# 三月 2018

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## 坐标法证明欧拉线定理

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

• 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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## 一个特殊行列式的两种计算方法

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

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

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

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

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

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

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

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

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

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## Linear Algebra and Its Applications,Solutions to Problem Set 5.1

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 ${1,-1,e^{\frac{2\pi}{3}},e^{\frac{\pi}{3}}}$. $\Box$

## Linear Algebra and Its Applications,Chapter 4,Review Exercises

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 ${\det A_n=n+1}$. $\Box$