7.2 Cyclic Decompositions and the Rational Form

这一节的内容不长但是证明很难。核心目的是证明VVV可以成为有限个cyclic space的直和。首先介绍了TTT-admissible的定义，是一个比invariant更强的定义，其能够保证多项式运算在子空间中有对应的分项（投影），Theorem 3是Cyclic Decomposition Theorem，与之前的Primary Decomposition Theorem相比，其说明VVV可以成为惟一的有限个TTT-admissible space的直和，且每个子空间都是一个cyclic space，其generator的TTT-annihilator是可以递归整除的。在本节的前文中，作者称这个定理是one of the deepest results in linear algebra，证明确实非常繁复。这一定理有一系列重要的推论，例如每一个TTT-admissible空间都有一个invariant的互补空间。Theorem 4是广义的Cayley-Hamiltion定理，在之前的最小多项式整除特征多项式的结论之上，还可以推出二者有相同的prime factors，且已知最小多项式就可以得出特征多项式。Theorem 5声明每个矩阵都相似于一个唯一的rational form的矩阵。

Exercises

1.Let TTT be the linear operator on F2F^2F2 which is represented in the standard ordered basis by the matrix [0010]\begin{bmatrix}0&0\\1&0\end{bmatrix}[0100]. Let α1=(0,1)\alpha_1=(0,1)α1=(0,1). Show that F2≠Z(α1;T)F^2\neq Z(\alpha_1;T)F2=Z(α1;T), and that there is no non-zero vector α2\alpha_2α2 in F2F^2F2 with Z(α2;T)Z(\alpha_2;T)Z(α2;T) disjoint from Z(α1;T)Z(\alpha_1;T)Z(α1;T).
Solution: We have
Tα1=[0010][01]=0T\alpha_1=\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=0Tα1=[0100][01]=0
thus pα1=xp_{\alpha_1}=xpα1=x, which means dim⁡Z(α1;T)=1\dim Z(\alpha_1;T)=1dimZ(α1;T)=1, so F2≠Z(α1;T)F^2\neq Z(\alpha_1;T)F2=Z(α1;T).
Suppose there is some α2=(a,b)≠0\alpha_2=(a,b)\neq 0α2=(a,b)=0 such that Z(α2;T)Z(\alpha_2;T)Z(α2;T) is disjoint from Z(α1;T)Z(\alpha_1;T)Z(α1;T), then dim⁡Z(α2;T)=1\dim Z(\alpha_2;T)=1dimZ(α2;T)=1, which means pα2=xp_{\alpha_2}=xpα2=x or Tα2=(a,0)=0T\alpha_2=(a,0)=0Tα2=(a,0)=0, so α2=(0,b)≠0\alpha_2=(0,b)\neq 0α2=(0,b)=0, but this means α2=bα1\alpha_2=b\alpha_1α2=bα1, which contradicts the hypothesis that Z(α2;T)Z(\alpha_2;T)Z(α2;T) is disjoint from Z(α1;T)Z(\alpha_1;T)Z(α1;T).

2.Let TTT be a linear operator on the finite-dimensional space VVV, and let RRR be the range of TTT.
( a ) Prove that RRR has a complementary TTT-invariant subspace if and only if RRR is independent of the null space NNN of TTT.
( b ) If RRR and NNN are independent, prove that NNN is the unique TTT-invariant subspace complementary to RRR.
Solution:
( a ) If RRR is independent of NNN, then from dim⁡R+dim⁡N=dim⁡V\dim R+\dim N=\dim VdimR+dimN=dimV we know that R⊕N=VR\oplus N=VR⊕N=V, and NNN is obviously TTT-invariant. Conversely, if RRR has a complementary TTT-invariant subspace R′R'R′, let β∈R′\beta\in R'β∈R′, then Tβ∈R′T\beta\in R'Tβ∈R′, but also Tβ∈RT\beta\in RTβ∈R, thus Tβ=0T\beta=0Tβ=0 and β∈N\beta\in Nβ∈N, so R′⊆NR'\subseteq NR′⊆N, since dim⁡R′=dim⁡N=dim⁡V−dim⁡R\dim R'=\dim N=\dim V-\dim RdimR′=dimN=dimV−dimR, we know R′=NR'=NR′=N and so R∩N={0}R\cap N=\{0\}R∩N={0}.
( b ) Let R′R'R′ be any TTT-invariant subspace complementary to RRR, from the prood of (a) we can see that R′=NR'=NR′=N, given RRR and NNN are independent.

3.Let TTT be the linear operator on R3R^3R3 which is represented in the standard ordered basis by the matrix
[200120003].\begin{bmatrix}2&0&0\\1&2&0\\0&0&3\end{bmatrix}.⎣⎡210020003⎦⎤.
Let WWW be the null space of T−2IT-2IT−2I. Prove that WWW has no complementary TTT-invariant subspace.
Solution: Assume there exists a TTT-invariant subspace W′W'W′ of R3R^3R3 such that R3=W⊕W′R^3=W{\oplus}W'R3=W⊕W′, then let β=ϵ1\beta=\epsilon_1β=ϵ1, we have (T−2I)β=ϵ2(T-2I)\beta=\epsilon_2(T−2I)β=ϵ2, since (T−2I)ϵ2=0(T-2I)\epsilon_2=0(T−2I)ϵ2=0 we see that (T−2I)β∈W(T-2I)\beta\in W(T−2I)β∈W. On the other hand, since β∈R3\beta\in R^3β∈R3, we can find α∈W,γ∈W′\alpha\in W,\gamma\in W'α∈W,γ∈W′ such that β=α+γ\beta=\alpha+\gammaβ=α+γ, so
(T−2I)β=(T−2I)α+(T−2I)γ∈W(T-2I)\beta=(T-2I)\alpha+(T-2I)\gamma\in W(T−2I)β=(T−2I)α+(T−2I)γ∈W
Since W′W'W′ is TTT-invariant, we see that (T−2I)γ=0(T-2I)\gamma=0(T−2I)γ=0 and (T−2I)β=(T−2I)α(T-2I)\beta=(T-2I)\alpha(T−2I)β=(T−2I)α, but α∈W\alpha\in Wα∈W means (T−2I)α=0(T-2I)\alpha=0(T−2I)α=0, but (T−2I)β=ϵ2(T-2I)\beta=\epsilon_2(T−2I)β=ϵ2, this is a contradiction.

4.Let TTT be the linear operator on F4F^4F4 which is represented in the standard ordered basis by the matrix
[c0001c0001c0001c].\begin{bmatrix}c&0&0&0\\1&c&0&0\\0&1&c&0\\0&0&1&c\end{bmatrix}.⎣⎢⎢⎡c1000c1000c1000c⎦⎥⎥⎤.
Let WWW be the null space of T−cIT-cIT−cI.
( a ) Prove that WWW is the subspace spanned by ϵ4\epsilon_4ϵ4.
( b ) Find the monic generators of the ideals S(ϵ4;W)S(\epsilon_4;W)S(ϵ4;W),S(ϵ3;W)S(\epsilon_3;W)S(ϵ3;W),S(ϵ2;W)S(\epsilon_2;W)S(ϵ2;W),S(ϵ1;W)S(\epsilon_1;W)S(ϵ1;W).
Solution:
( a ) A direct computation shows that the matrix of T−cIT-cIT−cI in the standard ordered basis is the matrix
[0000100001000010]\begin{bmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}⎣⎢⎢⎡0100001000010000⎦⎥⎥⎤
and we have (T−cI)(∑i=14aiϵi)=a1ϵ2+a2ϵ3+a3ϵ4(T-cI)(\sum_{i=1}^4a_i\epsilon_i)=a_1\epsilon_2+a_2\epsilon_3+a_3\epsilon_4(T−cI)(∑i=14aiϵi)=a1ϵ2+a2ϵ3+a3ϵ4, thus WWW consists of all vectors of the form aϵ4a\epsilon_4aϵ4.
( b ) As ϵ4\epsilon_4ϵ4 is already in WWW, we have f(T)ϵ4∈Wf(T)\epsilon_4\in Wf(T)ϵ4∈W for all f∈F[x]f\in F[x]f∈F[x], thus the monic generator of S(ϵ4;W)S(\epsilon_4;W)S(ϵ4;W) is 111.
We have Tϵ3=ϵ4T\epsilon_3=\epsilon_4Tϵ3=ϵ4, so the monic generator of S(ϵ3;W)S(\epsilon_3;W)S(ϵ3;W) is xxx. By the same logic, the monic generator of S(ϵ2;W)S(\epsilon_2;W)S(ϵ2;W) is x2x^2x2 and the monic generator of S(ϵ1;W)S(\epsilon_1;W)S(ϵ1;W) is x3x^3x3.

5.Let TTT be a linear operator on the vector space VVV over the field FFF. If fff is a polynomial over FFF and α∈V\alpha\in Vα∈V, let fα=f(T)αf\alpha=f(T)\alphafα=f(T)α. If V1,…,VkV_1,\dots,V_kV1,…,Vk are TTT-invariable subspaces and V=V1⊕⋯⊕VkV=V_1\oplus\cdots\oplus V_kV=V1⊕⋯⊕Vk, show that fV=fV1⊕⋯⊕fVkfV=fV_1\oplus\cdots\oplus fV_kfV=fV1⊕⋯⊕fVk.
Solution: For α∈V\alpha\in Vα∈V, we have α=α1+⋯+αk\alpha=\alpha_1+\cdots+\alpha_kα=α1+⋯+αk, in which αi∈Vi\alpha_i\in V_iαi∈Vi for i=1,…,ki=1,\dots,ki=1,…,k, so
fα=f(T)α=f(T)(α1+⋯+αk)=∑i=1kf(T)αi=∑i=1kfαif\alpha=f(T)\alpha=f(T)(\alpha_1+\cdots+\alpha_k)=\sum_{i=1}^kf(T)\alpha_i=\sum_{i=1}^kf\alpha_ifα=f(T)α=f(T)(α1+⋯+αk)=i=1∑kf(T)αi=i=1∑kfαi
this shows fV=fV1+⋯+fVkfV=fV_1+\cdots+ fV_kfV=fV1+⋯+fVk. To see the sum is a direct sum, let β∈fVi∩fVj\beta\in fV_i\cap fV_jβ∈fVi∩fVj with i≠ji\neq ji=j, then we can find β′∈Vi,γ′∈Vj\beta'\in V_i,\gamma'\in V_jβ′∈Vi,γ′∈Vj such that β=f(T)β′=f(T)γ′\beta=f(T)\beta'=f(T)\gamma'β=f(T)β′=f(T)γ′, since ViV_iVi and VjV_jVj are TTT-invariant, we have β∈Vi\beta\in V_iβ∈Vi and β∈Vj\beta\in V_jβ∈Vj, so β∈Vi∩Vj\beta\in V_i\cap V_jβ∈Vi∩Vj and β=0\beta=0β=0, this shows ViV_iVi and VjV_jVj are independent.

6.Let T,V,FT,V,FT,V,F be as in Exercise 5. Suppose α\alphaα and β\betaβ are vectors in VVV which have the same TTT-annihilator. Prove that, for any polynomial fff, the vectors fαf\alphafα and fβf\betafβ have the same TTT-annihilator.
Solution: Let ppp be the TTT-annihilator of both α\alphaα and β\betaβ. Suppose the TTT-annihilator of fαf\alphafα is qqq, then qfα=0qf\alpha=0qfα=0, which means qfqfqf is in the ideal generated by ppp, so we can find polynomial hhh such that qf=phqf=phqf=ph, this means qfβ=phβ=hpβ=0qf\beta=ph\beta=hp\beta=0qfβ=phβ=hpβ=0, thus the TTT-annihilator of fβf\betafβ divides qqq, with the same logic applying to the TTT-annihilator of fβf\betafβ, we see qqq divides the TTT-annihilator of fβf\betafβ, thus they are the same.

7.Find the minimal polynomials and the rational forms of each of the following real matrices.
[0−1−1100−100],[c0−10c1−11c],[cos⁡θsin⁡θ−sin⁡θcos⁡θ]\begin{bmatrix}0&-1&-1\\1&0&0\\-1&0&0\end{bmatrix},\quad \begin{bmatrix}c&0&-1\\0&c&1\\-1&1&c\end{bmatrix},\quad\begin{bmatrix}\cos\theta&\sin{\theta}\\{-\sin\theta}&{\cos\theta}\end{bmatrix}⎣⎡01−1−100−100⎦⎤,⎣⎡c0−10c1−11c⎦⎤,[cosθ−sinθsinθcosθ]
Solution: For the first matrix, we compute the characteristic polynomial
∣x11−1x010x∣=x3+x−x=x3\begin{vmatrix}x&1&1\\-1&x&0\\1&0&x\end{vmatrix}=x^3+x-x=x^3∣∣∣∣∣∣x−111x010x∣∣∣∣∣∣=x3+x−x=x3
and the minimal polynomial is also x3x^3x3. Thus the rational form of this matrix is
[000100010]\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}⎣⎡010001000⎦⎤
For the second matrix we compute the characteristic polynomial
∣x−c010x−c−11−1x−c∣=(x−c)[(x−c)2−1]−(x−c)=(x−c)[(x−c)2−2]=x3−3cx2+(3c2−2)x−c3+2c\begin{aligned}\begin{vmatrix}x-c&0&1\\0&x-c&-1\\1&-1&x-c\end{vmatrix}&=(x-c)[(x-c)^2-1]-(x-c)\\&=(x-c)[(x-c)^2-2]\\&=x^3-3cx^2+(3c^2-2)x-c^3+2c\end{aligned}∣∣∣∣∣∣x−c010x−c−11−1x−c∣∣∣∣∣∣=(x−c)[(x−c)2−1]−(x−c)=(x−c)[(x−c)2−2]=x3−3cx2+(3c2−2)x−c3+2c
and the minimal polynomial is also x3−3cx2+(3c2−2)x−c3+2cx^3-3cx^2+(3c^2-2)x-c^3+2cx3−3cx2+(3c2−2)x−c3+2c. Thus the rational form of this matrix is
[00c3−2c10−3c2+2013c]\begin{bmatrix}0&0&c^3-2c\\1&0&-3c^2+2\\0&1&3c\end{bmatrix}⎣⎡010001c3−2c−3c2+23c⎦⎤
For the third matrix we compute the characteristic polynomial
∣x−cos⁡θ−sin⁡θsin⁡θx−cos⁡θ∣=x2−2cos⁡θx+1\begin{vmatrix}x-\cos\theta&-\sin\theta\\\sin\theta&x-\cos\theta\end{vmatrix}=x^2-2\cos\theta x+1∣∣∣∣x−cosθsinθ−sinθx−cosθ∣∣∣∣=x2−2cosθx+1
and the minimal polynomial is also x2−2cos⁡θx+1x^2-2\cos\theta x+1x2−2cosθx+1. Thus the rational form of this matrix is [0−112cos⁡θ]\begin{bmatrix}0&-1\\1&2\cos\theta\end{bmatrix}[01−12cosθ].

8.Let TTT be the linear operator on R3R^3R3 which is represented in the standard basis by
[3−4−4−1322−4−3].\begin{bmatrix}3&-4&-4\\-1&3&2\\2&-4&-3\end{bmatrix}.⎣⎡3−12−43−4−42−3⎦⎤.
Find non-zero vectors α1,…,αr\alpha_1,\dots,\alpha_rα1,…,αr satisfying the conditions of Theorem 3.
Solution: We first compute the characteristic polynomial of TTT:
f=∣x−3441x−3−2−24x+3∣=∣x−3441x−3−202x−2x−1∣=(x−1)3f=\begin{vmatrix}x-3&4&4\\1&x-3&-2\\-2&4&x+3\end{vmatrix}=\begin{vmatrix}x-3&4&4\\1&x-3&-2\\0&2x-2&x-1\end{vmatrix}=(x-1)^3f=∣∣∣∣∣∣x−31−24x−344−2x+3∣∣∣∣∣∣=∣∣∣∣∣∣x−3104x−32x−24−2x−1∣∣∣∣∣∣=(x−1)3
Now the matrix of T−IT-IT−I is obviously not zero and the matrix of (T−I)2(T-I)^2(T−I)2 is
[2−4−4−1222−4−4][2−4−4−1222−4−4]=0\begin{bmatrix}2&-4&-4\\-1&2&2\\2&-4&-4\end{bmatrix}\begin{bmatrix}2&-4&-4\\-1&2&2\\2&-4&-4\end{bmatrix}=0⎣⎡2−12−42−4−42−4⎦⎤⎣⎡2−12−42−4−42−4⎦⎤=0
thus the minimal polynomial for TTT is p=(x−1)2p=(x-1)^2p=(x−1)2. Since Tϵ1=(3,−1,2)T\epsilon_1=(3,-1,2)Tϵ1=(3,−1,2) which is not a scalar multiple of ϵ1\epsilon_1ϵ1, Z(ϵ1;T)Z(\epsilon_1;T)Z(ϵ1;T) has dimension 2 and consists of all vectors
aϵ1+bTϵ1=a(1,0,0)+b(3,−1,2)=(a+3b,−b,2b)a\epsilon_1+bT\epsilon_1=a(1,0,0)+b(3,-1,2)=(a+3b,-b,2b)aϵ1+bTϵ1=a(1,0,0)+b(3,−1,2)=(a+3b,−b,2b)
So we can let α1=ϵ1\alpha_1=\epsilon_1α1=ϵ1, the vector α2\alpha_2α2 must be a characteristic vector of TTT which is not in Z(ϵ1;T)Z(\epsilon_1;T)Z(ϵ1;T), As we can see if α=(x1,x2,x3)\alpha=(x_1,x_2,x_3)α=(x1,x2,x3), then Tα=αT\alpha=\alphaTα=α means α\alphaα is in the form (2a+2b,a,b)(2a+2b,a,b)(2a+2b,a,b), let a=1,b=1a=1,b=1a=1,b=1 we see that we can make α2=(4,1,1)\alpha_2=(4,1,1)α2=(4,1,1).

9.Let AAA be the real matrix
A=[133313−3−3−5].A=\begin{bmatrix}1&3&3\\3&1&3\\-3&-3&-5\end{bmatrix}.A=⎣⎡13−331−333−5⎦⎤.
Find an invertible 3×33\times 33×3 real matrix PPP such that P−1APP^{-1}APP−1AP is in rational form.
Solution: First compute the characteristic polynomial for AAA
det⁡(xI−A)=∣x−1−3−3−3x−1−333x+5∣=∣x−1−3−3−3x−1−30x+2x+2∣=(x+2)(x2−2x+1−9+3x−3+9)=(x+2)2(x−1)\begin{aligned}\det (xI-A)&=\begin{vmatrix}x-1&-3&-3\\-3&x-1&-3\\3&3&x+5\end{vmatrix}=\begin{vmatrix}x-1&-3&-3\\-3&x-1&-3\\0&x+2&x+2\end{vmatrix}\\&=(x+2)(x^2-2x+1-9+3x-3+9)\\&=(x+2)^2(x-1)\end{aligned}det(xI−A)=∣∣∣∣∣∣x−1−33−3x−13−3−3x+5∣∣∣∣∣∣=∣∣∣∣∣∣x−1−30−3x−1x+2−3−3x+2∣∣∣∣∣∣=(x+2)(x2−2x+1−9+3x−3+9)=(x+2)2(x−1)
and since
(A+2I)(A−I)=[333333−3−3−3][033303−3−3−6]=0(A+2I)(A-I)=\begin{bmatrix}3&3&3\\3&3&3\\-3&-3&-3\end{bmatrix}\begin{bmatrix}0&3&3\\3&0&3\\-3&-3&-6\end{bmatrix}=0(A+2I)(A−I)=⎣⎡33−333−333−3⎦⎤⎣⎡03−330−333−6⎦⎤=0
the minimal polynomial for AAA is (x+2)(x−1)=x2+x−2(x+2)(x-1)=x^2+x-2(x+2)(x−1)=x2+x−2.
Since Aϵ1=(1,3,−3)A\epsilon_1=(1,3,-3)Aϵ1=(1,3,−3) is not a scalar multiple of ϵ1\epsilon_1ϵ1, one subspace can be (ϵ1,Aϵ1)(\epsilon_1,A\epsilon_1)(ϵ1,Aϵ1), which consists of vectors like (a+b,3b,−3b)(a+b,3b,-3b)(a+b,3b,−3b), choose a characteristic vector associated with the characteristic value −2-2−2, which may be (1,1,−2)(1,1,-2)(1,1,−2), then let
P=[1110310−3−2]P=\begin{bmatrix}1&1&1\\0&3&1\\0&-3&-2\end{bmatrix}P=⎣⎡10013−311−2⎦⎤
we have det⁡P=−3≠0\det P=-3\neq 0detP=−3=0, thus PPP is invertible, and
AP=[133313−3−3−5][1110310−3−2]=[11−23−3−2−334]AP=\begin{bmatrix}1&3&3\\3&1&3\\-3&-3&-5\end{bmatrix}\begin{bmatrix}1&1&1\\0&3&1\\0&-3&-2\end{bmatrix}=\begin{bmatrix}1&1&-2\\3&-3&-2\\-3&3&4\end{bmatrix}AP=⎣⎡13−331−333−5⎦⎤⎣⎡10013−311−2⎦⎤=⎣⎡13−31−33−2−24⎦⎤
the rational form of AAA is clearly [0201−1000−2]\begin{bmatrix}0&2&0\\1&-1&0\\0&0&-2\end{bmatrix}⎣⎡0102−1000−2⎦⎤, and we have
P[0201−1000−2]=[1110310−3−2][0201−1000−2]=[11−23−3−2−334]P\begin{bmatrix}0&2&0\\1&-1&0\\0&0&-2\end{bmatrix}=\begin{bmatrix}1&1&1\\0&3&1\\0&-3&-2\end{bmatrix}\begin{bmatrix}0&2&0\\1&-1&0\\0&0&-2\end{bmatrix}=\begin{bmatrix}1&1&-2\\3&-3&-2\\-3&3&4\end{bmatrix}P⎣⎡0102−1000−2⎦⎤=⎣⎡10013−311−2⎦⎤⎣⎡0102−1000−2⎦⎤=⎣⎡13−31−33−2−24⎦⎤
thus PPP is the matrix we need.

10.Let FFF be a subfield of the complex numbers and let TTT be the linear operator on F4F^4F4 which is represented in the standard ordered basis by the matrix
[200012000a2000b2].\begin{bmatrix}2&0&0&0\\1&2&0&0\\0&a&2&0\\0&0&b&2\end{bmatrix}.⎣⎢⎢⎡210002a0002b0002⎦⎥⎥⎤.
Find the characteristic polynomial for TTT. Consider the cases a=b=1a=b=1a=b=1; a=b=0a=b=0a=b=0; a=0,b=1a=0,b=1a=0,b=1. In each of these cases, find the minimal polynomial for TTT and non-zero vectors α1,…,αr\alpha_1,\dots,\alpha_rα1,…,αr which satisfy the conditions of Theorem 3.
Solution: The characteristic polynomial for TTT is (x−2)4(x-2)^4(x−2)4.
In the case a=b=1a=b=1a=b=1, the minimal polynomial for TTT is (x−2)4(x-2)^4(x−2)4, since for ϵ1=(1,0,0,0)\epsilon_1=(1,0,0,0)ϵ1=(1,0,0,0), we have Tϵ1=(1,1,0,0)T\epsilon_1=(1,1,0,0)Tϵ1=(1,1,0,0), T2ϵ1=(1,2,1,0)T^2\epsilon_1=(1,2,1,0)T2ϵ1=(1,2,1,0), T3ϵ1=(1,3,3,1)T^3\epsilon_1=(1,3,3,1)T3ϵ1=(1,3,3,1), we have α1=ϵ1\alpha_1=\epsilon_1α1=ϵ1.
In the case a=b=0a=b=0a=b=0 and a=0,b=1a=0,b=1a=0,b=1, the minimal polynomial for TTT is (x−2)2(x-2)^2(x−2)2. The non-zero vectors αi\alpha_iαi satisfing Theorem 3 can be found using techniques in the proof of Theorem 3.

11.Prove that if AAA and BBB are 3×33\times 33×3 matrices over a field FFF, a necessary and sufficient condition that AAA and BBB be similar over FFF is that they have the same characteristic polynomial and the same minimal polynomial. Give an example which shows that this is false for 4×44\times 44×4 matrices.
Solution: If AAA and BBB are similar, then both are similar to the same matrix RRR which is in rational form, thus the minimal polynomial for AAA and BBB are the same. Let it be ppp.
If deg⁡p=3\deg p=3degp=3, then it is the characteristic polynomial for both AAA and BBB.
If deg⁡p=2\deg p=2degp=2, then RRR must have the form [A100a]\begin{bmatrix}A_1&0\\0&a\end{bmatrix}[A100a], where A1A_1A1 is a 2×22\times 22×2 matrix in the rational form, then the characteristic polynomial for AAA and BBB are p(x−a)p(x-a)p(x−a).
If deg⁡p=1\deg p=1degp=1, then RRR must be diagonal, then it is apparent the characteristic polynomial for AAA and BBB are equal.
Conversely, if AAA and BBB have the same characteristic polynomial fff and the same minimal polynomial ppp. We find the unique matrix in the rational form that AAA and BBB are similar to, namely RAR_ARA and RBR_BRB, then
If deg⁡p=3\deg p=3degp=3, then we must have RA=RBR_A=R_BRA=RB.
If deg⁡p=2\deg p=2degp=2, then RA=[A100a]R_A=\begin{bmatrix}A_1&0\\0&a\end{bmatrix}RA=[A100a], RB=[A100b]R_B=\begin{bmatrix}A_1&0\\0&b\end{bmatrix}RB=[A100b], where A1A_1A1 is a 2×22\times 22×2 matrix in the rational form, and f/p=x−a=x−bf/p=x-a=x-bf/p=x−a=x−b means a=ba=ba=b, so RA=RBR_A=R_BRA=RB.
If deg⁡p=1\deg p=1degp=1, then RAR_ARA and RBR_BRB must be diagonal, since the characteristic polynomial for AAA and BBB are equal ,we have RA=RBR_A=R_BRA=RB.
Since AAA is similar to RAR_ARA and BBB is similar to RBR_BRB, we have AAA similar to BBB.
For a counterexample of 4×44\times 44×4 matrix, we let
A=[0−11211],B=[0−1120−112]A=\begin{bmatrix}0&-1\\1&2\\&&1\\&&&1\end{bmatrix},\quad B=\begin{bmatrix}0&-1\\1&2\\&&0&-1\\&&1&2\end{bmatrix}A=⎣⎢⎢⎡01−1211⎦⎥⎥⎤,B=⎣⎢⎢⎡01−1201−12⎦⎥⎥⎤
The characteristic polynomial of AAA and BBB is (x−1)4(x-1)^4(x−1)4, the minimal polynomial of AAA and BBB is (x−1)2(x-1)^2(x−1)2, but AAA and BBB are not similar.

12.Let FFF be a subfield of the field of complex numbers, and let AAA and BBB be n×nn\times nn×n matrices over FFF. Prove that if AAA and BBB are similar over the field of complex numbers, then they are similar over FFF.
Solution: The rational form of AAA is a matrix over FFF, thus a matrix over CCC, likewise for BBB. Thus if AAA and BBB are similar over the field of complex numbers, due to Theorem 5, the rational form of AAA is the same as BBB over CCC, which means AAA and BBB are similar to the same rational form over FFF, and the conclusion follows.

13.Let AAA be an n×nn\times nn×n matrix with complex entires. Prove that if every characteristic value of AAA is real, then AAA is similar to a matrix with real entries.
Solution: The characteristic polynomial for AAA contains only linear factors, and so is the minimal polynomial ppp for AAA, since every characteristic value of AAA is real, we see ppp consists only real coefficients.
Let TTT be the linear operator on FnF^nFn which is represented by AAA in the standard basis, then there is an ordered basis B\mathfrak BB for VVV such that
[T]B=[A1A2⋱Ar][T]_{\mathfrak B}=\begin{bmatrix}A_1\\&A_2\\&&{\ddots}\\&&&&A_r\end{bmatrix}[T]B=⎣⎢⎢⎡A1A2⋱Ar⎦⎥⎥⎤
where each AiA_iAi is the companion matrix of some polynomial pip_ipi, and p1=pp_1=pp1=p, and pi∣pp_i|ppi∣p for i=2,…,ri=2,\dots,ri=2,…,r, since ppp consists only real coefficients, so are all pip_ipi, which means all AiA_iAi have real entries, and so is [T]B[T]_{\mathfrak B}[T]B, apparently, AAA is similar to [T]B[T]_{\mathfrak B}[T]B.

14.Let TTT be a linear operator on the finite-dimensional space VVV. Prove that there exists a vector α∈V\alpha\in Vα∈V with this property. If fff is a polynomial and f(T)α=0f(T)\alpha=0f(T)α=0, then f(T)=0f(T)=0f(T)=0. (Such a vector α\alphaα is called a separating vector for the algebra of polynomials in TTT.) When TTT has a cyclic vector, give a direct proof that any cyclic vector is a separating vector for the algebra of polynomials in TTT.
Solution: We first prove if α\alphaα is a cyclic vector of TTT, then α\alphaα is a separating vector for the algebra of polynomials in TTT. Suppose dim⁡V=n\dim V=ndimV=n, then α,…,Tn−1α\alpha,\dots,T^{n-1}\alphaα,…,Tn−1α span VVV, so for any β∈V\beta\in Vβ∈V, we have β=g(T)α\beta=g(T)\alphaβ=g(T)α for some polynomial ggg. Now if fff is a polynomial and f(T)α=0f(T)\alpha=0f(T)α=0, we have
f(T)β=f(T)g(T)α=g(T)f(T)α=g(T)0=0f(T)\beta=f(T)g(T)\alpha=g(T)f(T)\alpha=g(T)0=0f(T)β=f(T)g(T)α=g(T)f(T)α=g(T)0=0
thus f(T)=0f(T)=0f(T)=0.
Now for any linear operator TTT on VVV, use the Cyclic Decomposition Theorem, we can write V=Z(α1;T)⊕⋯⊕Z(αr;T)V=Z(\alpha_1;T)\oplus\cdots\oplus Z(\alpha_r;T)V=Z(α1;T)⊕⋯⊕Z(αr;T), let α=α1+⋯+αr\alpha=\alpha_1+\cdots+\alpha_rα=α1+⋯+αr, then f(T)α=0f(T)\alpha=0f(T)α=0 means f(T)αi=0f(T)\alpha_i=0f(T)αi=0 since each Z(αi;T)Z(\alpha_i;T)Z(αi;T) is invariant under TTT, and VVV is the direct sum of all Z(αi;T)Z(\alpha_i;T)Z(αi;T). It follows that f(T)=0f(T)=0f(T)=0 on Z(αi;T)Z(\alpha_i;T)Z(αi;T) for i=1,…,ri=1,\dots,ri=1,…,r, which means f(T)=0f(T)=0f(T)=0 on VVV.

15.Let FFF be a subfield of the field of complex numbers, and let AAA be an n×nn\times nn×n matrix over FFF. Let ppp be the minimal polynomial for AAA. If we regard AAA as a matrix over CCC, then AAA has a minimal polynomial fff as an n×nn\times nn×n matrix over CCC. Use a theorem on linear equations to prove p=fp=fp=f. Can you also see how this follows from the cyclic decomposition theorem?
Solution: If we write p=c0+c1x+⋯+xkp=c_0+c_1x+\cdots+x^kp=c0+c1x+⋯+xk, then p(A)=0p(A)=0p(A)=0 means (c0,c1,…,1)(c_0,c_1,\dots,1)(c0,c1,…,1) is a solution for the system
x1I+⋯+xk+1Ak=0x_1I+\cdots+x_{k+1}A^k=0x1I+⋯+xk+1Ak=0
in the field FFF, and likewise, f=d0+d1x+⋯+xkf=d_0+d_1x+\cdots+x^kf=d0+d1x+⋯+xk is the polynomial which has coefficients (d0,d1,…,1)(d_0,d_1,\dots,1)(d0,d1,…,1) as a solution for the same system in the field CCC, then (d0,d1,…,1)(d_0,d_1,\dots,1)(d0,d1,…,1) is a solution in FFF due to the final remark in Sec 1.4. Thus both (c0,c1,…,1)(c_0,c_1,\dots,1)(c0,c1,…,1) and (d0,d1,…,1)(d_0,d_1,\dots,1)(d0,d1,…,1) are in the solution space for x1I+⋯+xk+1Ak=0x_1I+\cdots+x_{k+1}A^k=0x1I+⋯+xk+1Ak=0. To prove p=fp=fp=f, assume there is some ci≠dic_i\neq d_ici=di, then (c0,c1,…,ck−1)−(d0,d1,…,dk−1)(c_0,c_1,\dots,c_{k-1})-(d_0,d_1,\dots,d_{k-1})(c0,c1,…,ck−1)−(d0,d1,…,dk−1) is a non-trivial solution for the system
x1I+⋯+xkAk−1=0x_1I+\cdots+x_{k}A^{k-1}=0x1I+⋯+xkAk−1=0
Let hi=ci−dih_i=c_i-d_ihi=ci−di and h=h0+h1x+⋯+hk−1xk−1h=h_0+h_1x+\cdots+h_{k-1}x^{k-1}h=h0+h1x+⋯+hk−1xk−1, we see h(A)=0h(A)=0h(A)=0, but deg⁡h<deg⁡p\deg h<\deg pdegh<degp, a contradiction.
To get this result from the cyclic decomposition theorem, notice that by Exercise 12, AAA has the same rational form in FFF and CCC, and the first block matrix of the rational form of AAA is the companion matrix of ppp in FFF and fff in CCC, we have p=fp=fp=f.

16.Let AAA be an n×nn\times nn×n matrix with real entries such that A2+I=0A^2+I=0A2+I=0. Prove that nnn is even, and if n=2kn=2kn=2k, then AAA is similar over the field of real numbers to a matrix of the block form [0−II0]\begin{bmatrix}0&-I\\I&0\end{bmatrix}[0I−I0] where III is the k×kk\times kk×k identity matrix.
Solution: The minimal polynomial for AAA is x2+1x^2+1x2+1, by the generalized Cayley-Hamilton Theorem, the characteristic polynomial for AAA must be of the form f=(x2+1)kf=(x^2+1)^kf=(x2+1)k, so n=deg⁡fn=\deg fn=degf is even.
If n=2kn=2kn=2k, we know AAA is similar to one and only one matrix BBB in the rational form. If we write
B=[A1⋱Ar]B=\begin{bmatrix}A_1\\&\ddots\\&&A_r\end{bmatrix}B=⎣⎡A1⋱Ar⎦⎤
where each AiA_iAi is the companion matrix of pip_ipi, and pi+1p_{i+1}pi+1 divides pip_ipi, from the proof of Theorem 3 we know p1=x2+1p_1=x^2+1p1=x2+1, and the only possible polynomial which divides x2+1x^2+1x2+1 is x2+1x^2+1x2+1 and 111. Since 111 can only be the annihilator of zero vectors, we see that
B=[A1⋱Ak],Ai=[−11],i=1,…,kB=\begin{bmatrix}A_1\\&\ddots\\&&A_k\end{bmatrix}, \quad A_i=\begin{bmatrix}&-1\\1\end{bmatrix},i=1,\dots,kB=⎣⎡A1⋱Ak⎦⎤,Ai=[1−1],i=1,…,k
Let B={ϵ1,…,ϵn}\mathscr B=\{\epsilon_1,\dots,\epsilon_n\}B={ϵ1,…,ϵn} be a basis for RnR^nRn and TTT is the linear operator with [T]B=B[T]_{\mathscr B}=B[T]B=B, then
Tϵ2i−1=ϵ2i,Tϵ2i=−ϵ2i−1,i=1,…,kT\epsilon_{2i-1}=\epsilon_{2i},\quad T\epsilon_{2i}=-\epsilon_{2i-1},\quad i=1,\dots,kTϵ2i−1=ϵ2i,Tϵ2i=−ϵ2i−1,i=1,…,k
If we let αi=ϵ2i−1\alpha_i=\epsilon_{2i-1}αi=ϵ2i−1 for i=1,…,ki=1,\dots,ki=1,…,k and αi=ϵ2i−2k\alpha_i=\epsilon_{2i-2k}αi=ϵ2i−2k for i=k+1,…,ni=k+1,\dots,ni=k+1,…,n, then B′={α1,…,αn}\mathscr B'=\{\alpha_1,\dots,\alpha_n\}B′={α1,…,αn} is a basis for VVV, and we can verify [T]B′=[0−II0][T]_{\mathscr B'}=\begin{bmatrix}0&-I\\I&0\end{bmatrix}[T]B′=[0I−I0], which means BBB is similar to [0−II0]\begin{bmatrix}0&-I\\I&0\end{bmatrix}[0I−I0] and so is AAA.

17.Let TTT be a linear operator on a finite-dimensional vector space VVV. Suppose that
( a ) the minimal polynomial for TTT is a power of an irreducible polynomial;
( b ) the minimal polynomial is equal to the characteristic polynomial.
Show that no non-trivial TTT-invariant subspace has a complementary TTT-invariant subspace.
Solution: Let WWW be a non-trivial TTT-invariant subspace of VVV, assume there is W′W'W′ which is TTT-invariant such that W⊕W′=VW\oplus W'=VW⊕W′=V, let TW=UT_W=UTW=U and TW′=U′T_{W'}=U'TW′=U′, then the minimal polynomial ppp of TWT_WTW and p′p'p′ of TW′T_{W'}TW′ divide the minimal polynomial for TTT. Since the minimal polynomial for TTT is of the form qnq^nqn where qqq is irreducible, we have p=qrp=q^rp=qr and p′=qsp'=q^sp′=qs, where r+s≤nr+s\leq nr+s≤n. As WWW is non-trivial, we have r≥1r\geq 1r≥1.
Now if s≥1s\geq 1s≥1, we can get a contradiction by the following procedure: from (b) we know that TTT has a cyclic vector α\alphaα such that the TTT-annihilator of α\alphaα is qnq^nqn, and there is α1∈W,α2∈W′\alpha_1\in W,\alpha_2\in W'α1∈W,α2∈W′ such that α=α1+α2\alpha=\alpha_1+\alpha_2α=α1+α2, we let k=max⁡(r,s)k=\max(r,s)k=max(r,s), then 1≤k<n1\leq k<n1≤k<n, and qk(T)α1=qk(T)α2=0q^k(T)\alpha_1=q^k(T)\alpha_2=0qk(T)α1=qk(T)α2=0, which means qk(T)α=0q^k(T)\alpha=0qk(T)α=0, this is a contradiction.
Thus s=0s=0s=0, or the minimal polynomial for TW′T_{W'}TW′ is 111, which means

18.If TTT is a diagonalizable linear operator, then every TTT-invariant subspace has a complementary TTT-invariant subspace.
Solution: TTT is diagonalizable means if we let c1,…,ckc_1,\dots,c_kc1,…,ck be distinct characteristic values of TTT and let Vi=null (T−ciI)V_i=\text{null }(T-c_iI)Vi=null (T−ciI), then
V=V1⊕⋯⊕VkV=V_1\oplus\cdots\oplus V_kV=V1⊕⋯⊕Vk
Let WWW be a TTT-invariant subspace of VVV, then by Exercise 10 of Section 6.8, we have
W=(W∩V1)⊕⋯⊕(W∩Vk)W=(W\cap V_1)\oplus\cdots\oplus(W\cap V_k)W=(W∩V1)⊕⋯⊕(W∩Vk)
Consider W∩ViW\cap V_iW∩Vi, for any β∈W∩Vi\beta\in W\cap V_iβ∈W∩Vi, we have β∈Vi\beta\in V_iβ∈Vi, so Tβ=ciβT\beta=c_i\betaTβ=ciβ. Since W∩ViW\cap V_iW∩Vi is a subspace, we can find {α1,…,αri}\{\alpha_1,\dots,\alpha_{r_i}\}{α1,…,αri} to be a basis for it, then it can be extended to a basis for ViV_iVi, namely {α1,…,αsi}\{\alpha_1,\dots,\alpha_{s_i}\}{α1,…,αsi}, all of which are characteristic vectors associated with cic_ici. Let UiU_iUi be the space spanned by {αri+1,…,αsi}\{\alpha_{{r_i}+1},\dots,\alpha_{s_i}\}{αri+1,…,αsi}, then (W∩Vi)⊕Ui=Vi(W\cap V_i)\oplus U_i=V_i(W∩Vi)⊕Ui=Vi, let U=U1⊕⋯⊕UkU=U_1\oplus\cdots\oplus U_kU=U1⊕⋯⊕Uk, we see that V=W⊕UV=W\oplus UV=W⊕U, and as each UiU_iUi is invariant under TTT, so is UUU.

19.Let TTT be a linear operator on the finite-dimensional space VVV. Prove that TTT has a cyclic vector if and only if the following is true: Every linear operator UUU which commutes with TTT is a polynomial in TTT.
Solution: First suppose α\alphaα is a cyclic vector of TTT, then if dim⁡V=n\dim V=ndimV=n, we have {α,Tα,…,Tn−1α}\{\alpha,T\alpha,\dots,T^{n-1}\alpha\}{α,Tα,…,Tn−1α} being a basis for VVV. Given an operator UUU which commutes with TTT, we have
Uα=a0α+⋯+an−1Tn−1α=f(T)αU\alpha=a_0\alpha+\cdots+a_{n-1}T^{n-1}\alpha=f(T)\alphaUα=a0α+⋯+an−1Tn−1α=f(T)α
where f(x)=a0+a1x+⋯+an−1xn−1f(x)=a_0+a_1x+\cdots+a_{n-1}x^{n-1}f(x)=a0+a1x+⋯+an−1xn−1, notice that UTkα=TkUα=Tkf(T)α=f(T)Tkα,k=2,…n−1UT^k\alpha=T^kU\alpha=T^kf(T)\alpha=f(T)T^k\alpha,\quad k=2,\dots n-1UTkα=TkUα=Tkf(T)α=f(T)Tkα,k=2,…n−1
We can see that U=f(T)U=f(T)U=f(T) on a basis for VVV, thus on VVV.
Conversely, if every linear operator UUU which commutes with TTT is a polynomial in TTT, let the cyclic decomposition of VVV by TTT be
V=Z(α1;T)⊕⋯⊕Z(αr;T)V=Z(\alpha_1;T)\oplus\cdots\oplus Z(\alpha_r;T)V=Z(α1;T)⊕⋯⊕Z(αr;T)
and pip_ipi is the TTT-annihilator for αi\alpha_iαi with pi+1∣pip_{i+1}|p_ipi+1∣pi. Define UUU as follows: Uα=0U\alpha=0Uα=0 if α∈Z(α1;T)\alpha\in Z(\alpha_1;T)α∈Z(α1;T) and Uα=αU\alpha=\alphaUα=α if α∈Z(αi;T),i=2,…,r\alpha\in Z(\alpha_i;T),i=2,\dots,rα∈Z(αi;T),i=2,…,r. For any β∈V\beta\in Vβ∈V, we have β=β1+⋯+βr\beta=\beta_1+\cdots+\beta_rβ=β1+⋯+βr where each βi∈Z(αi;T)\beta_i\in Z(\alpha_i;T)βi∈Z(αi;T), so
UTβ=U(Tβ1+Tβ2+⋯+Tβr)=U(Tβ2+⋯+Tβr)=T(β2+⋯+βr)=T(Uβ2+⋯+Uβr)=TUβ\begin{aligned}UT\beta&=U(T\beta_1+T\beta_2+\cdots+T\beta_r)=U(T\beta_2+\cdots+T\beta_r)\\&=T(\beta_2+\cdots+\beta_r)=T(U\beta_2+\cdots+U\beta_r)=TU\beta\end{aligned}UTβ=U(Tβ1+Tβ2+⋯+Tβr)=U(Tβ2+⋯+Tβr)=T(β2+⋯+βr)=T(Uβ2+⋯+Uβr)=TUβ
Then UUU commutes with TTT, thus is a polynomial for TTT. Let U=q(T)U=q(T)U=q(T), since q(T)α1=0q(T)\alpha_1=0q(T)α1=0, we know p1∣qp_1|qp1∣q, which means pi∣qp_i|qpi∣q for i≥2i\geq 2i≥2, so αi=Uαi=q(T)αi=0\alpha_i=U\alpha_i=q(T)\alpha_i=0αi=Uαi=q(T)αi=0 for i≥2i\geq 2i≥2, which means Z(αi;T)={0}Z(\alpha_i;T)=\{0\}Z(αi;T)={0} for i≥2i\geq 2i≥2, so V=Z(α1;T)V=Z(\alpha_1;T)V=Z(α1;T) and TTT has a cyclic vector.

20.Let VVV be a finite-dimensional vector space over the field FFF, and let TTT be a linear operator on VVV. We ask when it is true that every non-zero vector in VVV is a cyclic vector for TTT. Prove that this is the case if and only if the characteristic polynomial for TTT is irreducible over FFF.
Solution: Let dim⁡V=n\dim V=ndimV=n. First suppose the characteristic polynomial fff for TTT is irreducible over FFF, then by the Generalized Cayley-Hamiltion Theorem, the minimal polynomial ppp for TTT is equal to fff and irreducible over FFF. For any nonzero vector α∈V\alpha\in Vα∈V, if α,Tα,…,Tn−1α\alpha,T\alpha,\dots,T^{n-1}\alphaα,Tα,…,Tn−1α is linearly dependent, then there is g∈F[x]g\in F[x]g∈F[x] with deg⁡g<n\deg g<ndegg<n such that g(T)α=0g(T)\alpha=0g(T)α=0, let pαp_{\alpha}pα be the TTT-annihilator for α\alphaα, then pα∣gp_{\alpha}|gpα∣g and deg⁡pα<n\deg p_{\alpha}<ndegpα<n, further from α≠0\alpha\neq 0α=0 we know deg⁡pα>1\deg p_{\alpha}>1degpα>1, since p(T)α=0p(T)\alpha=0p(T)α=0, we see pα∣pp_{\alpha}|ppα∣p, a contradiction to ppp being irreducible. Then it means α,Tα,…,Tn−1α\alpha,T\alpha,\dots,T^{n-1}\alphaα,Tα,…,Tn−1α is linearly independent, or Z(α;T)=VZ(\alpha;T)=VZ(α;T)=V.
Conversely, if every non-zero vector in VVV is a cyclic vector for TTT, and assume the characteristic polynomial fff for TTT is not irreducible over FFF, then if the minimal polynomial ppp for TTT is not the same as fff, we have deg⁡p<deg⁡f=n\deg p<\deg f=ndegp<degf=n, there is a vector α∈V\alpha\in Vα∈V such that the TTT-annihilator for α\alphaα is ppp, so Z(α;T)Z(\alpha;T)Z(α;T) has dimension deg⁡p\deg pdegp, which means α\alphaα is not a cyclic vector for TTT.
If p=fp=fp=f and f=ghf=ghf=gh where deg⁡g≥1,deg⁡h≥1\deg g\geq1,\deg h\geq1degg≥1,degh≥1, then it is apparent deg⁡h<n\deg h<ndegh<n, let h=h0+h1x+⋯+xkh=h_0+h_1x+\cdots+x^kh=h0+h1x+⋯+xk, there is a vector α∈V\alpha\in Vα∈V such that the TTT-annihilator for α\alphaα is ppp, thus g(T)h(T)α=0g(T)h(T)\alpha=0g(T)h(T)α=0, and β=g(T)α≠0\beta=g(T)\alpha\neq 0β=g(T)α=0. Notice that
h0β+h1Tβ+⋯+Tkβ=h(T)β=h(T)g(T)α=0h_0\beta+h_1T\beta+\cdots+T^k\beta=h(T)\beta=h(T)g(T)\alpha=0h0β+h1Tβ+⋯+Tkβ=h(T)β=h(T)g(T)α=0
this shows β,Tβ,…,Tkβ\beta,T\beta,\dots,T^k\betaβ,Tβ,…,Tkβ is linearly dependent, thus by Theorem 1, dim⁡Z(β;T)≤k=deg⁡h<n\dim Z(\beta;T)\leq k=\deg h<ndimZ(β;T)≤k=degh<n, so β\betaβ is not a cyclic vector for TTT.

21.Let AAA be an n×nn\times nn×n matrix with real entries. Let TTT be the linear operator on RnR^nRn which is represented by AAA in the standard ordered basis, and let UUU be the linear operator on CnC^nCn which is represented by AAA in the standard ordered basis. Use the result of Exercise 20 to prove the following: If the only subspace invariant under TTT are RnR^nRn and the zero subspace, then UUU is diagonalizable.
Solution: Since AAA is real, the characteristic polynomial for TTT and UUU are equal, both are f=det⁡(xI−A)f=\det(xI-A)f=det(xI−A). Now given any nonzero vector α∈Rn\alpha\in R^nα∈Rn, the cyclic space Z(α;T)Z(\alpha;T)Z(α;T) must be RnR^nRn since it is invariant under TTT and contains α\alphaα, so by Exercise 20, fff is irreducible over RRR, which means fff must be of the form x−cx-cx−c or x2+dx^2+dx2+d where d>0d>0d>0. Then in the field CCC, fff can be factored into prime factors and thus UUU is diagonalizable.

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7.2 Cyclic Decompositions and the Rational Form

Exercises

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