Linear Algebra (2ed) Hoffman & Kunze 6.3

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这一节主要是引入了一个新的定义：minimal polynomial。之前看过的教材中对此的定义是degree最低的能让T或者A为0的多项式，其实这个最低degree是有点概念性上的东西，但是这本书由于之前引入了ideal和generator，所以定义起来要严谨得多。比较容易证明的几个结论是：T和[T]_{\mathfrak B}有相同的minimal polynomial，相似的矩阵有相同的minimal polynomial.
后面的主要内容就是两个定理，Theorem 3说明特征多项式和最小多项式有相同的roots。Theorem 4即有名的Cayley-Hamilton定理，T的特征多项式可以annihilate T，因此最小多项式整除特征多项式，这一节中对此定理的证明用了行列式的方法。

Exercises

1.Let V be a finite-dimensional vector space. What is the minimal polynomial for the identity operator on V? What is the minimal polynomial for the zero operator?
Solution: The minimal polynomial for I is p=x-1, since p(I)=I-I=0. The mininal polynomial for 0 is p=x.

2.Let a,b,c be elements of a field F, and let A be the following 3\times 3 matrix over F:

\displaystyle{A=\begin{bmatrix}0&0&c\\1&0&b\\0&1&a\end{bmatrix}.}

Prove that the characteristic polynomial for A is x^3-ax^2-bx-c and that this is also the minimal polynomial for A.
Solution: We have

\displaystyle{\begin{aligned}\det (xI-A)&=\begin{vmatrix}x&0&-c\\-1&x&-b\\0&-1&x-a\end{vmatrix}=x\begin{vmatrix}x&-b\\-1&x-a\end{vmatrix}-c\begin{vmatrix}-1&x\\0&-1\end{vmatrix}\\&=x(x^2-ax-b)-c=x^3-ax^2-bx-c\end{aligned}}

To see this is also the minimal polynomial for A, notice that

\displaystyle{A^2=\begin{bmatrix}0&0&c\\1&0&b\\0&1&a\end{bmatrix}\begin{bmatrix}0&0&c\\1&0&b\\0&1&a\end{bmatrix}=\begin{bmatrix}0&c&ac\\0&b&c+ab\\1&a&b+a^2\end{bmatrix}}

thus for any polynomial f of degree 2, we write f=mx^2+nx+q, then

\displaystyle{f(A)=m\begin{bmatrix}0&c&ac\\0&b&c+ab\\1&a&b+a^2\end{bmatrix}+n\begin{bmatrix}0&0&c\\1&0&b\\0&1&a\end{bmatrix}+q\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}

assume f(A)=0, then f(A)_{31}=m=0,f(A)_{21}=n=0,f(A)_{11}=q=0, a contradiction to \deg f=2. Thus any polynomial of degree 2 or less cannot be the minimal polynomial for A.

3.Let A be the 4\times 4 matrix

\displaystyle{A=\begin{bmatrix}1&1&0&0\\-1&-1&0&0\\-2&-2&2&1\\1&1&-1&0\end{bmatrix}}

Show that the characteristic polynomial for A is x^2(x-1)^2 and that it is also the minimal polynomial.
Solution: We have

\displaystyle{\begin{aligned}\det (xI-A)&=\begin{vmatrix}x-1&-1&0&0\\1&x+1&0&0\\2&2&x-2&-1\\-1&-1&1&x\end{vmatrix}=\begin{vmatrix}x-1&-1\\1&x+1\end{vmatrix}\begin{vmatrix}x-2&-1\\1&x\end{vmatrix}\\&=(x^2-1+1)(x^2-2x+1)=x^2(x-1)^2\end{aligned}}

To see this is also the minimal polynomial for A, notice that

A^2=\begin{bmatrix}1&1&0&0\\-1&-1&0&0\\-2&-2&2&1\\1&1&-1&0\end{bmatrix}\begin{bmatrix}1&1&0&0\\-1&-1&0&0\\-2&-2&2&1\\1&1&-1&0\end{bmatrix}=\begin{bmatrix}0&0&0&0\\0&0&0&0\\-3&-3&3&2\\2&2&-2&-1\end{bmatrix}\\A^3=\begin{bmatrix}0&0&0&0\\0&0&0&0\\-3&-3&3&2\\2&2&-2&-1\end{bmatrix}\begin{bmatrix}1&1&0&0\\-1&-1&0&0\\-2&-2&2&1\\1&1&-1&0\end{bmatrix}=\begin{bmatrix}0&0&0&0\\0&0&0&0\\-4&-4&4&3\\3&3&-3&-2\end{bmatrix}

thus for any polynomial f of degree 3, write f=mx^3+nx^2+px+q, then

\displaystyle{f(A)=m\begin{bmatrix}0&0&0&0\\0&0&0&0\\-4&-4&4&3\\3&3&-3&-2\end{bmatrix}+n\begin{bmatrix}0&0&0&0\\0&0&0&0\\-3&-3&3&2\\2&2&-2&-1\end{bmatrix}+p\begin{bmatrix}1&1&0&0\\-1&-1&0&0\\-2&-2&2&1\\1&1&-1&0\end{bmatrix}+qI}

If f(A)=0, then f(A)_{12}=p=0, thus f(A)_{11}=p+q=0 means q=0, then f(A)_{31}=-4m-3n=0,f(A)_{41}=3m+2n=0, which means m=n=0, a contradiction. Thus any polynomial of degree 3 or less cannot be the minimal polynomial for A.

4.Is the matrix A of Exercise 3 similar over the field of complex numbers to a diagonal matrix?
Solution: We see the characteristic value of A are 0,1, it is easy to see \text{rank }A=3, thus \dim \text{null }(A-0I)=1, which means A cannot be similar to a diagonal matrix.

5.Let V be an n-dimensional vector space and let T be a linear operator on V. Suppose that there exists some positive integer k so that T^k=0. Prove that T^n=0.
Solution: When k\leq n the result is obvious. Now suppose k>n, from the intergers n,n+1,\cdots,k-1,k we can find one unique integer m such that T^{m-1}\neq 0 and T^m=0. By Cayley-Hamiltion Theorem we get f(T)=0, where f is the characteristic polynomial of T. Write f to be f=\sum_{i=0}^nc_ix^i, then

\displaystyle{f(T)=c_0I+c_1T+\cdots+c_nT^n=0}

if we multiple T^{m-1} on both sides, we get c_0T^{m-1}=0, thus c_0=0 and we reduce f(T) to c_1T+\cdots+c_nT^n. Multiple T^{m-2} we can get c_1=0, and continue this step we would eventually have c_nT^n=0, thus T^n=0 since c_n=1.

6.Find a 3\times 3 matrix for which the minimal polyomial is x^2.
Solution: A simple example would be A=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}.

7.Let n be a positive integer, and let V be the space of polynomials over R which have degree at most n (throw in the 0-polynomial). Let D be the differentiation operator on V. What is the minimal polynomial for D?
Solution: We can easily see D^{n+1}(f)=0 for all f\in V. To see x^{n+1} is the the minimal polynomial for D, assume there is g=x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0 which annihilate D, then

\displaystyle{(D^n+c_{n-1}D^{n-1}+\cdots+c_1D+c_0I)(f)=0,\quad \forall f\in V}

consider f=x^n, we have D^kf=\dfrac{n!}{(n-k)!}x^{n-k}, thus

\displaystyle{n!+c_{n-1}n!x+\cdots+c_1nx^{n-1}+c_0x^n=0}

let x=0 we get n!=0, a contradiction since n is a positive integer.

8.Let P be the operator on R^2 which projects each vector onto the x-axis, parallel to the y-axis:P(x,y)=(x,0). Show that P is linear. What is the minimal polynomial for P?
Solution: To see P is linear, notice that

\displaystyle{\begin{aligned}P(c(x_1,y_1)+(x_2,y_2))&=P(cx_1+y_1,cx_2+y_2)=(cx_1+y_1,0)\\&=c(x_1,0)+(y_1,0)=cP(x_1,y_1)+P(x_2,y_2)\end{aligned}}

The minimal polynomial for P is x^2-x.

9.Let A be an n\times n matrix with characteristic polynomial f=(x-c_1)^{d_1}\cdots(x-c_k)^{d_k}. Show that c_1d_1+\cdots+c_kd_k=\text{trace }(A).
Solution: There are no method to solve this problem using only contents before Section 6.3, in fact, later we can prove A is similar to an upper-triangular matrix with each c_i repeated d_i times, and the result follows since simlar matrices have the same trace.

10.Let V be the vector space of n\times n matrices over the fielf F. Let A be a fixed n\times n matrix. Let T be the linear operator on V defined by T(B)=AB. Show that the minimal polynomial for T is the minimal polynomial for A.
Solution: Let p be the minimal polynomial for T, thus p(T)=0. For B\in V we have T^k(B)=A^kB,k\in N, this means 0=p(T)(B)=p(A)B, since B is arbitrary we get p(A)=0. Similarly we have p(A)=0\Rightarrow p(T)=0, and the conclusion follows.

11.Let A and B be n\times n matrices over the field F. According to Exercise 9 in Section 6.2, the matrices AB and BA have the same characteristic values. Do they have the same characteristic polynomial? Do they have the same minimal polynomial?
Solution: To show they have the same characteristic polynomial we need to show \det (xI-AB)=\det (xI-BA). Define 2n\times 2n matrices C,D as

\displaystyle{C=\begin{bmatrix}xI&A\\B&I\end{bmatrix},\quad D=\begin{bmatrix}I&0\\-B&xI\end{bmatrix}}

then we see that

\displaystyle{CD=\begin{bmatrix}xI-AB&xA\\0&xI\end{bmatrix},\quad DC=\begin{bmatrix}xI&A\\0&xI-BA\end{bmatrix}}

Since \det (CD)=(\det C)(\det D)=\det (DC) and \det (xI)=x^n, we have

\displaystyle{\det (xI-AB)x^n=x^n\det (xI-BA)\implies \det (xI-AB)=\det (xI-BA)}

To see they need not have the same minimal polynomial, choose

\displaystyle{A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad B=\begin{bmatrix}0&0\\0&1\end{bmatrix}}

then AB=A while BA=0, thus the minimal polynomial of BA is x, which is not the same as that of AB.

Linear Algebra (2ed) Hoffman & Kunze 6.2

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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： V上的operator T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等价。这一定理有一个矩阵形式。

Exercises

1.In each of the following cases, let T be the linear operator on R^2 which is represented by the matrix A in the standard ordered basis for R^2, and let U be the linear operator on C^2 represented by A in the standard ordered basis. Find the characteristic polynomial for T and that for U, find the characteristic values of each operator, and for each such characteristic value c find a basis for the corresponding space of characteristic vectors.

\displaystyle{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}}

Solution:
For A=\begin{bmatrix}1&0\\0&0\end{bmatrix}, we have:
The characteristic polynomial for T is

\displaystyle{\det (xI-A)=\det \begin{bmatrix}x-1&0\\0&x\end{bmatrix}=x(x-1)}

the characteristic values of T is c_1=0,c_2=1;
the space of characteristic vectors with characteristic value c_1 is spanned by (0,1), the space of characteristic vectors with characteristic value c_2 is spanned by (1,0).
The results for U is the same as T.

For A=\begin{bmatrix}2&3\\-1&1\end{bmatrix}, we have:
The characteristic polynomial for T is

\displaystyle{\det (xI-A)=\det \begin{bmatrix}x-2&-3\\1&x-1\end{bmatrix}=x^2-3x+5}

There are no characteristic values of T.
The characteristic polynomial for U is

\displaystyle{\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)}

the characteristic values of U is c_1=\dfrac{3}{2}-\dfrac{\sqrt{11}}{2}i,c_2=\dfrac{3}{2}+\dfrac{\sqrt{11}}{2}i;
the space of characteristic vectors with characteristic value c_1 is spanned by \left(\dfrac{1}{2}-\dfrac{\sqrt{11}}{2},-1\right), the space of characteristic vectors with characteristic value c_2 is spanned by \left(\dfrac{1}{2}+\dfrac{\sqrt{11}}{2},-1\right).

For A=\begin{bmatrix}1&1\\1&1\end{bmatrix}, we have:
The characteristic polynomial for T is

\displaystyle{\det (xI-A)=\det \begin{bmatrix}x-1&-1\\-1&x-1\end{bmatrix}=x(x-2)}

the characteristic values of T is c_1=0,c_2=2;
the space of characteristic vectors with characteristic value c_1 is spanned by (1,-1), the space of characteristic vectors with characteristic value c_2 is spanned by (1,1).
The results for U is the same as T.

2.Let V be an n-dimensional vector space over F. What is the characteristic polynomial of the identity operator on V? What is the characteristic polynomial for the zero vector?
Solution: The matrix of the identity operator is I_n and the matrix of the zero operator is 0 under any basis, thus the characteristic polynomial of the identity operator is \det(xI-I)=(x-1)^n, the characteristic polynomial for the zero vector is \det(xI-0)=x^n.

3.Let A be an n\times n triangular matrix over the field F. Prove that the characteristic values of A are the diagonal entries of A, i.e., the scalars A_{ii}.
Solution: The matrix xI-A is also triangular, thus \det (xI-A)=\prod_{i=1}^n(x-A_{ii}), so all the values which can make \det (xI-A)=0 is the scalars A_{ii}.

4.Let T be the linear operator on R^3 which is represented in the standard ordered basis by the matrix

\displaystyle{\begin{bmatrix}-9&4&4\\-8&3&4\\-16&8&7\end{bmatrix}}

Prove that T is diagonalizable by exhibiting a basis for R^3, each vector of which is a characteristic vector of T.
Solution: We let the above matrix be A and compute

\displaystyle{\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}}

thus the characteristic values are -1 and 3. We have

\displaystyle{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}}

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

5.Let

\displaystyle{A=\begin{bmatrix}6&-3&-2\\4&-1&-2\\10&-5&-3\end{bmatrix}.}

Is A similar over the field R to a diagonal matrix? Is A similar over the field C to a diagonal matrix?
Solution: We have

\displaystyle{\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}}

In R, A has only one characteristic value c=2, and

\displaystyle{A-2I=\begin{bmatrix}4&-3&-2\\4&-3&-2\\10&-5&-5\end{bmatrix}.}

It is easy to see \text{rank }(A-2I)=2 and \text{null }(A-2I)=1\neq 3, thus A is not similar over R to a diagonal matrix.
When A is in C, \det (xI-A)=(x-2)(x-i)(x+i), thus A has 2,i,-i as its characteristic value, which means A is similar over the field C to a diagonal matrix.

6.Let T be the linear operator on R^4 which is represented in the standard ordered basis by the matrix

\displaystyle{\begin{bmatrix}0&0&0&0\\a&0&0&0\\0&b&0&0\\0&0&c&0\end{bmatrix}.}

Under what conditions on a,b,c is T diagonalizable?
Solution: Let A be the matrix above and the characteristic value of A is 0, by Exercise 3. Thus if T is diagonalizable, the null space of T shall have dimension 4, which means Ax=0 for all x\in R^4, thus for any x_1,x_2,x_3,x_4\in R,

\displaystyle{\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}

7.Let T be a linear operator on the n-dimensional vector space V, and suppose that T has n distinct characteristic values. Prove that T is diagonalizable.
Solution: Suppose T has c_1,\dots,c_n as characteristic values, in which c_i\neq c_j whenever i\neq j. Then for each c_i we can find an \alpha_i\neq 0 such that T\alpha_i=c_i\alpha_i, and \{\alpha_1,\dots,\alpha_n\} are linearly independent, thus a basis for V, which is the definition of diagonalizable operators.

8.Let A and B be n\times n matrices over the field F. Prove that if (I-AB) is invertible, then (I-BA) is invertible and

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

Solution: We have

\displaystyle{\begin{aligned}&\quad[I+B(I-AB)^{-1}A]\\&=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}}

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

9.Use the result of Exercise 8 to prove that, if A and B are n\times n matrices over the field F, then AB and BA have precisely the same characteristic values in F.
Solution: Let c be a characteristic value of AB, and \alpha\neq 0 such that AB\alpha=c\alpha.
If c=0, then ABX=0 has non-trivial solutions, which means AB is not invertible, so either A or B is not invertible, if A is not invertbile, we can find \beta\neq 0 such that A\beta=0, which means BA\beta=0, if A is invertble but B is not, then we can find \gamma\neq 0 such that B\gamma=0, as \gamma\neq 0 means A^{-1}\gamma\neq 0, we have BA(A^{-1}\gamma)=0, in both cases we proved 0 is a characteristic value of BA.
If c\neq 0, then we have (cI-AB)\alpha=0, or (I-\frac{1}{c}AB)\alpha=0, so (I-\frac{1}{c}AB) is not invertible, thus (I-\frac{1}{c}BA) is not invertible, which means (cI-BA)\beta=0 for some \beta\neq 0, so c is a characteristic value of BA.

10.Suppose that A is a 2\times 2 matrix with real entries which is symmetric(A^t=A). Prove that A is similar over R to a diagonal matrix.
Solution: Write A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, then from A^t=A we have b=c, and

\displaystyle{\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}}

So the polynomial of A can be totally factored.

11.Let N be a 2 \times 2 complex matrix such that N^2=0. Prove that either N=0 or N is similar over C to \begin{bmatrix}0&0\\1&0\end{bmatrix}.
Solution: If N\neq 0, at least one row vector of N is not zero, let this row vector of N be \alpha, then N\alpha\neq 0.
If c,d\in C such that c\alpha+dN\alpha=0, we shall have N(c\alpha+dN\alpha)=0, or cN\alpha=0, this means c=0 and d=0, so \{\alpha,N\alpha\} are linearly independent, thus the matrix P=[\alpha,N\alpha] is invertible. As

\displaystyle{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}}

We have N=P^{-1}\begin{bmatrix}0&0\\1&0\end{bmatrix}P.

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

\displaystyle{\begin{bmatrix}a&0\\0&b\end{bmatrix}\qquad \begin{bmatrix}a&0\\1&a\end{bmatrix}}

Solution: The characteristic polynomial of A is a polynomial of degree 2 over C, thus must be the form (x-a)(x-b) or (x-a)^2 for some a,b\in C. In the first case, a,b are characteristic values of A, and choose \alpha,\beta to be characteristic vectors of a and b, we shall have A\alpha=a\alpha,A\beta=b\beta, thus A is similar to \begin{bmatrix}a&0\\0&b\end{bmatrix} with the matrix P=[\alpha,\beta] and P^{-1}AP=\begin{bmatrix}a&0\\0&b\end{bmatrix}.
In the second case, a is the only possible characteristic value of A. let \alpha be a characteristic vector of a, then it can be extended to a basis \{\alpha,\beta\} of C^2, let \gamma:=(A-aI)\beta, so we can express \gamma=m\alpha+n\beta,m,n\in C.
Since (A-aI)\gamma=(A-aI)(m\alpha+n\beta)=n(A-aI)\beta=n\gamma, we shall have A\gamma=(a+n)\gamma, since a is the only possible characteristic value of A, we have either \gamma=0 or n=0, the first possibility means A-aI=0, if \gamma\neq 0 but n=0, then \gamma=m\alpha, so (A-aI)\gamma=0, which means (A-aI)^2=0. Use Exercise 11 we know A-aI is similar to \begin{bmatrix}0&0\\1&0\end{bmatrix}, so choose P invertible such that
A-aI=P^{-1}\begin{bmatrix}0&0\\1&0\end{bmatrix}P, we can finish the proof since

\displaystyle{A=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}P}.

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

\displaystyle{(Tf)(x)=\int_0^xf(t)dt.}

Prove that T has no characteristic values.
Solution: Assume T has a characteristic value c and a corresponding characteristic vector f\neq0, then

\displaystyle{(Tf)(x)=\int_0^xf(t)dt=cf(x)}

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

\displaystyle{f(x)\geq \frac{f(a)}{2},\quad a-\delta<x<a\implies\int_{a-\delta}^af(t)dt\geq \frac{f(a)}{2}\delta>0}

but \int_{a-\delta}^af(t)dt=\int_{0}^af(t)dt-\int_0^{a-\delta}f(t)dt=0, a contradiction. If f(a)<0 we can similarly get an contradiction. Thus the first case is not valid. We can only have c\neq 0 and f(0)=0.
Under the condition c\neq 0, we have, by the Fundamental Theorem of Calculus:

\displaystyle{\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}}

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

14.Let A be an n\times n diagonal matrix with characteristic polynomial

\displaystyle{(x-c_1)^{d_1}\dots(x-c_k)^{d_k}}

where c_1,\cdots,c_k are distinct. Let V be the space of n\times n matrices B such that AB=BA. Prove that the dimension of V is d_1^2+\cdots+d_k^2.
Solution: Notice that A is diagonal, thus

\displaystyle{A=\begin{bmatrix}c_1I_{d_1}&&\\&\ddots&\\&&c_kI_{d_k}\end{bmatrix}}

where I_{d_j} is the d_j\times d_j identity matrix, and we also have d_1+\cdots+d_k=n. Now write B\in V as

\displaystyle{B=\begin{bmatrix}B_{11}&\cdots&B_{1k}\\&\cdots&\\B_{k1}&\cdots&B_{kk}\end{bmatrix}}

where each B_{ij} is a d_i\times d_j block matrix, then if AB=BA, we have

\displaystyle{\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}}

doing the block matrix multiplication we have

\displaystyle{\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}}

Since c_1,\dots,c_k are distinct, a comparision shows that B_{ij}=0 whenever i\neq j, thus the blocks that can be arbitrary values in B are B_{ii},i=1,\cdots,k, each has d_j^2 entries, and the conclusion follows.

15.Let V be the space of n\times n matrices over F. Let A be a fixed n\times n matrix over F. Let T be the linear operator ‘left multiplication by A‘ on V. Is it true that A and T have the same characteristic values?
Solution: First if c is a characteristic value of T, then there is B\neq 0 such that T(B)=AB=cB, so (A-cI)B=0, since B\neq 0, at least one column of B is not zero, thus (A-cI)X=0 has non-trivial solutions and A-cI is not invertible, which means c is a characteristic value of A.
Conversely, let c be a characteristic value of A, and a corresponding characteristic vector \alpha\neq 0, let B=[\alpha,0,\cdots,0] be a n\times n matrix with the first column \alpha and the remaining columns 0, then AB=[A\alpha,0,\cdots,0]=cB, which means T(B)=cB, thus c is a characteristic value of T.

Definition and Theorems (Chapter 5)

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Definition. A ring is a set K, together with two operations (x,y)\to x+y and (x,y)\to xy satisfying
( a ) K is a commutative group under the operation (x,y)\to x+y (K is a commutative group under addition);
( b ) (xy)z=x(yz) (multiplication is associative);
( c ) x(y+z)=xy+xz;(y+z)x=yx+zx (the two distributive laws hold).
If xy=yx for all x,y\in K, we say that the ring K is commutative. If there is an element 1 in K such that 1x=x1=x for each x,K is said to be a ring with identity, and 1 is called the identity for K.

Definition. Let K be a commutative ring with identity, n a positive integer, and let D be a function which assigns to each n\times n matrix A over K a scalar D(A) in K. We say that D is n-linear if for each i,1\leq i\leq n, D is a linear function of the ith row when the other (n-1) rows are held fixed.

Lemma. A linear combination of n-linear functions is n-linear.

Definition. Let D be an n-linear function. We say D is alternating (or alternate) if the following two conditions are satisfied:
( a ) D(A)=0 whenever two rows of A are equal.
( b ) If A' is a matrix obtained from A by interchanging two rows of A, then D(A')=-D(A).

Definition. Let K be a commutative ring with identity, and let n be a positive integer. Suppose D is a function from n\times n matrices over K into K. We say that D is a determinant function if D is n-linear, alternating and D(I)=1.

Lemma. Let D be a 2-linear function with the property that D(A)=0 for all 2\times 2 matrices A over K having equal rows. Then D is alternating.

Lemma. Let D be an n-linear function on n\times n matrices over K. Suppose D has the property that D(A)=0 whenever two adjacent rows of A are equal. Then D is alternating.

Definition. If n>1 and A is an n\times n matrix over K, we let A(i|j) denote the (n-1)\times (n-1) matrix obtained by deleting the ith row and jth column of A. If D is an (n-1)-linear function and A is an n\times n matrix, we put D_{ij}(A)=D[A(i|j)].

Theorem 1. Let n>1 and let D be an alternating (n-1)-linear function on (n-1)\times (n-1) matrices over K. For each j,1\leq j\leq n, the function E_j defined by

\displaystyle{E_j(A)=\sum_{i=1}^n(-1)^{i+j}A_{ij}D_{ij}(A)}.

is an alternating n-linear function on n\times n matrices A. If D is a determinant function, so is each E_j.
Corollary. Let K be a commutative ring with identity and let n be a positive integer. There exists at least one determinant function on K^{n\times n}.

Theorem 2. Let K be a commutative ring with identity and let n be a positive integer. There is precisely one determinant function on the set of n\times n matrices over K, and it is the function \det defined by

\displaystyle{\det(A)=\sum_{\sigma}(\text{sgn }{\sigma})A(1,\sigma1)\cdots A(n,{\sigma}n)}

If D is any alternating n-linear function on K^{n\times n}, then for each n\times n matrix A

\displaystyle{D(A)=(\det A)D(I)}

Theorem 3. Let K be a commutative ring with identity, and let A and B be n\times n matrices over K. Then

\displaystyle{\det (AB)=(\det A)(\det B).}

Theorem 4. Let A be an n\times n matrix over K. Then A is invertible over K if and only if \det A is invertible over K. When A is invertible, the unique inverse for A is

\displaystyle{A^{-1}=(\det A)^{-1} \text{adj }A.}

In particular, an n\times n matrix over a field is invertible if and only if its determinant is different from zero.

Definition. The K-module V is called a free module if it has a basis. If V has a finite basis containing n elements, then V is called a free K-module with n generators.

Definition. The module V is finitely generated if it contains a finite subset which spans V. The rank of a finitely generated module is the smallest integer k that some k elements spans V.

Theorem 5. Let K be a commutative ring with identity. If V is a free K-module with n generators, then the rank of V is n.