三月 2018

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定理(欧拉线定理) 若点{O}、点{G}、点{H}分别是{\triangle ABC}的外心、重心、垂心,求证:

  • {\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}};
  • {\overrightarrow{GH}=2\overrightarrow{OG}}.

 

证明:

  • 使用坐标法.以{O}为原点,在平面{ABC}上建立平面直角坐标系.设 {\triangle ABC}的半径为{R},且点{A,B,C}的坐标依次为 {A(x_{A},y_{A})=(R\cos\alpha,R\sin\alpha)},{B(x_{B},y_{B})=(R\cos\beta,R\sin\beta)}, {C(x_{C},y_{C})=(R\cos\gamma,R\sin\gamma)}.先求垂心{H}的坐标{(x,y)}.因为向量{\overrightarrow{AH}\perp \overrightarrow{BC}},所以 {\overrightarrow{AH}\cdot\overrightarrow{BC}=0},由向量数量积的坐标表示,可 得

    \displaystyle (x-R\cos\alpha)(R\cos\gamma-R\cos\beta)+(y-R\sin\alpha)(R\sin\gamma-R\sin\beta)=0.

    整理可得

    \displaystyle (\cos\gamma-\cos\beta)x+(\sin\gamma-\sin\beta)y=(\cos\gamma-\cos\beta)R\cos\alpha+(\sin\gamma-\sin\beta)R\sin\alpha \ \ \ \ \ (1)

    同样由{\overrightarrow{BH}\perp \overrightarrow{CA}},可得

    \displaystyle (\cos\alpha-\cos\gamma)x+(\sin\alpha-\sin\gamma)y=(\cos\alpha-\cos\gamma)R\cos\beta+(\sin\alpha-\sin\gamma)R\sin\beta. \ \ \ \ \ (2)

    联立方程(1),(2),得到方程组.下面解该线性方程组.令行列式

    \displaystyle \begin{array}{rcl} \det A= \begin{vmatrix} \cos\gamma-\cos\beta & \sin\gamma-\sin\beta\\ \cos\alpha-\cos\gamma &\sin\alpha-\sin\gamma \end{vmatrix}&=& \begin{vmatrix} 2\sin \frac{\beta+\gamma}{2}\sin \frac{\beta-\gamma}{2}&2\cos \frac{\beta+\gamma}{2}\sin \frac{\gamma-\beta}{2}\\ 2\sin \frac{\gamma+\alpha}{2}\sin \frac{\gamma-\alpha}{2}&2\cos \frac{\gamma+\alpha}{2}\sin \frac{\alpha-\gamma}{2} \end{vmatrix}\\ &=&4\sin \frac{\beta-\gamma}{2}\sin \frac{\gamma-\alpha}{2} \begin{vmatrix} \sin \frac{\beta+\gamma}{2}&-2\cos \frac{\beta+\gamma}{2}\\ \sin \frac{\gamma+\alpha}{2}&-2\cos \frac{\gamma+\alpha}{2} \end{vmatrix}\\ &=&4\sin \frac{\alpha-\beta}{2}\sin \frac{\beta-\gamma}{2}\sin \frac{\gamma-\alpha}{2}. \end{array}

    \displaystyle \begin{array}{rcl} \det B_1&=& \begin{vmatrix} (\cos\gamma-\cos\beta)R\cos\alpha+(\sin\gamma-\sin\beta)R\sin\alpha &\sin\gamma-\sin\beta\\ (\cos\alpha-\cos\gamma)R\cos\beta+(\sin\alpha-\sin\gamma)R\sin\beta&\sin\alpha-\sin\gamma \end{vmatrix}\\ &=&R \begin{vmatrix} \cos(\alpha-\gamma)-\cos(\alpha-\beta)&\sin\gamma-\sin\beta\\ \cos(\beta-\alpha)-\cos(\beta-\gamma)&\sin\alpha-\sin\gamma \end{vmatrix}\\ &=&R \begin{vmatrix} 2\sin (\alpha-\frac{\beta+\gamma}{2})\sin \frac{\gamma-\beta}{2}& 2\cos \frac{\beta+\gamma}{2}\sin \frac{\gamma-\beta}{2}\\ 2\sin (\beta-\frac{\alpha+\gamma}{2})\sin \frac{\alpha-\gamma}{2}&2\cos \frac{\gamma+\alpha}{2}\sin \frac{\alpha-\gamma}{2} \end{vmatrix}\\ &=&4R\left[\sin(\alpha-\frac{\beta+\gamma}{2})\sin \frac{\gamma-\beta}{2}\cos \frac{\gamma+\alpha}{2}\sin \frac{\alpha-\gamma}{2}-\sin(\beta-\frac{\alpha+\gamma}{2})\sin \frac{\alpha-\gamma}{2}\cos \frac{\beta+\gamma}{2}\sin \frac{\gamma-\beta}{2}\right]\\ &=&4R\sin \frac{\alpha-\gamma}{2}\sin \frac{\gamma-\beta}{2}\left[\sin(\alpha-\frac{\beta+\gamma}{2})\cos \frac{\gamma+\alpha}{2}-\sin(\beta-\frac{\alpha+\gamma}{2})\cos \frac{\beta+\gamma}{2}\right] \end{array}

    \displaystyle \begin{array}{rcl} &=&4R\sin \frac{\alpha-\gamma}{2}\sin \frac{\gamma-\beta}{2}\left[\frac{\sin(\frac{3\alpha}{2}-\frac{\beta}{2})-\sin(\gamma+\frac{\beta-\alpha}{2})}{2}-\frac{\sin(\frac{3\beta}{2}-\frac{\alpha}{2})-\sin(\gamma-\frac{\beta-\alpha}{2})}{2}\right]\\ &=& 4R\sin \frac{\alpha-\gamma}{2}\sin \frac{\gamma-\beta}{2}\left[\frac{\sin(\frac{3\alpha}{2}-\frac{\beta}{2})-\sin (\frac{3\beta}{2}-\frac{\alpha}{2})}{2}+\frac{\sin(\gamma-\frac{\beta-\alpha}{2})-\sin(\gamma+\frac{\beta-\alpha}{2})}{2}\right]\\ &=&4R\sin \frac{\alpha-\gamma}{2}\sin \frac{\gamma-\beta}{2}\left[\cos \frac{\alpha+\beta}{2}\sin(\alpha-\beta)+\cos\gamma\sin \frac{\alpha-\beta}{2}\right]\\ &=& 4R\sin \frac{\alpha-\gamma}{2}\sin \frac{\gamma-\beta}{2}\left[2\cos \frac{\alpha+\beta}{2}\sin \frac{\alpha-\beta}{2}\cos \frac{\alpha-\beta}{2}+\cos\gamma\sin \frac{\alpha-\beta}{2}\right]\\ &=&4R\sin \frac{\alpha-\gamma}{2}\sin \frac{\gamma-\beta}{2}\sin \frac{\alpha-\beta}{2}\left[2\cos \frac{\alpha+\beta}{2}\cos \frac{\alpha-\beta}{2}+\cos\gamma\right]\\ &=&4R\sin \frac{\alpha-\gamma}{2}\sin \frac{\gamma-\beta}{2}\sin \frac{\alpha-\beta}{2}(\cos\alpha+\cos\beta+\cos\gamma)\\ &=&4R\sin \frac{\alpha-\beta}{2}\sin \frac{\beta-\gamma}{2}\sin \frac{\gamma-\alpha}{2}(\cos\alpha+\cos\beta+\cos\gamma) \end{array}

    由解线性方程组的Cramer法则,

    \displaystyle x=\frac{\det B_1}{\det A}=R\cos\alpha+R\cos\beta+R\cos\gamma=x_A+x_B+x_C

    由对称性,

    \displaystyle y=y_A+y_B+y_C.

    所以垂心{H}的坐标为{(x_A+x_B+x_C,y_A+y_B+y_C)}.所以{\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}.

  • 重心{G}的坐标为 {G(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3})}.所以 {\overrightarrow{OH}=3\overrightarrow{OG}},所以{\overrightarrow{GH}=2\overrightarrow{OG}}. \Box

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问题: 计算{n}阶方阵{A_{n}}的行列式

\displaystyle A_{n}=\begin{bmatrix} 1-\lambda&1&1&1&\cdots&1\\ 1&1-\lambda&1&1&\cdots&1\\ 1&1&1-\lambda&1&\cdots&1\\ 1&1&1&1-\lambda&\cdots&1\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&1&\cdots&1-\lambda \end{bmatrix}

 

解法1:{n}阶方阵

\displaystyle B_n= \begin{bmatrix} 1&1&1&1&\cdots&1\\ 1&1-\lambda&1&1&\cdots&1\\ 1&1&1-\lambda&1&\cdots&1\\ 1&1&1&1-\lambda&\cdots&1\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&1&\cdots&1-\lambda \end{bmatrix}

将矩阵{B_n}的第{2}行乘以{-1}加到第{1}行上,可得

\displaystyle B_n'= \begin{bmatrix} 0&\lambda&0&0&\cdots&0\\ 1&1-\lambda&1&1&\cdots&1\\ 1&1&1-\lambda&1&\cdots&1\\ 1&1&1&1-\lambda&\cdots&1\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&1&\cdots&1-\lambda \end{bmatrix}

将行列式{\det B_n'}按照第一行展开,可得递推关系

\displaystyle \det B_n'=-\lambda B_{n-1}'.

{\det B_2'=-\lambda}.因此{\det B_n'=(-\lambda)^{n-1}}.于是{\det B_n=\det B_n'=(-\lambda)^{n-1}}.

将矩阵{A_{n}}的第{2}行乘以{-1},加到第{1}行上,得到矩阵{A_{n}'}:

\displaystyle A_{n}'= \begin{bmatrix} -\lambda&\lambda&0&0&\cdots&0\\ 1&1-\lambda&1&1&\cdots&1\\ 1&1&1-\lambda&1&\cdots&1\\ 1&1&1&1-\lambda&\cdots&1\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&1&\cdots&1-\lambda \end{bmatrix}

由于{\det A_{n}=\det A_{n}'},我们只用求出{\det A_n'}.将行列式按照第一行 展开,可得

\displaystyle \det A_n'=-\lambda\det A_{n-1}-\lambda \det B_{n-1}

\displaystyle \det A_n=-\lambda \det A_{n-1}+(-\lambda)^{n-1}.

{\det A_2=\lambda^2-2\lambda}.因此{\det A_n=(-\lambda)^{n-1}(n-\lambda)}. \Box

解法2:{n}阶方阵{C_n}的各项元素都为{1}.则矩阵{C_n}的秩为{1},它的一个特征向 量是{ \begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix} },该特征向量对应的特征值为{n}.矩阵{C_n}的其余{n-1}个特征值都是{0}.

矩阵{C_n}的特征方程是{\det A_n=0},该特征方程是关于{\lambda}{n}次方程,该 {n}次方程的次数最高项的系数为{(-1)^{n}},它的{n}个根就是矩阵{C_n}的特征值,即分别为{n,0,0,\cdots,0}({n-1}{0}).因此{\det A_n}只能为{(-\lambda)^{n-1}(n-\lambda)}. \Box

Tags: , ,

Exercise 5.1.1 Find the eigenvalue and eigenvectors of the matrix {A= \begin{bmatrix} 1&-1\\ 2&4 \end{bmatrix} }.Verify that the trace equals the sum of the eigenvalues,and the determinant equals their product.

Solution: Suppose that the eigenvalue is {\lambda},then

\displaystyle \begin{vmatrix} 1-\lambda&-1\\ 2&4-\lambda \end{vmatrix}=0,

so {\lambda=2} or {\lambda=3}.Eigenvalue {2} correspond to the eigenvector {(1,-1)},eigenvalue {3} correspond to the eigenvector {(1,-2)}.The sum of the eigenvalue is {5},the trace of the determinant is also {1+4=5}.The product of the eigenvalue is {6},and the determinant of {A} is {4+2=6}. \Box

Exercise 5.1.2 With the same matrix {A},solve the differential equation {du/dt=Au},{u(0)= \begin{bmatrix} 0\\ 6 \end{bmatrix}. }What are the two pure exponential solutions?

Solution: Let {u= \begin{bmatrix} x\\ y \end{bmatrix} }.Then the differential equation {du/dt=Au} becomes

\displaystyle \begin{bmatrix} \frac{dx}{dt}\\ \frac{dy}{dt} \end{bmatrix}= \begin{bmatrix} 1&-1\\ 2&4 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}.

Now we find pure exponential solutions.Let {x=e^{\lambda t}a},{y=e^{\lambda t}b},then the differential equation becomes

\displaystyle \begin{bmatrix} \lambda e^{\lambda t}a\\ \lambda e^{\lambda t}b \end{bmatrix}= \begin{bmatrix} 1&-1\\ 2&4 \end{bmatrix} \begin{bmatrix} e^{\lambda t}a\\ e^{\lambda t}b \end{bmatrix},

simplify,we get

\displaystyle \begin{bmatrix} 1&-1\\ 2&4 \end{bmatrix} \begin{bmatrix} a\\ b \end{bmatrix}=\lambda \begin{bmatrix} a\\ b \end{bmatrix}.

When {\lambda=2},{ \begin{bmatrix} a\\ b \end{bmatrix}= \begin{bmatrix} k\\ -k \end{bmatrix} };When {\lambda=3},{ \begin{bmatrix} a\\ b \end{bmatrix}= \begin{bmatrix} k\\ -2k \end{bmatrix}. } So the two pure exponential solutions are { \begin{bmatrix} e^{\lambda t}\\ -e^{\lambda t} \end{bmatrix} } and { \begin{bmatrix} e^{\lambda t}\\ -2e^{\lambda t} \end{bmatrix}. }The general solution is

\displaystyle c_1 \begin{bmatrix} e^{\lambda t}\\ -e^{\lambda t} \end{bmatrix}+c_2 \begin{bmatrix} e^{\lambda t}\\ -2e^{\lambda t} \end{bmatrix}.

\Box

Exercise 5.1.3 If we shift to {A-7I},what are the eigenvalues and eigenvectors and how are they related to those of {A}?

\displaystyle B=A-7I= \begin{bmatrix} -6&-1\\ 2&-3 \end{bmatrix}.

 

Solution: The characteristic equation of matrix {A} is {\det (A-\lambda I)=0}.And the characteristic equation of matrix {A-7I} is {\det (A-(7+\lambda')I)=0}.So the eigenvalue of matrix {A-7I} is the corresponding eigenvalue of matrix {A} minus {7},which are {-5} and {-4}.

The eigenvectors of two matrices are same. \Box

Exercise 5.1.4 Solve {du/dt=Pu},when {P} is a projection:

\displaystyle \frac{du}{dt}= \begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2} \end{bmatrix}u~~~with~~~u(0)= \begin{bmatrix} 5\\ 3 \end{bmatrix}.

Part of {u(0)} increases exponentially while the nullspace part stays fixed.

Solution: Let {u(t)= \begin{bmatrix} x(t)\\ y(t) \end{bmatrix} },then the differential equation becomes

\displaystyle \begin{bmatrix} \frac{dx}{dt}\\ \frac{dy}{dt} \end{bmatrix}= \begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} \begin{bmatrix} x(t)\\ y(t) \end{bmatrix}.

Now we try to find the pure exponential solution.Suppose that { \begin{bmatrix} x(t)\\ y(t) \end{bmatrix}= \begin{bmatrix} e^{\lambda t}x(0)\\ e^{\lambda t}y(0) \end{bmatrix}, }then from the differential equation we know that

\displaystyle \begin{bmatrix} \lambda x(0)e^{\lambda t}\\ \lambda y(0)e^{\lambda t} \end{bmatrix}= \begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} \begin{bmatrix} x(0)e^{\lambda t}\\ y(0)e^{\lambda t} \end{bmatrix},

simplify it,we get

\displaystyle \begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} \begin{bmatrix} x(0)\\ y(0) \end{bmatrix}=\lambda \begin{bmatrix} x(0)\\ y(0) \end{bmatrix}.

When {\lambda=0},we could let { \begin{bmatrix} x(0)\\ y(0) \end{bmatrix}= \begin{bmatrix} 1\\ -1 \end{bmatrix}. }When {\lambda=1},we could let { \begin{bmatrix} x(0)\\ y(0) \end{bmatrix}= \begin{bmatrix} 1\\ 1\\ \end{bmatrix}. }So the general solution is

\displaystyle c_1 \begin{bmatrix} e^{\lambda t}\\ e^{\lambda t} \end{bmatrix}+c_2 \begin{bmatrix} e^{\lambda t}\\ -e^{\lambda t} \end{bmatrix}

\Box

Exercise 5.1.6 Give an example to show that the eigenvalues can be changed when a multiple of one row is subtracted from another. Why is a zero eigenvalue not changed by the steps of elimination?

Solution: Example:{A= \begin{bmatrix} 1&1\\ 1&1 \end{bmatrix} },subtract the second row from the first row,we get {B= \begin{bmatrix} 0&0\\ 1&1 \end{bmatrix} }.The eigenvalue of {A} are {0} and {2}.The eigenvalue of {B} are {0} and {1}.

A zero eigenvalue is not changed by the steps of elimination because,suppose the steps of elimination turn matrix {A} into matrix {B},and {B=MA},where {M} is an invertible matrix.{Ax=0x},so {Bx=(LA)x=L(Ax)=L(\mathbf{0})=\mathbf{0}=0x}.So {0} is also an eigenvalue of {B}. \Box

Exercise 5.1.7 Suppose that {\lambda} is an eigenvalue of {A},and {x} is its eigenvector:{Ax=\lambda x}.

  • Show that this same {x} is an eigenvector of {B=A-7I},and find the eigenvalue.This should confirm Exercise 3.
  • Assuming {\lambda\neq 0},show that {x} is also an eigenvector of {A^{-1}},and find the eigenvalue.

 

Solution: We solve the second part.{Ax=\lambda x},so {x=A^{-1}(\lambda x)=\lambda A^{-1}x},so {A^{-1}x=\frac{1}{\lambda }x}.The eigenvalue of {A^{-1}} is {\frac{1}{\lambda}}. \Box

Exercise 5.1.8 Show that the determinant equals the product of the eigenvalues by imagining that the characteristic polynomial is factored into

\displaystyle \det (A-\lambda I)=(\lambda_1-\lambda)(\lambda_2-\lambda)\cdots (\lambda_n-\lambda),

and making a clever choise of {\lambda}.

Solution: In the above formula,let {\lambda=0}. \Box

Exercise 5.1.9 Show that the trace equals the sum of the eigenvalues,in two steps.First,find the coefficient of {(-\lambda)^{n-1}} on the right side of equation (16).Next,find all the terms in

\displaystyle \det (A-\lambda I)=\det \begin{bmatrix} a_{11}-\lambda&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}-\lambda&\cdots&a_{2n}\\ \vdots&\vdots& &\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}-\lambda \end{bmatrix}

that involve {(-\lambda)^{n-1}}.They all come from the main diagonal!Find that coefficient of {(-\lambda)^{n-1}} and compare.

Exercise 5.1.10

  • Construct 2 by 2 matrices such that the eigenvalue of AB are not the products of the eigenvalues of A and B,and the eigenvalues of A+B are not the sums of the individual eigenvalues.
  • Verify,however,that the sum of the eigenvalues of {A+B} equals the sum of all the individual eigenvalues of {A} and {B},and similarly for products.Why is this true?

 

Solution:

  • The sum of all the eigenvalues of {A+B} is the trace of {A+B},the sum of all the individual eigenvalue of {A} is the trace of {A},the sum of all the individual eigenvalue of {B} is the trace of {B}.And

    \displaystyle trace(A+B)=trace(A)+trace(B).

    So the sum of all the eigenvalues of {A+B} equals the sum of all the individual eigenvalues of {A} and {B}.

    The product of all the eigenvalues of {AB} equals the determinant of {AB},the product of all the eigenvalues of {A} equals the determinant of {A},the product of all the eigenvalues of {B} equals the determinant of {B}.And

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

    So the product of all the eigenvalues of {AB} is the product of all the individual eigenvalues of {A} and {B}.

\Box

Exercise 5.1.12 Find the eigenvalues and eigenvectors of

\displaystyle A= \begin{bmatrix} 3&4\\ 4&-3 \end{bmatrix}~~and~~A= \begin{bmatrix} a&b\\ b&a \end{bmatrix}.

 

Solution:

  • The eigenvalues of {A} are {-1} and {7}.The corresponding eigenvectors are {(1,-1)} and {(1,1)}.
  • The eigenvalues of {A} are {a-b} and {a+b}.The corresponding eigenvectors are {(1,-1)} and {(1,1)}.

\Box

Exercise 5.1.13 If {B} has eigenvalues {1,2,3},{C} has eigenvalues {4,5,6},and {D} has eigenvalues {7,8,9},what are the eigenvalues of the {6} by {6} matrix {A= \begin{bmatrix} B&C\\ 0&D \end{bmatrix} } ?

Solution: {1,2,3,7,8,9}.This can be figured out by using determinant. \Box

Exercise 5.1.14 Find the rank and all four eigenvalues for both the matrix of ones and the checker board matrix:

\displaystyle A= \begin{bmatrix} 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1 \end{bmatrix}~~and~~C= \begin{bmatrix} 0&1&0&1\\ 1&0&1&0\\ 0&1&0&1\\ 1&0&1&0 \end{bmatrix}.

Which eigenvectors correspond to nonzero eigenvalues?

Solution: {rank(A)=1},{rank(C)=2}.The characteristic equation of matrix {A} is {\lambda^3(\lambda-4)=0}.So the eigenvalues of matrix {A} are {4,0,0,0}.

The characteristic equation of matrix {C} is {\lambda^4-4\lambda^2=0}.So the eigenvalues of matrix {C} are {0,0,2,-2}. \Box

Exercise 5.1.15 What are the rank and eigenvalues when {A} and {C} in the previous exercise are {n} by {n}?Remember that the eigenvalue {\lambda=0} is repeated {n-r} times.

Solution: {rank(A)=1}.One eigenvector of matrix {A} is { \begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix} },the corresponding eigenvalue of this eigenvector is {n}.So the eigenvalues of {A} are {n,0,0,\cdots,0}(There are {n-1} zeros). \Box

{rank(C)=2}.One eigenvector of matrix {C} is { \begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix}, }another eigenvector of {C} is { \begin{bmatrix} -1\\ 1\\ \vdots\\ (-1)^n \end{bmatrix} }.The corresponding eigenvalues of these two vectors are{\frac{n}{2}-\frac{1}{4}[1+(-1)^{n+1}]} and {-\frac{n}{2}+\frac{1}{4}[1+(-1)^{n+1}]} respectively.So the eigenvalues of {A} are {\frac{n}{2}-\frac{1}{4}[1+(-1)^{n+1}]},{-\frac{n}{2}+\frac{1}{4}[1+(-1)^{n+1}]},{0,\cdots,0}(There are {n-2} zeros). \Box

Exercise 5.1.15 What are the rank and eigenvalues when {A} and {C} in the previous exercise are {n} by {n}?Remember that the eigenvalue {\lambda=0} is repeated {n-r} times.

Solution: {rank(A)=1}.One eigenvector of matrix {A} is { \begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix} },the corresponding eigenvalue of this eigenvector is {n}.So the eigenvalues of {A} are {n,0,0,\cdots,0}(There are {n-1} zeros).

{rank(C)=2}.One eigenvector of matrix {C} is { \begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix}, }another eigenvector of {C} is { \begin{bmatrix} -1\\ 1\\ \vdots\\ (-1)^n \end{bmatrix} }.The corresponding eigenvalues of these two vectors are{\frac{n}{2}-\frac{1}{4}[1+(-1)^{n+1}]} and {-\frac{n}{2}+\frac{1}{4}[1+(-1)^{n+1}]} respectively.So the eigenvalues of {A} are {\frac{n}{2}-\frac{1}{4}[1+(-1)^{n+1}]},{-\frac{n}{2}+\frac{1}{4}[1+(-1)^{n+1}]},{0,\cdots,0}(There are {n-2} zeros). \Box

Exercise 5.1.17 Choose the third row of the “companion matrix”

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

so that its characteristic polynomial {|A-\lambda I|} is {-\lambda^3+4\lambda^2+5\lambda+6}.

Solution:

\displaystyle A= \begin{bmatrix} 0&1&0\\ 0&0&1\\ 6&5&4\\ \end{bmatrix}

\Box

Exercise 5.1.18 Suppose {A} has eigenvalues {0,3,5} with independent eigenvectors {u,v,w}.

  • Give a basis for the nullspace and a basis for the column space.
  • Find a particular solution to {Ax=v+w}.Find all solutions.
  • Show that {Ax=u} has no solution.(If it had a solution,then {\underline{~~~~~~}} would be in the column space.)

 

Solution:

  • A basis for the nullspace is {u}.A basis for the column space is {\{v,w\}}.
  • A particular solution is {\frac{1}{3}v+\frac{1}{5}w}.All solutions are of the form {\frac{1}{3}v+\frac{1}{5}w+ku},where {k} is an arbitrary real number.
  • Otherwise,{u} would be in the column space,this is an contradiction to the fact that {u,v,w} are linearly independent.

\Box

Exercise 5.1.19 The powers {A^k} of this matrix {A} approaches a limit as {k\rightarrow\infty}:

\displaystyle A= \begin{bmatrix} .8&.3\\ .2&.7 \end{bmatrix},A^2= \begin{bmatrix} .70&.45\\ .30&.55 \end{bmatrix},~~and~~A^{\infty}= \begin{bmatrix} .6&.6\\ .4&.4 \end{bmatrix}.

The matrix {A^2} is halfway between {A} and {A^{\infty}}.Explain why {A^2=\frac{1}{2}(A+A^{\infty})} from the eigenvalues and eigenvectors of these three matrices.

Solution: The eigenvalues of {A} are {1} and {\frac{1}{2}}.The corresponding eigenvectors are { \begin{bmatrix} 3\\ 2 \end{bmatrix} } and { \begin{bmatrix} 1\\ -1 \end{bmatrix}. }

So the eigenvalues of {A^2} are {1} and {\frac{1}{4}}.The corresponding eigenvectors are { \begin{bmatrix} 3\\ 2 \end{bmatrix} } and { \begin{bmatrix} 1\\ -1 \end{bmatrix} }.

And the eigenvalues of {A^{\infty}} are {1} and {0},the corresponding eigenvectors are { \begin{bmatrix} 3\\ 2 \end{bmatrix} } and { \begin{bmatrix} 1\\ -1 \end{bmatrix} }.

So {A^2=\frac{1}{2}(A+A^{\infty})}. \Box

Exercise 5.1.25 From the unit vector {u= \begin{bmatrix} \frac{1}{6}\\ \frac{1}{6}\\ \frac{3}{6}\\ \frac{5}{6} \end{bmatrix} },construct the rank-1 projection matrix {P=uu^T}.

  • Show that {Pu=u}.Then {u} is an eigenvector with {\lambda=1}.
  • If {v} is perpendicular to {u} show that {Pv=}zero vector.Then {\lambda=0}.
  • Find three independent eigenvectors of {P} all with eigenvalue {\lambda=0}.

 

Solution:

  • {Pu=(uu^T)u=u(u^Tu)=1u=u}.
  • {Pv=(uu^T)v=u(u^Tv)=0u=}zero vector.
  • { \begin{bmatrix} -3\\ -5\\ 1\\ 1 \end{bmatrix} },{ \begin{bmatrix} -5\\ -3\\ 1\\ 1 \end{bmatrix} },{ \begin{bmatrix} -15\\ 0\\ 5\\ 0 \end{bmatrix} }.

\Box

Exercise 5.1.26 Solve {\det (Q-\lambda I)=0} by the quadratic formula,to reach {\lambda=\cos\theta\pm i\sin\theta}:

\displaystyle Q= \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix}~~~rotates~ the~ xy-plane~ by~ the~ angle~ \theta.

Find the eigenvectors of {Q} by solving {(Q-\lambda I)x=0}.Use {i^2=-1}.

Solution:

\displaystyle \begin{vmatrix} \cos\theta-\lambda&-\sin\theta\\ \sin\theta&\cos\theta-\lambda \end{vmatrix}=0

,so {\lambda=\cos\theta\pm i\sin\theta}.The corresponding eigenvectors are {(-i,1)} and {(i,1)}. \Box

Exercise 5.1.27 Every permutation matrix leaves {x=(1,1,\cdots,1)} unchanged.Then {\lambda=1}.Find two more {\lambda}‘s for these permutations:

\displaystyle P= \begin{bmatrix} 0&1&0\\ 0&0&1\\ 1&0&0 \end{bmatrix}~~and~~P= \begin{bmatrix} 0&0&1\\ 0&1&0\\ 1&0&0 \end{bmatrix}.

 

Solution:

  • {\lambda_2=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}},{\lambda_3=\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}}.
  • {\lambda_2=-1}.

\Box

Exercise 5.1.28 If {A} has {\lambda_1=4} and {\lambda_2=5},then {\det (A-\lambda I)=(\lambda-4)(\lambda-5)=\lambda^2-9\lambda+20}.Find three matrices that have trace {a+d=9},determinant {20},and {\lambda=4,5}.

Solution: All the matrices of the form { \begin{bmatrix} 4&x\\ 0&5 \end{bmatrix} } have trace {9},determinant {20},and characteristic {4,5}. \Box

Exercise 5.1.29 A {3} by {3} matrix {B} is known to have eigenvalue {0,1,2}.This information is enough to find three of these:

  • the rank of {B},
  • the determinant of {B^TB},
  • the eigenvalues of {B^TB},and
  • the eigenvalues of {(B+I)^{-1}.}

 

Solution:

  • {2}.
  • {\det B^TB=(\det B)^2=0}.
  • the eigenvalues of {B+I} are {1,2,3}.So the eigenvalues of {(B+I)^{-1}} are {1,\frac{1}{2},\frac{1}{3}}.

\Box

Exercise 5.1.30 Choose the second row of {A= \begin{bmatrix} 0&1\\ *&* \end{bmatrix} } so that {A} has eigenvalues {4} and {7}.

Solution:

\displaystyle A= \begin{bmatrix} 0&1\\ -28 &11 \end{bmatrix}.

\Box

Exercise 5.1.31 Choose {a,b,c},so that {\det (A-\lambda I)=9\lambda-\lambda^3}.Then the eigenvalues are {-3,0,3}:

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

 

Solution:

\displaystyle A= \begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&9&0 \end{bmatrix}

\Box

Exercise 5.1.32 Construct any {3} by {3} Markov matrix {M}:positive entries down each column add to {1}.If {e=(1,1,1)},verify that {M^Te=e}.By Problem 11,{\lambda=1} is also an eigenvalue of {M}.Challenge:A {3} by {3} singular Markov matrix with trace {\frac{1}{2}} has eigenvalues {\lambda=\underline{- \frac{1}{2}}}.

Exercise 5.1.34 This matrix is singular with rank {1}.Find three {\lambda}‘s and three eigenvectors:

\displaystyle A= \begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix} \begin{bmatrix} 2&1&2 \end{bmatrix}= \begin{bmatrix} 2&1&2\\ 4&2&4\\ 2&1&2 \end{bmatrix}.

 

Solution: {\lambda_1=6},the corresponding eigenvector is { \begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix} }.The rest two eigenvalues are {0} and {0}. \Box

Exercise 5.1.37 When {a+b=c+d},show that {(1,1)} is an eigenvector and find both eigenvalues:

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

 

Solution: The eigenvalues are {a+b} and {a-c}. \Box

Exercise 5.1.38 If we exchange rows {1} and {2} and columns {1} and {2},the eigenvalues don’t change.Find eigenvalues of {A} and {B} for {\lambda=11}.Rank one gives {\lambda_2=\lambda_3=0}.

\displaystyle A= \begin{bmatrix} 1&2&1\\ 3&6&3\\ 4&8&4 \end{bmatrix}~~and~~B=PAP^T= \begin{bmatrix} 6&3&3\\ 2&1&1\\ 8&4&4 \end{bmatrix}.

 

Solution:

\displaystyle A= \begin{bmatrix} 1\\ 3&\\ 4 \end{bmatrix} \begin{bmatrix} 1&2&1 \end{bmatrix}.

The eigenvector of {A} corresponding to eigenvalue {11} is { \begin{bmatrix} 1\\ 3\\ 4 \end{bmatrix} }.

\displaystyle B= \begin{bmatrix} 3\\ 1\\ 4 \end{bmatrix} \begin{bmatrix} 2&1&1 \end{bmatrix}.

The eigenvector of {B} corresponding to eigenvalue {11} is { \begin{bmatrix} 3\\ 1\\ 4 \end{bmatrix} }. \Box

Exercise 5.1.39 Challenge problem:Is there a real {2} by {2} matris(other than {I})with {A^3=I}?Its eigenvalues must satisfy {\lambda^3=I}.They can be {e^{\frac{2\pi i}{3}}} and {e^{\frac{-2\pi i}{3}}}.What trace and determinant would this give?Construct {A}.

Solution:

\displaystyle A= \begin{bmatrix} e^{\frac{2\pi}{3}}&0\\ 0&e^{\frac{2\pi}{3}} \end{bmatrix},or~A=\begin{bmatrix} e^{-\frac{2\pi}{3}}&0\\ 0&e^{-\frac{2\pi}{3}} \end{bmatrix}

\Box

Exercise 5.1.40 There are six {3} by {3} permutation matrices {P}.What numbers can be the determinants of {P}?What numbers can be pivots?What numbers can be the trace of {P}?What four numbers can be eigenvalues of {P}?

Solution: All the six {3} by {3} permutation matrices are

\displaystyle \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}, \begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0\\ \end{bmatrix}, \begin{bmatrix} 0&1&0\\ 1&0&0\\ 0&0&1 \end{bmatrix},

\displaystyle \begin{bmatrix} 0&1&0\\ 0&0&1\\ 1&0&0 \end{bmatrix}, \begin{bmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{bmatrix}, \begin{bmatrix} 0&0&1\\ 0&1&0\\ 1&0&0 \end{bmatrix}.

{1} and {-1} can be the determinant of {P}.The pivots are {1,1,1}.The eigenvalues of {P} can be {1,-1,e^{\frac{2\pi}{3}},e^{\frac{\pi}{3}}}. \Box

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Exercise 4.9 If {P_1} is an even permutation matrix and {P_2} is odd,prove that {\det (P_1+P_2)=0}.

Solution:

\displaystyle \begin{array}{rcl} \det (P_1+P_2)&=&\det P_1(I+P_{odd}) \\&=&\det (I+P_{odd}) \\&=&\det P_{odd}(P_{odd}^{-1}+I) \\&=&-\det (I+P_{odd}^{-1}) \\&=&-\det (I+P_{odd}^T) \\&=&-\det (I+P_{odd})^T \\&=&-\det (I+P_{odd}) \\&=&-\det (P_1+P_2), \end{array}

where {P_{odd}=P_1^{-1}P_2} is an odd permutation matrix.So {\det (P_1+P_2)=0}. \Box

Exercise 4.12 In analogy with the previous exercise,what is the equation for {(x,y,z)} to be on the plane through {(2,0,0),(0,2,0)},and {(0,0,4)}?It involves a {4} by {4} determinant.

Solution:

\displaystyle \begin{vmatrix} x&y&z&1\\ 2&0&0&1\\ 0&2&0&1\\ 0&0&4&1 \end{vmatrix}=0.

\Box

Exercise 4.13 If the points {(x,y,z)},{(2,1,0)} and {(1,1,1)} lie on a plane through the origin,what determinant is zero?Are the vectors {(1,0,-1),(2,1,0),(1,1,1)} independent?

Solution:

\displaystyle \begin{vmatrix} x&y&z\\ 2&1&0\\ 1&1&1 \end{vmatrix}=0.

The vectors {(1,0,-1),(2,1,0),(1,1,1)} are dependent because

\displaystyle \begin{vmatrix} 1&0&-1\\ 2&1&0\\ 1&1&1 \end{vmatrix}=0.

\Box

Exercise 4.16 The circular shift permutes {(1,2,\cdots,n)} into {(2,3,\cdots,1)}.What is the corresponding permutation matrix {P},and(depending on {n})what is its determinant?

Solution: Denote the corresponding permutation matrix by {A_n}.Then the {i(1\leq i\leq n-1)}th row of {A} is the {i+1} th row of the identity matrix {I_n},and the {n} th row of {A} is the first row of {I_n}.

{\det A_n=(-1)^{n-1}}. \Box

Exercise 4.17 Find the determinant of A=eye(5)+ones(5) and if possible eye(n)+ones(n).(They are Matlab commands)

Solution: Denote eye(n)+ones(n) by {A_n},and let matrix {B_n} equals to {A_n} except that the element {a_{11}} changed from {2} to {1}.

By extracting the second row of {A_n} from the first row of {A_n},then applying cofactor expansion with regard to the first row of {A_{n}},we get a recursive relation:

\displaystyle \det A_n=\det A_{n-1}+\det B_{n-1}.

And from the cofactor expansion we can find that

\displaystyle \det B_n=\det A_n-\det A_{n-1}.

So

\displaystyle \det A_n=2\det A_{n-1}-\det A_{n-2},

and {\det A_1=2,\det A_2=3}.So {\det A_n=n+1}. \Box

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