6.2 Characteristic Values

这一章的名字叫Elementary Canonical Forms，翻译过来应该叫初级的规范形式，围绕着如何将一个linear operator以最简单的形式表现出来，或找一个最简单的矩阵under some specific basis。本节讲的是特征值，特征值与对角化紧密联系。

本节中，定义特征值和特征向量之后，还定义了characteristic space associated with 特征值。Theorem 1说明，特征值c和T-cI是singular以及T-cI行列式为0是等价的，由此定义了矩阵的特征值。这个定理比较重要，因其提供了特征值判断转化为operator可逆性判断和行列式判断的方法，后面证明相似矩阵有相同特征多项式时，用的是行列式性质。

后面是关于对角化的问题，对角化的定义是T有一个characteristic vector basis。通过两个Lemma可以证明Theorem 2： VVV上的operator TTT可对角化和特征多项式的完全分解，以及属于不同特征值的特征空间维数之和等于dim⁡V\dim VdimV等价。这一定理有一个矩阵形式。

Exercises

1.In each of the following cases, let TTT be the linear operator on R2R^2R2 which is represented by the matrix AAA in the standard ordered basis for R2R^2R2, and let UUU be the linear operator on C2C^2C2 represented by AAA in the standard ordered basis. Find the characteristic polynomial for TTT and that for UUU, find the characteristic values of each operator, and for each such characteristic value ccc find a basis for the corresponding space of characteristic vectors.

A=[1000],A=[23−11],A=[1111]A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad A=\begin{bmatrix}2&3\\-1&1\end{bmatrix},\quad A=\begin{bmatrix}1&1\\1&1\end{bmatrix}A=[10​00​],A=[2−1​31​],A=[11​11​]

Solution:

For A=[1000]A=\begin{bmatrix}1&0\\0&0\end{bmatrix}A=[10​00​], we have:

The characteristic polynomial for TTT is

det⁡(xI−A)=det⁡[x−100x]=x(x−1)\det (xI-A)=\det \begin{bmatrix}x-1&0\\0&x\end{bmatrix}=x(x-1)det(xI−A)=det[x−10​0x​]=x(x−1)

the characteristic values of TTT is c1=0,c2=1c_1=0,c_2=1c1​=0,c2​=1;

the space of characteristic vectors with characteristic value c1c_1c1​ is spanned by (0,1)(0,1)(0,1), the space of characteristic vectors with characteristic value c2c_2c2​ is spanned by (1,0)(1,0)(1,0).

The results for UUU is the same as TTT.

For A=[23−11]A=\begin{bmatrix}2&3\\-1&1\end{bmatrix}A=[2−1​31​], we have:

The characteristic polynomial for TTT is det⁡(xI−A)=det⁡[x−2−31x−1]=x2−3x+5\det (xI-A)=\det \begin{bmatrix}x-2&-3\\1&x-1\end{bmatrix}=x^2-3x+5det(xI−A)=det[x−21​−3x−1​]=x2−3x+5

There are no characteristic values of TTT.

The characteristic polynomial for UUU is det⁡(xI−A)=det⁡[x−2−31x−1]=(x−32+112i)(x−32−112i)\det (xI-A)=\det \begin{bmatrix}x-2&-3\\1&x-1\end{bmatrix}=\left(x-\frac{3}{2}+\frac{\sqrt{11}}{2}i\right)\left(x-\frac{3}{2}-\frac{\sqrt{11}}{2}i\right)det(xI−A)=det[x−21​−3x−1​]=(x−23​+211​​i)(x−23​−211​​i)

the characteristic values of UUU is c1=32−112i,c2=32+112ic_1=\dfrac{3}{2}-\dfrac{\sqrt{11}}{2}i,c_2=\dfrac{3}{2}+\dfrac{\sqrt{11}}{2}ic1​=23​−211​​i,c2​=23​+211​​i;

the space of characteristic vectors with characteristic value c1c_1c1​ is spanned by (12−112,−1)\left(\dfrac{1}{2}-\dfrac{\sqrt{11}}{2},-1\right)(21​−211​​,−1), the space of characteristic vectors with characteristic value c2c_2c2​ is spanned by (12+112,−1)\left(\dfrac{1}{2}+\dfrac{\sqrt{11}}{2},-1\right)(21​+211​​,−1).

For A=[1111]A=\begin{bmatrix}1&1\\1&1\end{bmatrix}A=[11​11​], we have:

The characteristic polynomial for TTT is det⁡(xI−A)=det⁡[x−1−1−1x−1]=x(x−2)\det (xI-A)=\det \begin{bmatrix}x-1&-1\\-1&x-1\end{bmatrix}=x(x-2)det(xI−A)=det[x−1−1​−1x−1​]=x(x−2)

the characteristic values of TTT is c1=0,c2=2c_1=0,c_2=2c1​=0,c2​=2;

the space of characteristic vectors with characteristic value c1c_1c1​ is spanned by (1,−1)(1,-1)(1,−1), the space of characteristic vectors with characteristic value c2c_2c2​ is spanned by (1,1)(1,1)(1,1).

The results for UUU is the same as TTT.

2.Let VVV be an nnn-dimensional vector space over FFF. What is the characteristic polynomial of the identity operator on VVV? What is the characteristic polynomial for the zero vector?

Solution: The matrix of the identity operator is InI_nIn​ and the matrix of the zero operator is 000 under any basis, thus the characteristic polynomial of the identity operator is det⁡(xI−I)=(x−1)n\det(xI-I)=(x-1)^ndet(xI−I)=(x−1)n, the characteristic polynomial for the zero vector is det⁡(xI−0)=xn\det(xI-0)=x^ndet(xI−0)=xn.

3.Let AAA be an n×nn\times nn×n triangular matrix over the field FFF. Prove that the characteristic values of AAA are the diagonal entries of AAA, i.e., the scalars AiiA_{ii}Aii​.

Solution: The matrix xI−AxI-AxI−A is also triangular, thus det⁡(xI−A)=∏i=1n(x−Aii)\det (xI-A)=\prod_{i=1}^n(x-A_{ii})det(xI−A)=∏i=1n​(x−Aii​), so all the values which can make det⁡(xI−A)=0\det (xI-A)=0det(xI−A)=0 is the scalars AiiA_{ii}Aii​.

4.Let TTT be the linear operator on R3R^3R3 which is represented in the standard ordered basis by the matrix

[−944−834−1687]\begin{bmatrix}-9&4&4\\-8&3&4\\-16&8&7\end{bmatrix}⎣⎡​−9−8−16​438​447​⎦⎤​

Prove that TTT is diagonalizable by exhibiting a basis for R3R^3R3, each vector of which is a characteristic vector of TTT.

Solution: We let the above matrix be AAA and compute

det⁡(xI−A)=∣x+9−4−48x−3−416−8x−7∣=∣x+9−4−48x−3−40−2x−2x+1∣=∣x+9−12−48x−11−400x+1∣=(x+1)∣x+9−128x−11∣=(x+1)2(x−3)\begin{aligned}\det (xI-A)&=\left| \begin{matrix}x+9&-4&-4\\8&x-3&-4\\16&-8&x-7\end{matrix}\right|=\left| \begin{matrix}x+9&-4&-4\\8&x-3&-4\\0&-2x-2&x+1\end{matrix}\right|\\&=\left| \begin{matrix}x+9&-12&-4\\8&x-11&-4\\0&0&x+1\end{matrix}\right|=(x+1)\left|\begin{matrix}x+9&-12\\8&x-11\end{matrix}\right|\\&=(x+1)^2(x-3)\end{aligned}det(xI−A)​=∣∣∣∣∣∣​x+9816​−4x−3−8​−4−4x−7​∣∣∣∣∣∣​=∣∣∣∣∣∣​x+980​−4x−3−2x−2​−4−4x+1​∣∣∣∣∣∣​=∣∣∣∣∣∣​x+980​−12x−110​−4−4x+1​∣∣∣∣∣∣​=(x+1)∣∣∣∣​x+98​−12x−11​∣∣∣∣​=(x+1)2(x−3)​

thus the characteristic values are −1-1−1 and 333. We have

A+I=[−844−844−1688],A−3I=[−1244−804−1684]A+I=\begin{bmatrix}-8&4&4\\-8&4&4\\-16&8&8\end{bmatrix},\quad A-3I=\begin{bmatrix}-12&4&4\\-8&0&4\\-16&8&4\end{bmatrix}A+I=⎣⎡​−8−8−16​448​448​⎦⎤​,A−3I=⎣⎡​−12−8−16​408​444​⎦⎤​

The null space of A+IA+IA+I is spanned by (1,0,2),(1,2,0)(1,0,2),(1,2,0)(1,0,2),(1,2,0) and the null space of A−3IA-3IA−3I is spanned by (1,1,2)(1,1,2)(1,1,2), thus the basis we search for is consisting of (1,0,2),(1,2,0),(1,1,2)(1,0,2),(1,2,0),(1,1,2)(1,0,2),(1,2,0),(1,1,2).

5.Let

A=[6−3−24−1−210−5−3].A=\begin{bmatrix}6&-3&-2\\4&-1&-2\\10&-5&-3\end{bmatrix}.A=⎣⎡​6410​−3−1−5​−2−2−3​⎦⎤​.

Is AAA similar over the field RRR to a diagonal matrix? Is AAA similar over the field CCC to a diagonal matrix?

Solution: We have

det⁡(xI−A)=∣x−632−4x+12−105x+3∣=∣x−22−x0−4x+12−105x+3∣=∣x−200−4x−32−10−5x+3∣=(x−2)(x2+1)\begin{aligned}\det (xI-A)&=\left| \begin{matrix}x-6&3&2\\-4&x+1&2\\-10&5&x+3\end{matrix}\right|=\left| \begin{matrix}x-2&2-x&0\\-4&x+1&2\\-10&5&x+3\end{matrix}\right|\\&=\left| \begin{matrix}x-2&0&0\\-4&x-3&2\\-10&-5&x+3\end{matrix}\right|=(x-2)(x^2+1)\end{aligned}det(xI−A)​=∣∣∣∣∣∣​x−6−4−10​3x+15​22x+3​∣∣∣∣∣∣​=∣∣∣∣∣∣​x−2−4−10​2−xx+15​02x+3​∣∣∣∣∣∣​=∣∣∣∣∣∣​x−2−4−10​0x−3−5​02x+3​∣∣∣∣∣∣​=(x−2)(x2+1)​

In RRR, AAA has only one characteristic value c=2c=2c=2, and

A−2I=[4−3−24−3−210−5−5].A-2I=\begin{bmatrix}4&-3&-2\\4&-3&-2\\10&-5&-5\end{bmatrix}.A−2I=⎣⎡​4410​−3−3−5​−2−2−5​⎦⎤​.

It is easy to see rank(A−2I)=2\text{rank }(A-2I)=2rank(A−2I)=2 and null(A−2I)=1≠3\text{null }(A-2I)=1\neq 3null(A−2I)=1​=3, thus AAA is not similar over RRR to a diagonal matrix.

When AAA is in CCC, det⁡(xI−A)=(x−2)(x−i)(x+i)\det (xI-A)=(x-2)(x-i)(x+i)det(xI−A)=(x−2)(x−i)(x+i), thus AAA has 2,i,−i2,i,-i2,i,−i as its characteristic value, which means AAA is similar over the field CCC to a diagonal matrix.

6.Let TTT be the linear operator on R4R^4R4 which is represented in the standard ordered basis by the matrix

[0000a0000b0000c0].\begin{bmatrix}0&0&0&0\\a&0&0&0\\0&b&0&0\\0&0&c&0\end{bmatrix}.⎣⎢⎢⎡​0a00​00b0​000c​0000​⎦⎥⎥⎤​.

Under what conditions on a,b,ca,b,ca,b,c is TTT diagonalizable?

Solution: Let AAA be the matrix above and the characteristic value of AAA is 000, by Exercise 3. Thus if TTT is diagonalizable, the null space of TTT shall have dimension 444, which means Ax=0Ax=0Ax=0 for all x∈R4x\in R^4x∈R4, thus for any x1,x2,x3,x4∈Rx_1,x_2,x_3,x_4\in Rx1​,x2​,x3​,x4​∈R,

[0000a0000b0000c0][x1x2x3x4]=[0ax1bx2cx3]=0⟹a=b=c=0\begin{bmatrix}0&0&0&0\\a&0&0&0\\0&b&0&0\\0&0&c&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}0\\ax_1\\bx_2\\cx_3\end{bmatrix}=0 {\implies}a=b=c=0⎣⎢⎢⎡​0a00​00b0​000c​0000​⎦⎥⎥⎤​⎣⎢⎢⎡​x1​x2​x3​x4​​⎦⎥⎥⎤​=⎣⎢⎢⎡​0ax1​bx2​cx3​​⎦⎥⎥⎤​=0⟹a=b=c=0

7.Let TTT be a linear operator on the nnn-dimensional vector space VVV, and suppose that TTT has nnndistinctcharacteristic values. Prove that TTT is diagonalizable.

Solution: Suppose TTT has c1,…,cnc_1,\dots,c_nc1​,…,cn​ as characteristic values, in which ci≠cjc_i\neq c_jci​​=cj​ whenever i≠ji\neq ji​=j. Then for each cic_ici​ we can find an αi≠0\alpha_i\neq 0αi​​=0 such that Tαi=ciαiT\alpha_i=c_i\alpha_iTαi​=ci​αi​, and {α1,…,αn}\{\alpha_1,\dots,\alpha_n\}{α1​,…,αn​} are linearly independent, thus a basis for VVV, which is the definition of diagonalizable operators.

8.Let AAA and BBB be n×nn\times nn×n matrices over the field FFF. Prove that if (I−AB)(I-AB)(I−AB) is invertible, then (I−BA)(I-BA)(I−BA) is invertible and

(I−BA)−1=I+B(I−AB)−1A.(I-BA)^{-1}=I+B(I-AB)^{-1}A.(I−BA)−1=I+B(I−AB)−1A.

Solution: We have

[I+B(I−AB)−1A](I−BA)=I−BA+B(I−AB)−1A−B(I−AB)−1ABA=I−BA+B(I−AB)−1(I−AB)A=I−BA+BA=I\begin{aligned}&\quad[I+B(I-AB)^{-1}A](I-BA)\\&=I-BA+B(I-AB)^{-1}A-B(I-AB)^{-1}ABA\\&=I-BA+B(I-AB)^{-1}(I-AB)A\\&=I-BA+BA=I\end{aligned}​[I+B(I−AB)−1A](I−BA)=I−BA+B(I−AB)−1A−B(I−AB)−1ABA=I−BA+B(I−AB)−1(I−AB)A=I−BA+BA=I​

Thus I+B(I−AB)−1AI+B(I-AB)^{-1}AI+B(I−AB)−1A is a left inverse for I−BAI-BAI−BA, which means I−BAI-BAI−BA is invertible.

9.Use the result of Exercise 8 to prove that, if AAA and BBB are n×nn\times nn×n matrices over the field FFF, then ABABAB and BABABA have precisely the same characteristic values in FFF.

Solution: Let ccc be a characteristic value of ABABAB, and α≠0\alpha\neq 0α​=0 such that ABα=cαAB\alpha=c\alphaABα=cα.

If c=0c=0c=0, then ABX=0ABX=0ABX=0 has non-trivial solutions, which means ABABAB is not invertible, so either AAA or BBB is not invertible, if AAA is not invertbile, we can find β≠0\beta\neq 0β​=0 such that Aβ=0A\beta=0Aβ=0, which means BAβ=0BA\beta=0BAβ=0, if AAA is invertble but BBB is not, then we can find γ≠0\gamma\neq 0γ​=0 such that Bγ=0B\gamma=0Bγ=0, as γ≠0\gamma\neq 0γ​=0 means A−1γ≠0A^{-1}\gamma\neq 0A−1γ​=0, we have BA(A−1γ)=0BA(A^{-1}\gamma)=0BA(A−1γ)=0, in both cases we proved 000 is a characteristic value of BABABA.

If c≠0c\neq 0c​=0, then we have (cI−AB)α=0(cI-AB)\alpha=0(cI−AB)α=0, or (I−1cAB)α=0(I-\frac{1}{c}AB)\alpha=0(I−c1​AB)α=0, so (I−1cAB)(I-\frac{1}{c}AB)(I−c1​AB) is not invertible, thus (I−1cBA)(I-\frac{1}{c}BA)(I−c1​BA) is not invertible, which means (cI−BA)β=0(cI-BA)\beta=0(cI−BA)β=0 for some β≠0\beta\neq 0β​=0, so ccc is a characteristic value of BABABA.

10.Suppose that AAA is a 2×22\times 22×2 matrix with real entries which is symmetric(At=AA^t=AAt=A). Prove that AAA is similar over RRR to a diagonal matrix.

Solution: Write A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix}A=[ac​bd​], then from At=AA^t=AAt=A we have b=cb=cb=c, and

det⁡(xI−A)=∣x−a−b−bx−d∣=(x−a)(x−d)−b2=x2−(a+d)x+ad−b2=(x−a+d2)2−(b2+(a−d)24)=(x−a+d2+b2+(a−d)24)(x−a+d2−b2+(a−d)24)\begin{aligned}\det(xI-A)&=\left|\begin{matrix}x-a&-b\\-b&x-d\end{matrix}\right|=(x-a)(x-d)-b^2\\&=x^2-(a+d)x+ad-b^2\\&=\left(x-\frac{a+d}{2}\right)^2-\left(b^2+\frac{(a-d)^2}{4}\right)\\&=\left(x-\frac{a+d}{2}+\sqrt{b^2+\frac{(a-d)^2}{4}}\right)\left(x-\frac{a+d}{2}-\sqrt{b^2+\frac{(a-d)^2}{4}}\right)\end{aligned}det(xI−A)​=∣∣∣∣​x−a−b​−bx−d​∣∣∣∣​=(x−a)(x−d)−b2=x2−(a+d)x+ad−b2=(x−2a+d​)2−(b2+4(a−d)2​)=(x−2a+d​+b2+4(a−d)2​​)(x−2a+d​−b2+4(a−d)2​​)​

So the polynomial of AAA can be totally factored.

11.Let NNN be a 2×22 \times 22×2 complex matrix such that N2=0N^2=0N2=0. Prove that either N=0N=0N=0 or NNN is similar over CCC to[0010]\begin{bmatrix}0&0\\1&0\end{bmatrix}[01​00​].

Solution: If N≠0N\neq 0N​=0, at least one row vector of NNN is not zero, let this row vector of NNN be α\alphaα, then Nα≠0N\alpha\neq 0Nα​=0.

If c,d∈Cc,d\in Cc,d∈C such that cα+dNα=0c\alpha+dN\alpha=0cα+dNα=0, we shall have N(cα+dNα)=0N(c\alpha+dN\alpha)=0N(cα+dNα)=0, or cNα=0cN\alpha=0cNα=0, this means c=0c=0c=0 and d=0d=0d=0, so {α,Nα}\{\alpha,N\alpha\}{α,Nα} are linearly independent, thus the matrix P=[α,Nα]P=[\alpha,N\alpha]P=[α,Nα] is invertible. As

NP=[Nα,N2α]=[Nα,0]=[α,Nα][0010]=P[0010]NP=[N\alpha,N^2\alpha]=[N\alpha,0]=[\alpha,N\alpha]\begin{bmatrix}0&0\\1&0\end{bmatrix}=P\begin{bmatrix}0&0\\1&0\end{bmatrix}NP=[Nα,N2α]=[Nα,0]=[α,Nα][01​00​]=P[01​00​]

We have N=P−1[0010]PN=P^{-1}\begin{bmatrix}0&0\\1&0\end{bmatrix}PN=P−1[01​00​]P.

12.Use the result of Exercise 11 to prove the following: If AAA is a 2×22\times 22×2 matrix with complex entries, then AAA is similar over CCC to a matrix of one of the two types

[a00b][a01a]\begin{bmatrix}a&0\\0&b\end{bmatrix}\qquad \begin{bmatrix}a&0\\1&a\end{bmatrix}[a0​0b​][a1​0a​]

Solution: The characteristic polynomial of AAA is a polynomial of degree 2 over CCC, thus must be the form (x−a)(x−b)(x-a)(x-b)(x−a)(x−b) or (x−a)2(x-a)^2(x−a)2 for some a,b∈Ca,b\in Ca,b∈C. In the first case, a,ba,ba,b are characteristic values of AAA, and choose α,β\alpha,\betaα,β to be characteristic vectors of aaa and bbb, we shall have Aα=aα,Aβ=bβA\alpha=a\alpha,A\beta=b\betaAα=aα,Aβ=bβ, thus AAA is similar to [a00b]\begin{bmatrix}a&0\\0&b\end{bmatrix}[a0​0b​] with the matrix P=[α,β]P=[\alpha,\beta]P=[α,β] and P−1AP=[a00b]P^{-1}AP=\begin{bmatrix}a&0\\0&b\end{bmatrix}P−1AP=[a0​0b​].

In the second case, aaa is the only possible characteristic value of AAA. let α\alphaα be a characteristic vector of aaa, then it can be extended to a basis {α,β}\{\alpha,\beta\}{α,β} of C2C^2C2, let γ:=(A−aI)β\gamma:=(A-aI)\betaγ:=(A−aI)β, so we can express γ=mα+nβ,m,n∈C\gamma=m\alpha+n\beta,m,n\in Cγ=mα+nβ,m,n∈C.

Since (A−aI)γ=(A−aI)(mα+nβ)=n(A−aI)β=nγ(A-aI)\gamma=(A-aI)(m\alpha+n\beta)=n(A-aI)\beta=n\gamma(A−aI)γ=(A−aI)(mα+nβ)=n(A−aI)β=nγ, we shall have Aγ=(a+n)γA\gamma=(a+n)\gammaAγ=(a+n)γ, since aaa is the only possible characteristic value of AAA, we have either γ=0\gamma=0γ=0 or n=0n=0n=0, the first possibility means A−aI=0A-aI=0A−aI=0, if γ≠0\gamma\neq 0γ​=0 but n=0n=0n=0, then γ=mα\gamma=m\alphaγ=mα, so (A−aI)γ=0(A-aI)\gamma=0(A−aI)γ=0, which means (A−aI)2=0(A-aI)^2=0(A−aI)2=0. Use Exercise 11 we know A−aIA-aIA−aI is similar to [0010]\begin{bmatrix}0&0\\1&0\end{bmatrix}[01​00​], so choose PPP invertible such that

A−aI=P−1[0010]PA-aI=P^{-1}\begin{bmatrix}0&0\\1&0\end{bmatrix}PA−aI=P−1[01​00​]P, we can finish the proof since

A=P−1[0010]P+aI=P−1([0010]+aI)P=P−1[a01a]PA=P^{-1}\begin{bmatrix}0&0\\1&0\end{bmatrix}P+aI=P^{-1}\left(\begin{bmatrix}0&0\\1&0\end{bmatrix}+aI\right)P=P^{-1}\begin{bmatrix}a&0\\1&a\end{bmatrix}PA=P−1[01​00​]P+aI=P−1([01​00​]+aI)P=P−1[a1​0a​]P.

13.Let VVV be the vector space of all functions from RRR into RRR which are continuous, i.e., the space of continuous real-valued functions on the real line. Let TTT be the linear operator on VVV defined by

(Tf)(x)=∫0xf(t)dt.(Tf)(x)=\int_0^xf(t)dt.(Tf)(x)=∫0x​f(t)dt.

Prove that TTT has no characteristic values.

Solution: Assume TTT has a characteristic value ccc and a corresponding characteristic vector f≠0f\neq0f​=0, then

(Tf)(x)=∫0xf(t)dt=cf(x)(Tf)(x)=\int_0^xf(t)dt=cf(x)(Tf)(x)=∫0x​f(t)dt=cf(x)

Let x=0x=0x=0 we see cf(0)=0cf(0)=0cf(0)=0, so c=0c=0c=0 or f(0)=0f(0)=0f(0)=0, the first case means ∫0xf(t)dt=0\int_0^xf(t)dt=0∫0x​f(t)dt=0 for all x∈Rx\in Rx∈R, but since f≠0f\neq 0f​=0 we can find a∈R,f(a)≠0a\in R,f(a)\neq 0a∈R,f(a)​=0, we suppose f(a)>0f(a)>0f(a)>0, then as fff is continuous, there is some δ>0\delta>0δ>0 such that

f(x)≥f(a)2,a−δ<x<a⟹∫a−δaf(t)dt≥f(a)2δ>0f(x)\geq \frac{f(a)}{2},\quad a-\delta<x<a\implies\int_{a-\delta}^af(t)dt\geq \frac{f(a)}{2}\delta>0f(x)≥2f(a)​,a−δ<x<a⟹∫a−δa​f(t)dt≥2f(a)​δ>0

but ∫a−δaf(t)dt=∫0af(t)dt−∫0a−δf(t)dt=0\int_{a-\delta}^af(t)dt=\int_{0}^af(t)dt-\int_0^{a-\delta}f(t)dt=0∫a−δa​f(t)dt=∫0a​f(t)dt−∫0a−δ​f(t)dt=0, a contradiction. If f(a)<0f(a)<0f(a)<0 we can similarly get an contradiction. Thus the first case is not valid. We can only have c≠0c\neq 0c​=0 and f(0)=0f(0)=0f(0)=0.

Under the condition c≠0c\neq 0c​=0, we have, by the Fundamental Theorem of Calculus:

1c=f(x)∫0xf(t)dt⟹ln⁡(∫0xf(t)dt)=xc⟹∫0xf(t)dt=ex/c\frac{1}{c}=\frac{f(x)}{\int_0^xf(t)dt}\implies \ln\left(\int_0^xf(t)dt\right)=\frac{x}{c}\implies \int_0^xf(t)dt=e^{x/c}c1​=∫0x​f(t)dtf(x)​⟹ln(∫0x​f(t)dt)=cx​⟹∫0x​f(t)dt=ex/c

Now let x=0x=0x=0, we get 0=10=10=1, a contradiction.

14.Let AAA be an n×nn\times nn×n diagonal matrix with characteristic polynomial

(x−c1)d1…(x−ck)dk(x-c_1)^{d_1}\dots(x-c_k)^{d_k}(x−c1​)d1​…(x−ck​)dk​

where c1,⋯,ckc_1,\cdots,c_kc1​,⋯,ck​ are distinct. Let VVV be the space of n×nn\times nn×n matrices BBB such that AB=BAAB=BAAB=BA. Prove that the dimension of VVV is d12+⋯+dk2d_1^2+\cdots+d_k^2d12​+⋯+dk2​.

Solution: Notice that AAA is diagonal, thus

A=[c1Id1⋱ckIdk]A=\begin{bmatrix}c_1I_{d_1}&&\\&\ddots&\\&&c_kI_{d_k}\end{bmatrix}A=⎣⎡​c1​Id1​​​⋱​ck​Idk​​​⎦⎤​

where IdjI_{d_j}Idj​​ is the dj×djd_j\times d_jdj​×dj​ identity matrix, and we also have d1+⋯+dk=nd_1+\cdots+d_k=nd1​+⋯+dk​=n. Now write B∈VB\in VB∈V as

B=[B11⋯B1k⋯Bk1⋯Bkk]B=\begin{bmatrix}B_{11}&\cdots&B_{1k}\\&\cdots&\\B_{k1}&\cdots&B_{kk}\end{bmatrix}B=⎣⎡​B11​Bk1​​⋯⋯⋯​B1k​Bkk​​⎦⎤​

where each BijB_{ij}Bij​ is a di×djd_i\times d_jdi​×dj​ block matrix, then if AB=BAAB=BAAB=BA, we have

[c1Id1⋱ckIdk][B11⋯B1k⋯Bk1⋯Bkk]=[B11⋯B1k⋯Bk1⋯Bkk][c1Id1⋱ckIdk]\begin{bmatrix}c_1I_{d_1}&&\\&\ddots&\\&&c_kI_{d_k}\end{bmatrix}\begin{bmatrix}B_{11}&\cdots&B_{1k}\\&\cdots&\\B_{k1}&\cdots&B_{kk}\end{bmatrix}=\begin{bmatrix}B_{11}&\cdots&B_{1k}\\&\cdots&\\B_{k1}&\cdots&B_{kk}\end{bmatrix}\begin{bmatrix}c_1I_{d_1}&&\\&\ddots&\\&&c_kI_{d_k}\end{bmatrix}⎣⎡​c1​Id1​​​⋱​ck​Idk​​​⎦⎤​⎣⎡​B11​Bk1​​⋯⋯⋯​B1k​Bkk​​⎦⎤​=⎣⎡​B11​Bk1​​⋯⋯⋯​B1k​Bkk​​⎦⎤​⎣⎡​c1​Id1​​​⋱​ck​Idk​​​⎦⎤​

doing the block matrix multiplication we have

[c1B11⋯c1B1k⋯ckBk1⋯ckBkk]=[c1B11⋯ckB1k⋯c1Bk1⋯ckBkk]\begin{bmatrix} c_1B_{11}&\cdots&c_1B_{1k}\\&\cdots&\\c_kB_{k1}&\cdots&c_kB_{kk}\end{bmatrix}=\begin{bmatrix} c_1B_{11}&\cdots&c_kB_{1k}\\&\cdots&\\c_1B_{k1}&\cdots&c_kB_{kk}\end{bmatrix}⎣⎡​c1​B11​ck​Bk1​​⋯⋯⋯​c1​B1k​ck​Bkk​​⎦⎤​=⎣⎡​c1​B11​c1​Bk1​​⋯⋯⋯​ck​B1k​ck​Bkk​​⎦⎤​

Since c1,…,ckc_1,\dots,c_kc1​,…,ck​ are distinct, a comparision shows that Bij=0B_{ij}=0Bij​=0 whenever i≠ji\neq ji​=j, thus the blocks that can be arbitrary values in BBB are Bii,i=1,⋯,kB_{ii},i=1,\cdots,kBii​,i=1,⋯,k, each has dj2d_j^2dj2​ entries, and the conclusion follows.

15.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 be the linear operator 'left multiplication by AAA’ on VVV. Is it true that AAA and TTT have the same characteristic values?

Solution: First if ccc is a characteristic value of TTT, then there is B≠0B\neq 0B​=0 such that T(B)=AB=cBT(B)=AB=cBT(B)=AB=cB, so (A−cI)B=0(A-cI)B=0(A−cI)B=0, since B≠0B\neq 0B​=0, at least one column of BBB is not zero, thus (A−cI)X=0(A-cI)X=0(A−cI)X=0 has non-trivial solutions and A−cIA-cIA−cI is not invertible, which means ccc is a characteristic value of AAA.

Conversely, let ccc be a characteristic value of AAA, and a corresponding characteristic vector α≠0\alpha\neq 0α​=0, let B=[α,0,⋯,0]B=[\alpha,0,\cdots,0]B=[α,0,⋯,0] be a n×nn\times nn×n matrix with the first column α\alphaα and the remaining columns 000, then AB=[Aα,0,⋯,0]=cBAB=[A\alpha,0,\cdots,0]=cBAB=[Aα,0,⋯,0]=cB, which means T(B)=cBT(B)=cBT(B)=cB, thus ccc is a characteristic value of TTT.