6.4 Invariant Subspaces

这一节除了讨论invariant subspaces外，还有很多其他扩展性的内容。例如EXAMPLE 8说明：和TTT可交换的operator的range和null space都是在TTT下invariant的。对于invariant subspace WWW，限制在某一subspace WWW上的operator TWT_WTW可以定义，通过对WWW和TTT的矩阵形式讨论，可以得到TWT_WTW的特征多项式、最小多项式都整除TTT的特征多项式、最小多项式这一结论。EXAMPLE 10实际是对Theorem 2在另一个层面的讨论。
下一部分内容是本节的核心，首先定义Conductor，用TTT取值时将某一向量α\alphaα送入WWW的所有多项式集合S(α;W)S(\alpha;W)S(α;W)是一个ideal，其generator称为TTT-conductor of α\alphaα into WWW。每一个TTT-conductor都整除TTT的最小多项式，因为最小多项式将α\alphaα送入0。利用conductor性质可以证明Lemma，继而证明Theorem 5，即TTT有三角化矩阵的充要条件是TTT的最小多项式可以被totally factored，或者说是一系列线性多项式的乘积。这一定理的推论说明，在algebraically closed的域上所有矩阵都与三角阵相似。Theorem 6的结论是对角化的充要条件是TTT的最小多项式是一次线性多项式的乘积，或者说没有重复根。这一定理可以用于判断是否对角化：在得到特征多项式后，直接计算factor一次幂的乘积所导致的operator是不是zero operator。
Theorem 5提供了一个新的对（algebraically closed field上）Cayley-Hamilton定理的证明，因为任何一个operatorTTT都有一组ordered basis{α1,⋯,αn}\{\alpha_1,\cdots,\alpha_n\}{α1,⋯,αn}下的上三角矩阵AAA，故TTT的特征多项式是f=(x−A11)⋯(x−Ann)f=(x-A_{11})\cdots(x-A_{nn})f=(x−A11)⋯(x−Ann)，由于T−AiiIT-A_{ii}IT−AiiI将αi\alpha_iαi送入span {α1,⋯,αi−1}\text{span }\{\alpha_1,\cdots,\alpha_{i-1}\}span {α1,⋯,αi−1}中，因此f(T)=0f(T)=0f(T)=0.

Exercises

1.Let TTT be the linear operator on R2R^2R2, the matrix of which in the standard ordered basis is A=[1−122]A=\begin{bmatrix}1&-1\\2&2\end{bmatrix}A=[12−12].
( a ) Prove that the only subspaces of R2R^2R2 invariant under TTT are R2R^2R2 and the zero subspace.
( b ) If UUU is the linear operator on C2C^2C2, the matrix of which in the standard ordered basis is AAA, show that UUU has 111-dimensional invariant subspaces.
Solution:
( a ) The characteristic polynomial of TTT is
det⁡(xI−A)=∣x−11−2x−2∣=x2−3x+4\det (xI-A)=\begin{vmatrix}x-1&1\\-2&x-2\end{vmatrix}=x^2-3x+4det(xI−A)=∣∣∣∣x−1−21x−2∣∣∣∣=x2−3x+4
Thus AAA has no characteristic value on RRR.
( b ) The characteristic polynomial of UUU is
det⁡(xI−A)=∣x−11−2x−2∣=(x−32+72i)(x−32−72i)\det (xI-A)=\begin{vmatrix}x-1&1\\-2&x-2\end{vmatrix}=\left(x-\frac{3}{2}+\frac{\sqrt{7}}{2}i\right)\left(x-\frac{3}{2}-\frac{\sqrt{7}}{2}i\right)det(xI−A)=∣∣∣∣x−1−21x−2∣∣∣∣=(x−23+27i)(x−23−27i)
Thus UUU has characteristic value 32−72i\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}i23−27i, one characteristic vector associated with this value is (−4,1+7i)(-4,1+\sqrt{7}i)(−4,1+7i). The subspace spanned by this vector is invariant under UUU.

2.Let WWW be an invariant subspace for TTT. Prove that the minimal polynomial for the restriction operator TWT_WTW divides the minimal polynomial for TTT, without referring to matrices.
Solution: Let ppp be the minimal polynomial for TTT, then p(T)=0p(T)=0p(T)=0, using the conclusion from Lemma we know that WWW is invariant under p(T)p(T)p(T), so for any α∈W\alpha\in Wα∈W, we have p(TW)α=p(T)α=0p(T_W)\alpha=p(T)\alpha=0p(TW)α=p(T)α=0, so p(TW)=0p(T_W)=0p(TW)=0, which means the minimal polynomial for TWT_WTW divides ppp.

3.Let ccc be a characteristic value of TTT and let WWW be the space of characteristic vectors associated with the characteristic value ccc. What is the restriction operator TWT_WTW?
Solution: It is the operator which multiplies all vector in WWW by ccc. Since for any α∈W\alpha \in Wα∈W we have Tα=cαT\alpha=c\alphaTα=cα.

4.Let
A=[0102−222−32].A=\begin{bmatrix}0&1&0\\2&-2&2\\2&-3&2\end{bmatrix}.A=⎣⎡0221−2−3022⎦⎤.
Is AAA similar over the field of real numbers to a triangular matrix? If so, find such a triangular matrix.
Solution: We shall compute the minimal polynomial for AAA. First the characteristic polynomial for AAA is
det⁡(xI−A)=∣x−10−2x+2−2−23x−2∣=x∣x+2−23x−2∣+∣−2−2−2x−2∣=x(x2−4+6)−2x=x3\begin{aligned}\det (xI-A)&=\begin{vmatrix}x&-1&0\\-2&x+2&-2\\-2&3&x-2\end{vmatrix}=x\begin{vmatrix}x+2&-2\\3&x-2\end{vmatrix}+\begin{vmatrix}-2&-2\\-2&x-2\end{vmatrix}\\&=x(x^2-4+6)-2x=x^3\end{aligned}det(xI−A)=∣∣∣∣∣∣x−2−2−1x+230−2x−2∣∣∣∣∣∣=x∣∣∣∣x+23−2x−2∣∣∣∣+∣∣∣∣−2−2−2x−2∣∣∣∣=x(x2−4+6)−2x=x3
Thus AAA is similar to a triangular matrix, and the minimal polynomial is x3x^3x3. The solution space of AX=0AX=0AX=0 is spanned by α1=(1,0,−1)\alpha_1=(1,0,-1)α1=(1,0,−1). Choose (1,1,1)(1,1,1)(1,1,1) and we see α2=A(1,1,1)T=(1,2,1)\alpha_2=A(1,1,1)^T=(1,2,1)α2=A(1,1,1)T=(1,2,1) satisfies Aα2=2α1A\alpha_2=2\alpha_1Aα2=2α1, now α3=(1,1,1)\alpha_3=(1,1,1)α3=(1,1,1) is valid since Aα3=α2A\alpha_3=\alpha_2Aα3=α2. Thus we have
A[α1,α2,α3]=[0,2α1,α2]=[α1,α2,α3][020001000]A[\alpha_1,\alpha_2,\alpha_3]=[0,2\alpha_1,\alpha_2]=[\alpha_1,\alpha_2,\alpha_3]\begin{bmatrix}0&2&0\\0&0&1\\0&0&0\end{bmatrix}A[α1,α2,α3]=[0,2α1,α2]=[α1,α2,α3]⎣⎡000200010⎦⎤

5.Every matrix AAA such that A2=AA^2=AA2=A is similar to a diagonal matrix.
Solution: If A=0A=0A=0 or A=IA=IA=I then AAA is already a diagonal matrix. Suppose A≠0,A≠IA\neq 0,A\neq IA=0,A=I, we have A(A−I)=0A(A-I)=0A(A−I)=0 and so the minimal polynomial of AAA is x(x−1)x(x-1)x(x−1), and the conclusion follows from Theorem 6.

6.Let TTT be a diagonalizable linear operator on the nnn-dimensional vector space VVV, and let WWW be a subspace which is invariant under TTT. Prove that the restriction operator TWT_WTW is diagonalizable.
Solution: If TTT is diagonalizable, then the minimal polynomial for TTT has the form p=(x−c1)⋯(x−ck)p=(x-c_1)\cdots(x-c_k)p=(x−c1)⋯(x−ck), since the minimal polynomial for TWT_WTW divides the minimal polynomial for TTT, it must have the form
(x−cj1)⋯(x−cji),j1,⋯,ji∈{1,⋯,k}(x-c_{j_1})\cdots(x-c_{j_i}),\quad j_1,\cdots,j_i\in \{1,\cdots,k\}(x−cj1)⋯(x−cji),j1,⋯,ji∈{1,⋯,k}
and the conclusion follows from Theorem 6.

7.Let TTT be a linear operator on a finite-dimensional vector space over the field of complex numbers. Prove that TTT is diagonalizable if and only if TTT is annihilated by some polynomial over CCC which has distinct roots.
Solution: If TTT is a diagonalizable linear operator, then the minimal polynomial for TTT is a product of distinct linear factors. Conversely, if TTT is annihilated by some polynomial ppp over CCC which has distinct roots, then the minimal polynomial must has no distinct roots since it divides this ppp, and the conclusion follows from Theorem 6.

8.Let TTT be a linear operator on VVV. If every subspace of VVV is invariant under TTT, then TTT is a scalar multiple of the identity operator.
Solution: Let {α1,⋯,αn}\{\alpha_1,\cdots,\alpha_n\}{α1,⋯,αn} be a basis for VVV, then the space spanned by αi\alpha_iαi is a subspace of VVV, thus invariant under TTT, which means we have Tαi=kiαiT\alpha_i=k_i\alpha_iTαi=kiαi for i=1,⋯,ni=1,\cdots,ni=1,⋯,n. Now assume ki≠kjk_i\neq k_jki=kj for some i,ji,ji,j, then consider the subspace WWW spanned by αi+αj\alpha_i+\alpha_jαi+αj, we have T(αi+αj)=kiαi+kjαj∉WT(\alpha_i+\alpha_j)=k_i\alpha_i+k_j\alpha_j\notin WT(αi+αj)=kiαi+kjαj∈/W, a contradiction.

9.Let TTT be the indefinite integral operator (Tf)(x)=∫0xf(t)dt(Tf)(x)=\int_0^xf(t)dt(Tf)(x)=∫0xf(t)dt on the space of continuous functions on the interval [0,1][0,1][0,1]. Is the space of polynomial functions invariant under TTT? The space of differentiable functions? The space of functions which vanish at x=12x=\frac{1}{2}x=21?
Solution: The space of polynomial functions are invariant under TTT, since if p=∑i=0ncixip=\sum_{i=0}^nc_ix^ip=∑i=0ncixi, then T(p)=∑i=1n+1ci−1ixiT(p)=\sum_{i=1}^{n+1}\dfrac{c_{i-1}}{i}x^iT(p)=∑i=1n+1ici−1xi is still a polynomial function.
The space of differentiable functions are invariant under TTT, since if fff is differentiable, then fff is continuous and integrable, thus T(f)T(f)T(f) is differentiable and [(Tf)(x)]′=f(x)[(Tf)(x)]'=f(x)[(Tf)(x)]′=f(x).
The space of functions which vanish at x=12x=\frac{1}{2}x=21 is not invariant under TTT. Consider f(x)=x−12f(x)=x-\frac{1}{2}f(x)=x−21, we have (Tf)(x)=12(x2−x)(Tf)(x)=\frac{1}{2}(x^2-x)(Tf)(x)=21(x2−x) and (Tf)(12)=−18≠0(Tf)(\frac{1}{2})=-\frac{1}{8}\neq 0(Tf)(21)=−81=0.

10.Let AAA be a 3×33\times 33×3 matrix with real entries. Prove that, if AAA is not similar over RRR to a triangular matrix, then AAA is similar over CCC to a diagonal matrix.
Solution: By Theorem 5, if AAA is not similar over RRR to a triangular matrix, the minimal polynomial of AAA must be of the form
x2+ax+bor (x2+ax+b)(x−c),a2<4bx^2+ax+b\text{ or }(x^2+ax+b)(x-c),\quad a^2<4bx2+ax+b or (x2+ax+b)(x−c),a2<4b
Thus if we consider AAA over CCC, the minimal polynomial of AAA must have no distinct roots, which means AAA is diagonalizable.

11.True or false? If the triangular matrix AAA is similar to a diagonal matrix, then AAA is already diagonal.
Solution: False. Since if we let A=[1102]A=\begin{bmatrix}1&1\\0&2\end{bmatrix}A=[1012], and P=[1101]P=\begin{bmatrix}1&1\\0&1\end{bmatrix}P=[1011], then P−1=[1−101]P^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}P−1=[10−11] and P−1AP=[1002]P^{-1}AP=\begin{bmatrix}1&0\\0&2\end{bmatrix}P−1AP=[1002].

12.Let TTT be a linear operator on a finite-dimensional vector space over an algebraically closed field FFF. Let fff be a polynomial over FFF. Prove that ccc is a characteristic value of f(T)f(T)f(T) if and only if c=f(t)c=f(t)c=f(t), where ttt is a characteristic value of TTT.
Solution: Since TTT is over an algebraically closed field FFF, there is an ordered basis B\mathfrak BB of the vector space under which TTT has an upper triangular matrix
[T]B=[a1⋯∗⋱⋮an]⟹[f(T)]B=[f(a1)⋯∗⋱⋮f(an)][T]_{\mathfrak B}=\begin{bmatrix}a_1&\cdots&*\\&\ddots&\vdots\\&&a_n\end{bmatrix}\implies [f(T)]_{\mathfrak B}=\begin{bmatrix}f(a_1)&\cdots&*\\&\ddots&\vdots\\&&f(a_n)\end{bmatrix}[T]B=⎣⎢⎡a1⋯⋱∗⋮an⎦⎥⎤⟹[f(T)]B=⎣⎢⎡f(a1)⋯⋱∗⋮f(an)⎦⎥⎤
thus we have det⁡(xI−T)=∏i=1n(x−ai)\det (xI-T)=\prod_{i=1}^n(x-a_i)det(xI−T)=∏i=1n(x−ai) and det⁡(xI−f(T))=∏i=1n(x−f(ai))\det (xI-f(T))=\prod_{i=1}^n(x-f(a_i))det(xI−f(T))=∏i=1n(x−f(ai)), thus ccc is a characteristic value of f(T)f(T)f(T) if and only if c=f(ai)c=f(a_i)c=f(ai), and aia_iai is a characteristic value of TTT since det⁡(aiI−T)=0\det (a_iI-T)=0det(aiI−T)=0.

13.Let VVV be the space of n×nn\times nn×n matrices over FFF. Let AAA be a fixed n×nn\times nn×n matrix over FFF. Let TTT and UUU be the linear operators on VVV defined by T(B)=AB,U(B)=AB−BAT(B)=AB,U(B)=AB-BAT(B)=AB,U(B)=AB−BA.
( a ) True or false? If AAA is diagonalizable (over FFF), then TTT is diagonalizable.
( b ) True or false? If AAA is diagonalizable, then UUU is diagonalizable.
Solution:
( a ) True, since if AAA is diagonalizable, then the minimal polynomial of AAA has the form p=(x−c1)⋯(x−ck)p=(x-c_1)\cdots(x-c_k)p=(x−c1)⋯(x−ck), now for any matrix BBB we have
p(T)(B)=(T−c1I)⋯(T−ckI)(B)=(A−c1I)⋯(A−ckI)(B)=p(A)(B)=0\begin{aligned}p(T)(B)&=(T-c_1I)\cdots(T-c_kI)(B)\\&=(A-c_1I)\cdots(A-c_kI)(B)\\&=p(A)(B)=0\end{aligned}p(T)(B)=(T−c1I)⋯(T−ckI)(B)=(A−c1I)⋯(A−ckI)(B)=p(A)(B)=0
thus the minimal polynomial of TTT divides ppp, which means the minimal polynomial of TTT has no distinct roots, thus TTT is diagonalizable.
( b ) True. If AAA is diagonalizable, then there is invertible PPP such that P−1AP=D=diag (d1,…,dn)P^{-1}AP=D=\text{diag }(d_1,\dots,d_n)P−1AP=D=diag (d1,…,dn). Let Ep,qE^{p,q}Ep,q be the matrix which has only 111 in the pppth row and the qqqth column, then {Ep,q,p,q=1,…,n}\{E^{p,q},p,q=1,\dots,n\}{Ep,q,p,q=1,…,n} form a basis for VVV. Since (DEp,q−Ep,qD)ij=∑k=1nDikEkjp,q−∑k=1nEikp,qDkj(DE^{p,q}-E^{p,q}D)_{ij}=\sum_{k=1}^nD_{ik}E^{p,q}_{kj}-\sum_{k=1}^nE^{p,q}_{ik}D_{kj}(DEp,q−Ep,qD)ij=∑k=1nDikEkjp,q−∑k=1nEikp,qDkj, we see that the only nonzero item of the matrix DEp,q−Ep,qDDE^{p,q}-E^{p,q}DDEp,q−Ep,qD is (DEp,q−Ep,qD)pq=(dp−dq)(DE^{p,q}-E^{p,q}D)_{pq}=(d_p-d_q)(DEp,q−Ep,qD)pq=(dp−dq), which means DEp,q−Ep,qD=(dp−dq)Ep,qDE^{p,q}-E^{p,q}D=(d_p-d_q)E^{p,q}DEp,q−Ep,qD=(dp−dq)Ep,q. Let Fp,q=PEp,qP−1F^{p,q}=PE^{p,q}P^{-1}Fp,q=PEp,qP−1, then {Fp,q,p,q=1,…,n}\{F^{p,q},p,q=1,\dots,n\}{Fp,q,p,q=1,…,n} is also a basis for VVV, since PPP is invertible. Now we have
U(Fp,q)=AFp,q−Fp,qA=APEp,qP−1−PEp,qP−1A=(PP−1)APEp,qP−1−PEp,qP−1A(PP−1)=P(P−1AP)Ep,qP−1−PEp,q(P−1AP)P−1=P[DEp,q−Ep,qD]P−1=(dp−dq)PEp,qP−1=(dp−dq)Fp,q\begin{aligned}U(F^{p,q})&=AF^{p,q}-F^{p,q}A=APE^{p,q}P^{-1}-PE^{p,q}P^{-1}A\\&=(PP^{-1})APE^{p,q}P^{-1}-PE^{p,q}P^{-1}A(PP^{-1})\\&=P(P^{-1}AP)E^{p,q}P^{-1}-PE^{p,q}(P^{-1}AP)P^{-1}\\&=P[DE^{p,q}-E^{p,q}D]P^{-1}\\&=(d_p-d_q)PE^{p,q}P^{-1}=(d_p-d_q)F^{p,q}\end{aligned}U(Fp,q)=AFp,q−Fp,qA=APEp,qP−1−PEp,qP−1A=(PP−1)APEp,qP−1−PEp,qP−1A(PP−1)=P(P−1AP)Ep,qP−1−PEp,q(P−1AP)P−1=P[DEp,q−Ep,qD]P−1=(dp−dq)PEp,qP−1=(dp−dq)Fp,q
This means Fp,qF^{p,q}Fp,q is a characteristic vector for UUU for any p,q=1,…,np,q=1,\dots,np,q=1,…,n, thus UUU is diagonalizable.

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6.4 Invariant Subspaces

Exercises

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