, 那么在 什么条件下, 他才会成为该方程的解? yAAA 1)( −′′ 是不是方程的解? A is nm× and has rank m so we know that nm ≤ , and AA′ is a square matrix of mm× rankA=m , rank( nmrankAAA ≤=≤′ ) , So if nm ≠ ,then rank( AA′ )<n , 1)( −′AA does mot exist m and yAAA ′′ −1)( isn’t a solution if yxA = If m=n , and rankA=m , so A is nonsingular , then we have rank( AA′ )=rank(A)=m , and A yAAA ′′ −1)( =A yyAAA =′′ −− 11 )( that is yAAA ′′ −1)( is a solution , RankA=M ∴ Rank( AA′ )=m AA′ is monsingular and 1)( −′AA exists , so we have yyAAAA =′′ −1)( , that is , yAAA 1)( −′′ is a solution of yxA = , PROBLEMS OF CHAPTER 4 4.1 An oscillation can be generated by 一个振荡器可由下式描述: XX − = 01 10 试证其解为: Show that its solution is )0( cossin sincos )( X tt tt tX − = Proof: )0()0()( 01 10 XeXetX t At −== ,the eigenvalues of A are j,-j; Let λββλ 10)( +=h .If teh λλ =)( ,then on the spectrum of A,then tjtejjh tjtejjh jt jt sincos)( sincos)( 10 10 −==−=− +==+= −ββ ββ then t t sin cos 1 0 = = β β so − = − + =+= tt tt ttAIAh cossin sincos 01 10 sin 10 01 cos)( 10 ββ )0( cossin sincos )0()0()( 01 10 X tt tt XeXetX t At − === − 4.2 Use two different methods to find the unit-step response of用两种方法求下面系统的单位阶 跃响应: UXX + −− = 1 1 22 10 [ ]XY 32= Answer: assuming the initial state is zero state. method1:we use (3.20) to compute − + ++ = + − =− − − s s sss s AsI 2 12 22 1 22 1 )( 2 1 1 then tAt e ttt ttt AsILe −−− −− + =−= sincossin2 sinsincos ))(( 11 and )22( 5 )22( 5)()()( 22 1 ++ = ++ =−= − sssss ssBUAsICsY then tety t sin5)( −= for t>=0 method2: t tt tt Att tA t tA te tete tete C BeeCABdeC dBueCty − −− −− −− − = −− −+ −− = −== = ∫ ∫ sin5 1sin3cos 1sin2cos 01 5.01 )( )()( 01 0 )( 0 )( t tt t t for t>=0 4.3 Discretize the state equation in Problem 4.2 for T=1 and T=π .离散化习题 4.3中的状态方 程,T分别取 1和π Answer: ][][][ ][][]1[ 0 kDUkCXkY kBUdekXekX T AAT += +=+ ∫ αα For T=1,use matlab: [ab,bd]=c2d(a,b,1) ab =0.5083 0.3096 -0.6191 -0.1108 bd =1.0471 -0.1821 [ ] ][32][ ][ 1821.0 0471.1 ][ 1108.06191.0 3096.05083.0 ]1[ kXkY kUkXkX = − + −− =+ for T=π ,use matlab: [ab,bd]=c2d(a,b,3.1415926) ab =-0.0432 0.0000 -0.0000 -0.0432 bd =1.5648 -1.0432 [ ] ][32][ ][ 0432.1 5648.1 ][ 0432.00 00432.0 ]1[ kXkY kUkXkX = − + − − =+ 4.4 Find the companion-form and modal-form equivalent equations of求系统的等价友形和模式 规范形。 UXX + −− − = 1 0 1 220 101 002 [ ]XY 011 −= Answer: use [ab ,bb,cb,db,p]=canon(a,b,c,d,’companion) We get the companion form ab = 0 0 -4 1 0 -6 0 1 -4 bb = 1 0 0 cb = 1 -4 8 db =0 p =1.0000 1.0000 0 0.5000 0.5000 -0.5000 0.2500 0 -0.2500 UXX + − − − = 0 0 1 410 601 400 [ ]XY 841 −= use use [ab ,bb,cb,db,p]=canon(a,b,c,d) we get the modal form ab = -1 1 0 -1 -1 0 0 0 -2 bb = -3.4641 0 1.4142 cb = 0 -0.5774 0.7071 db = 0 p = -1.7321 -1.7321 -1.7321 0 1.7321 0 1.4142 0 0 UXX − + − −− − = 4142.1 0 4641.3 200 011 011 [ ]XY 7071.05774.00 −= 4.5 Find an equivalent state equation of the equation in Problem 4.4 so that all state variables have their larest magnitudes roughly equal to the largest magnitude of the output.If all signals are required to lie inside 10± volts and if the input is a step function with magnitude a,what is the permissible largest a?找出习题 4.4中方程的等价状态方程使所有状态变量的最大量几乎等于 输出的最大值。如果所有的信号需要在 10± 伏以内,输入为大小为 a 的的阶跃函数。求所 允许的最大的 a值。 Answer: first we use matlab to find its unit-step response .we type a=[-2 0 0;1 0 1;0 -2 -2]; b=[1 0 1]';c=[1 -1 0]; d=0; [y,x,t]=step(a,b,c,d) plot(t,y,'.',t,x(:,1),'*',t,x(:,2),'-.',t,x(:,3),'--') 0 1 2 3 4 5 6 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 x2 x1 x3 y so from the fig above.we know max(|y|)=0.55, max(|x1|=0.5;max(|x2|)=1.05,and max(|x3|)=0.52 for unit step input.Define ,,5.0, 332211 xxxxxx === then UXX + −− − = 1 0 1 240 5.005.0 002 [ ]XY 021 −= the largest permissible a is 10/0.55=18.2 4.6 Consider U b b XX + = 2 1 0 0 λ λ [ ]XccY 11= where the overbar denotes complex conjugate.Verify that the equation can be transformed into XCyuBXAX 1=+= with [ ])Re(2)Re(2 1 010 11111 cbcbCBA λλλλλ −= = +− = ,by using the transformation XQX = with − − = 11 11 bb bb Q λ λ 。试验证以上变换成立。 Answer: let XQX = with − − =
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