Linear Algebra and Its Applications_Strang

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几个离散动力系统的应用题

(练习5.3.3)Bernadelli研究这一类甲虫，“它只能活3年，而且在第3年繁殖”.如果年龄一岁的组以$\frac{1}{2}$的概率存活，两岁的组以$\frac{1}{3}$概率存活，而$3$岁的甲虫生6个雌虫后死去，相应的矩阵为
$$A= \begin{bmatrix} 0&0&6\\ \frac{1}{2}&0&0\\ 0&\frac{1}{3}&0 \end{bmatrix}.$$

$$\begin{vmatrix} -\lambda&0&6\\ \frac{1}{2}&-\lambda&0\\ 0&\frac{1}{3}&-\lambda \end{vmatrix}=0,$$

(练习5.3.4)假设有三个大型运货卡车中心.每个月中,在波士顿和在洛杉矶的卡车的一半开往芝加哥，而其余一半留在原地.而在芝加哥的卡车分成相等的两半分别去波士顿和洛杉矶.给出$3\times 3$转移矩阵并求相应于特征值$1$的稳定状态.

$$\begin{bmatrix} x_{n+1}\\ y_{n+1}\\ z_{n+1} \end{bmatrix}= \begin{bmatrix} \frac{1}{2}&0&\frac{1}{2}\\ 0&\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}&0 \end{bmatrix} \begin{bmatrix} x_n\\ y_n\\ z_n \end{bmatrix}$$

$$A=\begin{bmatrix} \frac{1}{2}&0&\frac{1}{2}\\ 0&\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}&0 \end{bmatrix},$$

(练习5.3.5)假定有一种传染病，在每个月内，健康人的一半会染上病，而病人的$\frac{1}{4}$会死亡.求相应Markov过程的稳定状态.
$$\begin{bmatrix} d_{k+1}\\ s_{k+1}\\ w_{k+1} \end{bmatrix}= \begin{bmatrix} 1&\frac{1}{4}&0\\ 0&\frac{3}{4}&\frac{1}{2}\\ 0&0&\frac{1}{2} \end{bmatrix} \begin{bmatrix} d_k\\ s_k\\ w_k \end{bmatrix}.$$

(练习5.3.8)对方程组$V_{n+1}=\alpha(V_n+W_n)$和$W_{n+1}=\alpha(V_n+W_n)$,什么样的$\alpha$值产生不稳定性？

$$\begin{pmatrix} V_{n+1}\\ W_{n+1} \end{pmatrix}= \begin{pmatrix} \alpha&\alpha\\ \alpha&\alpha \end{pmatrix} \begin{pmatrix} V_n\\ W_n \end{pmatrix}.$$

$$A= \begin{pmatrix} \alpha&\alpha\\ \alpha&\alpha \end{pmatrix}$$

推导正规方程

1.用几何法推导正规方程

Gilbert Strang的《线性代数及其应用》(第二版,侯自新等翻译)第119页推导了关于多变元最小二乘法的正规方程.在此，我使用自己更加能接受的方式重新推导正规方程.

$$\mathbf{b}’=x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n},$$

$$\begin{cases} [\mathbf{b}-(x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n})]\cdot \mathbf{v}_{1}=0\\ [\mathbf{b}-(x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n}]\cdot \mathbf{v}_{2}=0\\ \vdots\\ [\mathbf{b}-(x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n}]\cdot \mathbf{v}_{n}=0. \end{cases}$$

$$\begin{cases} (x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n})\cdot \mathbf{v}_{1}=\mathbf{b}\cdot \mathbf{v}_{1}\\ (x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n})\cdot \mathbf{v}_{2}=\mathbf{b}\cdot \mathbf{v}_{2}\\ \vdots\\ (x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n})\cdot \mathbf{v}_{n}=\mathbf{b}\cdot \mathbf{v}_{n}\\ \end{cases}.$$

$$\begin{pmatrix} \mathbf{v}_{1}\cdot\mathbf{v}_{1}&\mathbf{v}_{2}\cdot \mathbf{v}_{1}&\cdots&\mathbf{v}_{n}\cdot\mathbf{v}_{1}\\ \mathbf{v}_{1}\cdot \mathbf{v}_{2}&\mathbf{v}_{2}\cdot \mathbf{v}_{2}&\cdots& \mathbf{v}_{n}\cdot \mathbf{v}_{2}\\ \vdots&\vdots&\cdots&\vdots\\ \mathbf{v}_{1}\cdot \mathbf{v}_{n}&\mathbf{v}_{2}\cdot \mathbf{v}_{n}&\cdots& \mathbf{v}_{n}\cdots \mathbf{v}_{n} \end{pmatrix} \begin{pmatrix} x_{1}\\ x_{2}\\ \vdots\\ x_{n} \end{pmatrix}= \begin{pmatrix} \mathbf{b}\cdot \mathbf{v}_{1}\\ \mathbf{b}\cdot \mathbf{v}_{2}\\ \vdots\\ \mathbf{b}\cdot \mathbf{v}_{n} \end{pmatrix}.$$

$$A^{T}Ax=A^{T}b,$$

$$A=\begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\cdots&\vdots\\ a_{mn}&a_{m2}&\cdots&a_{mn} \end{pmatrix},x= \begin{pmatrix} x_{1}\\ x_{2}\\ \vdots\\ x_{n} \end{pmatrix}.$$

2.用微分法推导正规方程

\begin{align*}
y&=||Ax-b||^{2}
\\&=(Ax-b)^{T}(Ax-b)
\\&=((Ax)^{T}-b^{T})(Ax-b)
\\&=(Ax)^TAx-(Ax)^Tb-b^TAx+b^Tb.
\end{align*}

$$dy=dy_1-dy_2-dy_3+dy_4,$$

\begin{align*}
dy_1&=[A(x+dx)]^TA(x+dx)-(Ax)^TAx
\\&=(x+dx)^TA^TA(x+dx)-x^TA^TAx
\\&=[x^{T}+(dx)^{T}]A^TA(x+dx)-x^TA^TAx
\end{align*}
\begin{align*}
\\&=(x^{T}A^{T}A-b^{T}A)dx+(dx)^T(A^{T}Ax-A^{T}b)
\\&=M^{T}+M,
\end{align*}

$$M^{T}+M=0$$

\label{eq:1}
A^TAx=A^Tb

矩阵AB=BA的充要条件

Gilbert Strang著Introduction to Linear Algebra第5版第296页有一个加框的结论:

$A$ and $B$ share the same $n$ independent eigenvectors if and only if $AB=BA$.

Hi professor,

I enjoy your linear algebra book.But on page 296 of the fifth edition of Introduction to linear algebra,there is a boxed conclusion:

$A$ and $B$ share the same $n$ independent eigenvectors if and only if $AB=BA$.

I think by default this conclusion means:

(1)If $A$ and $b$ share the same n indepent eigenvectors,then $AB=BA$.

(2)If $AB=BA$,then $A$ and $B$ share the same $n$ independent eigenvectors.

Obviously (2) is not correct.

In the forth edition of the same book(Page 305),the conclusion is:

Suppose both $A$ and $B$ can be diagonalized,they share the same eigenvector matrix $S$ if and only if $AB=BA$.

In which the prerequisite “$A$ and $B$ can be diagonalized” is added,whichI think is better.

Suppose both $A$ and $B$ can be diagonalized,they share the same eigenvector matrix $S$ if and only if $AB=BA$.

$$AB=SD_1D_2S^{-1}=SD_2D_1S^{-1}=BA.$$

$$(AB)\mathbf{x}=A(B\mathbf{x})=A(\lambda \mathbf{x})=\lambda A\mathbf{x}=(BA)\mathbf{x}=B(A\mathbf{x}),$$

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

Linear Algebra and Its Applications,Solutions to Problem Set 4.4

Exercise 4.4.1 Find the determinant and all nine cofactors ${C_{ij}}$ of this triangular matrix:

$\displaystyle A= \begin{bmatrix} 1&2&3\\ 0&4&0\\ 0&0&5 \end{bmatrix}.$

Form ${C^T}$ and verify that ${AC^T=(\det A)I}$.What is ${A^{-1}}$?

Solution:

$\displaystyle C= \begin{bmatrix} 20&0&0\\ -10&5&0\\ -12&0&4 \end{bmatrix},C^T= \begin{bmatrix} 20&-10&-12\\ 0&5&0\\ 0&0&4 \end{bmatrix}.$

$\displaystyle AC^T= \begin{bmatrix} 20&0&0\\ 0&20&0\\ 0&0&20 \end{bmatrix}=(\det A)I.$

$\displaystyle A^{-1}=\frac{C^T}{\det A}= \begin{bmatrix} 1&-\frac{1}{2}&-\frac{3}{5}\\ 0&\frac{1}{4}&0\\ 0&0&\frac{1}{5} \end{bmatrix}.$

$\Box$

Exercise 4.4.2 Use the cofactor matrix ${C}$ to invert these symmetric matrices:

$\displaystyle A= \begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{bmatrix}~~~~~and~~~~~B= \begin{bmatrix} 1&1&1\\ 1&2&2\\ 1&2&3 \end{bmatrix}.$

Solution:

$\displaystyle A^{-1}=\frac{C_{A}^T}{\det A}=\frac{ \begin{bmatrix} 3&2&1\\ 2&4&2\\ 1&2&3 \end{bmatrix} }{4}= \begin{bmatrix} \frac{3}{4}&\frac{1}{2}&\frac{1}{4}\\ \frac{1}{2}&1&\frac{1}{2}\\ \frac{1}{4}&\frac{1}{2}&\frac{3}{4} \end{bmatrix}.$

$\displaystyle B^{-1}=\frac{C_B^T}{\det B}=\frac{ \begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&1 \end{bmatrix} }{1}= \begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&1 \end{bmatrix}.$

$\Box$

Exercise 4.4.3 Find ${x,y}$,and ${z}$ by Cramer’s Rule in equation (4):

$\displaystyle \begin{cases} ax+by=1\\ cx+dy=0 \end{cases}~~~~~and~~~~~ \begin{cases} x+4y-z=1\\ x+y+z=0\\ 2x+3z=0 \end{cases}.$

Solution:

• ${x=\frac{ \begin{vmatrix} 1&b\\ 0&d \end{vmatrix} }{ \begin{vmatrix} a&b\\ c&d \end{vmatrix} }=\frac{d}{ad-bc}}$,${y=\frac{ \begin{vmatrix} a&1\\ c&0 \end{vmatrix} }{ \begin{vmatrix} a&b\\ c&d \end{vmatrix} }=\frac{-c }{ad-bc}}$.
• $\displaystyle \begin{array}{rcl} x&=&\frac{ \begin{vmatrix} 1&4&-1\\ 0&1&1\\ 0&0&3 \end{vmatrix} }{ \begin{vmatrix} 1&4&-1\\ 1&1&1\\ 2&0&3 \end{vmatrix} }= \frac{3}{1}=3.y=\frac{ \begin{vmatrix} 1&1&-1\\ 1&0&1\\ 2&0&3 \end{vmatrix} }{ \begin{vmatrix} 1&4&-1\\ 1&1&1\\ 2&0&3 \end{vmatrix} }=\frac{-1}{1}=-1.\\z&=&\frac{ \begin{vmatrix} 1&4&1\\ 1&1&0\\ 2&0&0 \end{vmatrix} }{ \begin{vmatrix} 1&4&-1\\ 1&1&1\\ 2&0&3 \end{vmatrix} }=\frac{-2}{1}=-2. \end{array}$

$\Box$

Exercise 4.4.4

• Find the determinant when a vector ${x}$ replaces column ${j}$ of the identity(consider ${x_j=0}$ as a separate case):

$\displaystyle if~~~~~M= \begin{bmatrix} 1& &x_1& & \\ &1&\cdot& &\\ & &x_j& & &\\ & &\cdot&1&\\ & &x_n& &1 \end{bmatrix}~~~~~then~~~~~\det M=\underline{~~~~~~~~~~}.$

• If ${Ax=b}$,show that ${AM}$ is the matrix ${B_j}$ in equation (4),with ${b}$ in column ${j}$.
• Derive Cramer’s rule by taking determinants in ${AM=B_j}$.

Solution:

• ${x_j}$.This is a direct consequence of the big formula.Or this can be deduced by the elimination of rows of determinants.
• This is a direct consequence of matrix multiplication.
• ${AM=B_j}$,so ${\det A\det M=\det B_j}$,so ${x_j=\det M=\frac{\det B_j}{\det A}}$.

$\Box$

Remark  This exercise provides a geometrical explaination to Cramer’s Rule.

Exercise 4.4.5

• Draw the triangle with vertices ${A=(2,2)}$,${B=(-1,3)}$,and ${C=(0,0)}$.By regarding it as half of a parallelogram,explain why its area equals

$\displaystyle area(ABC)=\frac{1}{2}\det \begin{bmatrix} 2&2\\ -1&3 \end{bmatrix}.$

• Move the third vertex to ${C=(1,-4)}$ and justify the formula

$\displaystyle area(ABC)=\frac{1}{2}\det \begin{bmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{bmatrix}=\frac{1}{2}\det \begin{bmatrix} 2&2&1\\ -1&3&1\\ 1&-4&1 \end{bmatrix}.$

Solution:

• $\displaystyle \frac{1}{2}\det \begin{bmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{bmatrix}=\frac{1}{2}\det \begin{bmatrix} x_1&y_1&1\\ x_2-x_1&y_2-y_1&0\\ x_3-x_2&y_3-y_2&0 \end{bmatrix}=\frac{1}{2}\det \begin{bmatrix} x_2-x_1&y_2-y_1\\ x_3-x_2&y_3-y_2 \end{bmatrix}.$

$\Box$

Remark:The second result has a solid geometric meaning.

Exercise 4.4.7 Predict in advance,and confirm by elimination,the pivot entries of

$\displaystyle A= \begin{bmatrix} 2&1&2\\ 4&5&0\\ 2&7&0 \end{bmatrix}~~~and~~~B= \begin{bmatrix} 2&1&2\\ 4&5&3\\ 2&7&0 \end{bmatrix}.$

Solution:

• ${2,3,6}$.
• ${2,3,0}$.

$\Box$

Exercise 4.4.8 Find all the odd permutations of the numbers ${\{1,2,3,4\}}$.They come from an odd number of exchanges and lead to ${\det P=-1}$.

Solution: They are

$\displaystyle (1,2,4,3),(1,3,2,4),(1,4,3,2),$

$\displaystyle (2,1,3,4),(2,3,4,1),(2,4,1,3),$

$\displaystyle (3,1,4,2),(3,2,1,4),(3,4,2,1),$

$\displaystyle (4,1,2,3),(4,2,3,1),(4,3,1,2).$

$\Box$

Exercise 4.4.9 Suppose the permutation ${P}$ takes ${(1,2,3,4,5)}$ to ${(5,4,1,2,3)}$.

• What does ${P^2}$ do to ${(1,2,3,4,5)}$?
• What does ${P^{-1}}$ do to ${(1,2,3,4,5)}$?

Solution:

• The matrix of the permutation ${P}$ is

$\displaystyle P=\begin{bmatrix} 0&0&0&0&1\\ 0&0&0&1&0\\ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0 \end{bmatrix}$

So

$\displaystyle P^2= \begin{bmatrix} 0&0&1&0&0\\ 0&1&0&0&0\\ 0&0&0&0&1\\ 0&0&0&1&0\\ 1&0&0&0&0 \end{bmatrix}.$

So ${P^2}$ takes ${(1,2,3,4,5)}$ to ${(3,2,5,4,1)}$.

• The matrix of the permutation ${P^{-1}}$ is

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

So ${P^{-1}}$ takes ${(1,2,3,4,5)}$ to ${(3,4,5,2,1)}$.

$\Box$

Exercise 4.4.11 Prove that if you keep multiplying ${A}$ by the same permutation matrix ${P}$,the first row eventually comes back to its original place.

Solution: For an ${n}$-tuple ${(1,2,\cdots,n)}$,suppose that ${P(1)=i_1}$,${P(i_1)=i_2}$,${P(i_2)=i_3,\cdots}$,where ${i_k(k\in \mathbf{N}^{+})\in \{1,2,\cdots,n\}}$,and ${\forall m\neq n}$,${i_m\neq i_n}$.So there must exists ${l\in \mathbf{N}^{+}}$,such that ${P(i_{l})=1}$.Then ${P^{l+1}(1)=1}$. $\Box$

Exercise 4.4.12 If ${A}$ is a ${5}$ by ${5}$ matrix with all ${|a_{ij}|\leq 1}$,then ${\det A\leq \underline{~~~~~}}$.Volumns or the big formula or pivots should give some upper bound on the determinant.

Solution: Consider the volume.One answer is ${(\sqrt{5})^5=25 \sqrt{5}}$,according to Hadamard’s inequality. $\Box$

Exercise 4.4.13 Solve these linear equations by Cramer’s Rule ${x_{j}=\frac{\det B_j}{\det A}}$:

• $\displaystyle \begin{array}{rcl} 2x_1+5x_2&=&1\\ x_1+4x_2&=&2 \end{array}$

• $\displaystyle \begin{array}{rcl} 2x_1+x_2&=&1\\ x_1+2x_2+x_3&=&70\\ x_2+2x_3&=&0 \end{array}$

Solution:

• $\displaystyle x_1= \frac{ \begin{vmatrix} 1&5\\ 2&4 \end{vmatrix} }{ \begin{vmatrix} 2&5\\ 1&4 \end{vmatrix} }=\frac{-6}{3}=-2$

• $\displaystyle x_1=\frac{ \begin{vmatrix} 1&1&0\\ 70&2&1\\ 0&1&2 \end{vmatrix} }{ \begin{vmatrix} 2&1&0\\ 1&2&1\\ 0&1&2 \end{vmatrix} }=\frac{-137}{4},x_2=\frac{ \begin{vmatrix} 2&1&0\\ 1&70&1\\ 0&0&2 \end{vmatrix} }{ \begin{vmatrix} 2&1&0\\ 1&2&1\\ 0&1&2 \end{vmatrix} }=\frac{139}{2},x_3=\frac{ \begin{vmatrix} 2&1&1\\ 1&2&70\\ 0&1&0 \end{vmatrix} }{ \begin{vmatrix} 2&1&0\\ 1&2&1\\ 0&1&2 \end{vmatrix} }=\frac{-139}{4}.$

$\Box$

Exercise 4.4.14 Use Cramer’s Rule to solve for ${y}$(only).Call the ${3}$ by ${3}$ determinant ${D}$:

• $\displaystyle \begin{array}{rcl} ax+by&=&1\\ cx+dy&=&0 \end{array}$

• $\displaystyle \begin{array}{rcl} ax+by+cz&=&1\\ dx+ey-fz&=&0\\ gx+hy+iz&=&0 \end{array}$

Solution:

• $\displaystyle x=\frac{ \begin{vmatrix} 1&b\\ 0&d \end{vmatrix} }{ \begin{vmatrix} a&b\\ c&d \end{vmatrix} }=\frac{d}{ad-bc},y=\frac{ \begin{vmatrix} a&1\\ c&0 \end{vmatrix} }{ \begin{vmatrix} a&b\\ c&d \end{vmatrix} }=\frac{-c }{ad-bc}.$

• $\displaystyle x=\frac{ \begin{vmatrix} 1&b&c\\ 0&e&-f\\ 0&h&i \end{vmatrix} }{\det D}=\frac{ei+hf}{\det D},y=\frac{ \begin{vmatrix} a&1&c\\ d&0&-f\\ g&0&i \end{vmatrix} }{\det D}=\frac{-di-gf}{\det D},z=\frac{ \begin{vmatrix} a&b&1\\ d&e&0\\ g&h&0 \end{vmatrix} }{\det D}=\frac{fh-eg}{\det D}.$

$\Box$

Exercise 4.4.18 Find ${A^{-1}}$ from the cofactor formula ${C^T/\det A}$.Use symmetry in part (b):

$\displaystyle (a)~~~~~A= \begin{bmatrix} 1&2&0\\ 0&3&0\\ 0&4&1 \end{bmatrix},~~~~~(b)~~~~~A= \begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{bmatrix}.$

Solution:

• $\displaystyle A^{-1}=\frac{C^T}{\det A}=\frac{ \begin{bmatrix} 3&-2&0\\ 0&1&0\\ 0&-4&3 \end{bmatrix} }{3}= \begin{bmatrix} 1&-\frac{2}{3}&0\\ 0&\frac{1}{3}&0\\ 0&-\frac{4}{3}&1 \end{bmatrix}.$

• $\displaystyle A^{-1}=\frac{C^T}{\det A}=\frac{ \begin{bmatrix} 3&2&1\\ 2&4&2\\ 1&2&3 \end{bmatrix} }{4}= \begin{bmatrix} \frac{3}{4}&\frac{1}{2}&\frac{1}{4}\\ \frac{1}{2}&1&\frac{1}{2}\\ \frac{1}{4}&\frac{1}{2}&\frac{3}{4} \end{bmatrix}.$

$\Box$

Exercise 4.4.19 If all the cofactors are zero,how do you know that ${A}$ has no inverse?If none of the cofactors are zero,is ${A}$ sure to be invertible.

Solution: If all the cofactors are zero,then ${\det A=0}$,which means ${A}$ has no inverse.If none of the cofactors are zero,${A}$ is not sure to be invertible.An example:${A= \begin{bmatrix} 1&1\\ 1&1 \end{bmatrix}. }$ $\Box$

Exercise 4.4.20 Find the cofactors of ${A}$ and multiply ${AC^T}$ to find ${\det A}$:

$\displaystyle A= \begin{bmatrix} 1&1&4\\ 1&2&2\\ 1&2&5 \end{bmatrix},C= \begin{bmatrix} 6&-3&0\\ \cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot\\ \end{bmatrix},and~AC^T=\underline{~~~~~~}.$

If you change that corner entry from ${4}$ to ${100}$,why is ${\det A}$ unchanged?

Solution:

$\displaystyle C^T= \begin{bmatrix} 6&\cdot&\cdot\\ -3&\cdot&\cdot\\ 0&\cdot&\cdot \end{bmatrix}$

$\displaystyle AC^T= \begin{bmatrix} 3&0&0\\ 0&3&0\\ 0&0&3 \end{bmatrix},$

so ${\det A=3}$.If the corner entry is changed from ${4}$ to ${100}$,${\det A}$ is unchanged,because the corresponding cofactor is always ${0}$. $\Box$

Exercise 4.4.21 Suppose ${\det A=1}$ and you know all the cofactors.How can you find ${A}$?

Solution: When all the cofactors are known,${C^T}$ can be determined.Then from ${A^{-1}=\frac{C^T}{\det A}=C^T}$,we can know ${A^{-1}}$.Then ${A}$ is known by finding the inverse of ${A^{-1}}$. $\Box$

Exercise 4.4.22 From the formula ${AC^T=(\det A)I}$ show that ${\det C=(\det A)^{n-1}}$.

Solution: ${AC^T=(\det A)I\Rightarrow \det A\det C^T=\det ((\det A) I))=(\det A)^n}$.So ${\det C=\det C^T=(\det A)^{n-1}}$. $\Box$

Exercise 4.4.23 If you know all ${16}$ cofactors of a ${4}$ by ${4}$ invertible matrix ${A}$,how would you find ${A}$?

Solution: ${C^{T}}$ is known.From the previous exercise,${\det C^T=(\det A)^{n-1}}$,so ${\det A}$ is known.Then from ${A^{-1}=\frac{C^T}{\det A}}$,we can find ${A^{-1}}$.Then ${A}$ is known by finding the inverse of ${A^{-1}}$. $\Box$

Exercise 4.4.24 If all entries of ${A}$ are integers,and ${\det A=\pm 1}$,prove that all entries of ${A^{-1}}$ are integers.Give a ${2}$ by ${2}$ example.

Solution: ${A^{-1}=\frac{C^T}{\det A}=\pm C^T}$,so all entries of ${A^{-1}}$ are integers.Example:${A= \begin{bmatrix} 1&n\\ 0&1 \end{bmatrix} }$,${A^{-1}= \begin{bmatrix} 1&-n\\ 0&1 \end{bmatrix} }$. $\Box$

Exercise 4.4.25 ${L}$ is lower triangular and ${S}$ is symmetric.Assume they are invertible:

$\displaystyle L= \begin{bmatrix} a&0&0\\ b&c&0\\ d&e&f \end{bmatrix},S= \begin{bmatrix} a&b&d\\ b&c&e\\ d&e&f \end{bmatrix}.$

• Which three cofactors of ${L}$ are zero?Then ${L^{-1}}$ is lower triangular.
• Which three pairs of cofactors of ${S}$ are equal?Then ${S^{-1}}$ is symmetric.

Solution:

• ${b,d,e}$.
• ${b,d,e}$.

$\Box$

Exercise 4.4.26 For ${n=5}$ the matrix ${C}$ contains ${\underline{25}}$ cofactors and each ${4}$ by ${4}$ cofactor contains ${\underline{4!=24}}$ terms and each term needs ${\underline{3}}$ multiplications.Compare with ${5^3=125}$ for the Gauss-Jordan computation of ${A^{-1}}$.

Exercise 4.4.27

• Find the area of the parallelogram with edges ${v=(3,2)}$ and ${w=(1,4)}$.
• Find the area of the triangle with sides ${v,w,}$and ${v+w}$.Draw it.
• Find the area of the triangle with sides ${v,w,}$and ${w-v}$.Draw it.

Solution:

• $\displaystyle \begin{vmatrix} 3&2\\ 1&4 \end{vmatrix}=10.$

• ${\frac{1}{2} \begin{vmatrix} 3&2\\ 1&4 \end{vmatrix}=5. }$Picture omitted.
• ${\frac{1}{2} \begin{vmatrix} 3&2\\ 1&4 \end{vmatrix}=5. }$Picture Omitted.

$\Box$

Exercise 4.4.28 A box has edges from ${(0,0,0)}$ to ${(3,1,1),(1,3,1),}$and ${(1,1,3)}$.Find its volumn and also find the area of each parallelogram face.

Solution: The volumn of the box is

$\displaystyle \begin{vmatrix} 3&1&1\\ 1&3&1\\ 1&1&3 \end{vmatrix}=20.$

The area of each parallelogram face is

$\displaystyle \sqrt{ \begin{vmatrix} 3&1\\ 1&3 \end{vmatrix}^2+ \begin{vmatrix} 3&1\\ 1&1 \end{vmatrix}^2+ \begin{vmatrix} 1&1\\ 3&1 \end{vmatrix}^2 }=6 \sqrt{2},$

$\displaystyle \sqrt{ \begin{vmatrix} 1&3\\ 1&1 \end{vmatrix}^2+ \begin{vmatrix} 3&1\\ 1&3 \end{vmatrix}^2+ \begin{vmatrix} 1&1\\ 1&3 \end{vmatrix}^2 }=6 \sqrt{2},$

$\displaystyle \sqrt{ \begin{vmatrix} 3&1\\ 1&1 \end{vmatrix}^2+ \begin{vmatrix} 3&1\\ 1&3 \end{vmatrix}^2+ \begin{vmatrix} 1&1\\ 1&3 \end{vmatrix}^2 }=6 \sqrt{2}.$

In order to compute the area of each parallelogram face,we can also compute

$\displaystyle \sqrt{\det A^TA}=\sqrt{\det \left(\begin{bmatrix} 3&1&1\\ 1&3&1 \end{bmatrix} \begin{bmatrix} 3&1\\ 1&3\\ 1&1 \end{bmatrix}\right)}= 6 \sqrt{2}$

etc.$\Box$

Exercise 4.4.35 An ${n}$– dimensional cube has how many corners?How many edges?How many ${n-1}$-dimensional faces?The ${n}$– cube whose edges are the rows of ${2I}$ has volumn ${\underline{~~~~~~~~~~}}$.A hypercube computer has parallel processors at the corners with connections along the edeges.

Solution: An ${n}$-dimensional cube has ${2^n}$ corners.Now we count the number of edges of an ${n}$– dimensional cube.Suppose an ${n}$-dimensional cube has ${f(n)}$ edeges,then ${f(n+1)=2f(n)+2^n}$,and ${f(1)=1}$.So ${f(n)=n2^{n-1}}$.

The ${n}$-cube whose edges are the rows of ${2I}$ has volumn ${2^n}$. $\Box$

Exercise 4.4.36 The triangle with corners ${(0,0),(1,0),(0,1)}$ has area ${\frac{1}{2}}$.The pyramid with four corners ${(0,0,0),(0,1,0),(0,0,1)}$ has volumn ${\underline{~~~~~~}}$ .The pyramid in ${\mathbf{R}^4}$ with five corners at ${(0,0,0,0)}$ and the rows of ${I}$ has what volumn?

Solution:

• $\displaystyle \frac{1}{6} \begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{vmatrix}=\frac{1}{6}.$

• According to this post,the answer is ${\frac{1}{4!}=\frac{1}{24}}$.

$\Box$

Exercise 4.4.37 Polar coordinates satisfy ${x=r\cos\theta}$ and ${y=r\sin\theta}$.Polar area ${J drd\theta}$ includes ${J}$:

$\displaystyle J= \begin{vmatrix} \frac{\partial x}{\partial r}&\frac{\partial x}{\partial\theta}\\ \frac{\partial y}{\partial r}&\frac{\partial y}{\partial\theta} \end{vmatrix}= \begin{vmatrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta \end{vmatrix}.$

The two columns are orthogonal.Their lengths are ${\underline{1,r}}$.Thus ${J=\underline{r}}$.

Exercise 4.4.38 Spherical coordinates ${\rho,\phi,\theta}$ give ${x=\rho \sin\phi\cos\theta}$,${y=\rho\sin\phi\sin\theta}$,${z=\rho\cos\phi}$.Find the Jacobian matrix of ${9}$ partial derivatives:${\frac{\partial x}{\partial\rho}}$,${\frac{\partial x}{\partial \phi}}$,${\frac{\partial x}{\partial\theta}}$ are in row ${1}$.Simplify its determinant to ${J=\rho^2\sin\phi}$.Then ${dV=\rho^2\sin\phi d\rho d\phi d\theta}$.

Solution: The Jacobian matrix is

$\displaystyle \begin{bmatrix} \frac{\partial x}{\partial \rho}&\frac{\partial x}{\partial \phi}&\frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial \rho}&\frac{\partial y}{\partial \phi}&\frac{\partial y}{\partial \theta}\\ \frac{\partial z}{\partial \rho}&\frac{\partial z}{\partial \phi}&\frac{\partial z}{\partial \theta} \end{bmatrix}= \begin{bmatrix} \sin\phi\cos\theta&\rho\cos\phi\cos\theta&-\rho\sin\phi\sin\theta\\ \sin\phi\sin\theta&\rho\cos\phi\sin\theta&\rho\sin\phi\cos\theta\\ \cos\phi&-\rho\sin\phi&0 \end{bmatrix}$

It is an orthogonal matrix,and it has positive sign(According to geometric meaning).So ${J=\rho^2\sin\phi}$. $\Box$

Exercise 4.4.39 The matrix that connects ${r,\theta}$ to ${x,y}$ is in Problem 37.Invert that matrix:

$\displaystyle J^{-1}= \begin{vmatrix} \frac{\partial r}{\partial x}&\frac{\partial r}{\partial y}\\ \frac{\partial \theta}{\partial x}&\frac{\partial \theta}{\partial y} \end{vmatrix}= \begin{vmatrix} \cos\theta&\sin\theta\\ -\frac{\sin\theta}{r}&\frac{\cos\theta}{r} \end{vmatrix}.$

It is surprising that ${\frac{\partial r}{\partial x}=\frac{\partial x}{\partial r}}$.The product ${JJ^{-1}=I}$ gives the chain rule

$\displaystyle \frac{\partial x}{\partial x}=\frac{\partial x}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial x}{\partial \theta}\frac{\partial\theta}{\partial x}=1.$

Exercise 4.4.41 Let ${P=(1,0,-1)}$,${Q=(1,1,1)}$,and ${R=(2,2,1)}$.Choose ${S}$ so that ${PQRS}$ is a parallelogram,and compute its area.Choose ${T,U,V}$ so that ${OPQRSTUV}$ is a tilted box,and compute its volumn.

Solution: Let ${S=(x,y,z)}$,then

$\displaystyle \overrightarrow{PS}=\overrightarrow{QR},$

i.e.,

$\displaystyle (x,y,z)-(1,0,-1)=(2,2,1)-(1,1,1)\iff (x-1,y,z+1)=(1,1,0).$

so ${S=(x,y,z)=(2,1,-1)}$.${\overrightarrow{PQ}=(0,1,2)}$,${\overrightarrow{PS}=(1,1,0)}$.Let matrix ${A= \begin{bmatrix} 0&1&2\\ 1&1&0 \end{bmatrix}, }$then the area of the parallelogram is ${\sqrt{\det AA^T}=3}$.

In order that ${OPQRSTUV}$ be a tilted box,let ${T=(0,1,2)}$,${U(1,2,2)}$,${V(1,1,0)}$.Then

$\displaystyle \overrightarrow{OP}=(1,0,-1),\overrightarrow{OT}=(0,1,2),\overrightarrow{OV}=(1,1,0).$

So the volumn of the tilted box is

$\displaystyle \left|\begin{vmatrix} 1&0&-1\\ 0&1&2\\ 1&1&0 \end{vmatrix}\right|=1.$

$\Box$

Exercise 4.4.42 Suppose ${(x,y,z),(1,1,0)}$,and ${(1,2,1)}$ lie on a plane through the origin.What determinant is zero?What equation does this give for the plane?

Solution: The determinant

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

So the equation of the plane is ${x-y+z=0}$. $\Box$

Exercise 4.4.43 Suppose ${(x,y,z)}$ is a linear combination of ${(2,3,1)}$ and ${(1,2,3)}$.What determinant is zero?What equation does this give for the plane of all combinations?

Solution: The determinant

$\displaystyle \begin{vmatrix} x&y&z\\ 2&3&1\\ 1&2&3 \end{vmatrix}=0.$

So the equation of the plane is ${7x-4y+z=0}$. $\Box$

Exercise 4.4.44 If ${Ax=(1,0,\cdots,0)}$ show how Cramer’s Rule gives ${x=}$ first column of ${A^{-1}}$.

Solution: We know that ${AA^{-1}=I}$,the first column of ${I}$ is ${ \begin{bmatrix} 1\\ 0\\ \vdots\\ 0 \end{bmatrix}. }$According to the rule of matrix multiplication,${x}$ is the first column of ${A^{-1}}$.

We can also use Cramer’s Rule.Replace the ${i}$th column of matrix ${A}$ by ${ \begin{bmatrix} 1\\ 0\\ \vdots\\ 0 \end{bmatrix} }$,we get matrix ${B_i}$.According to Cramer’s Rule,

$\displaystyle x= \begin{bmatrix} \frac{\det B_1}{\det A}\\ \frac{\det B_2}{\det A}\\ \vdots\\ \frac{\det B_n}{\det A} \end{bmatrix}= \begin{bmatrix} \frac{C_{11}}{\det A}\\ \frac{C_{12}}{\det A}\\ \vdots\\ \frac{C_{1n}}{\det A} \end{bmatrix},$

where ${C_{1j}}$ is a cofactor of matrix ${A}$.According to the formula ${A^{-1}=\frac{C^T}{\det A}}$,${x}$ is the first column of ${A^{-1}}$. $\Box$

Linear Algebra and Its Applications,Solutions to Problem Set 4.3

Exercise 4.3.22 Prove that ${4}$ is the largest determinant for a ${3}$ by ${3}$ matrix of ${1}$s and ${-1}$s.

Solution: Let matrix ${A= \begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix} }$,and every element of ${A}$ is either ${1}$ or ${-1}$.The cofactor expansion of ${\det A}$ is

$\displaystyle \det A=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13},$

Every element is either ${1}$ or ${-1}$,so ${C_{11},C_{12},C_{13}\in \{-2,0,2\}}$.So ${\det A\in \{-6,-4,-2,0,2,4,6\}}$.

Now we prove that ${\det A\neq 6}$.Otherwise,if ${\det A=6}$,then ${|C_{11}|=|C_{12}|=|C_{13}|=2}$.So determinants

$\displaystyle \begin{vmatrix} a_{21}&a_{22}\\ a_{31}&a_{32} \end{vmatrix}, \begin{vmatrix} a_{21}&a_{23}\\ a_{31}&a_{33} \end{vmatrix}$

are equal numbers or opposite numbers.

• 1. When ${ \begin{vmatrix} a_{21}&a_{22}\\ a_{31}&a_{32} \end{vmatrix}= \begin{vmatrix} a_{21}&a_{23}\\ a_{31}&a_{33} \end{vmatrix} }$,by the linearity of the determinant,${ \begin{vmatrix} a_{21}&a_{22}-a_{23}\\ a_{31}&a_{32}-a_{33}\end{vmatrix}=0 }$.So ${|a_{22}-a_{23}|=|a_{32}-a_{33}|}$.Both ${|a_{22}-a_{23}|}$ and ${|a_{32}-a_{33}|}$ are even numbers ${0}$ or ${2}$.
• 1.1 When ${|a_{22}-a_{23}|=|a_{32}-a_{33}|= 2}$,the sign of ${a_{22},a_{23}}$ must be opposite,and the sign of ${a_{32},a_{33}}$ are also opposite.In such case,${ \begin{bmatrix} a_{22}&a_{23} \end{bmatrix}, \begin{bmatrix} a_{32}&a_{33} \end{bmatrix} }$are linearly dependent vectors,which means that ${ \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}=0 }$,contradiction.
• 1.2 When ${|a_{22}-a_{23}|=|a_{32}-a_{33}|=0}$,${a_{22}=a_{23}}$ and ${a_{32}=a_{33}}$.In such case,${ \begin{bmatrix} a_{22}&a_{23} \end{bmatrix}, \begin{bmatrix} a_{32}&a_{33} \end{bmatrix} }$are linearly dependent vectors,which means that ${ \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}=0 }$,contradiction.
• 2. When ${ \begin{vmatrix} a_{21}&a_{22}\\ a_{31}&a_{32} \end{vmatrix}= -\begin{vmatrix} a_{21}&a_{23}\\ a_{31}&a_{33} \end{vmatrix} }$,by the linearity of the determinant,${ \begin{vmatrix} a_{21}&a_{22}+a_{23}\\ a_{31}&a_{32}+a_{33} \end{vmatrix}=0 }$.So ${|a_{22}+a_{23}|=|a_{32}+a_{33}|}$.Both ${|a_{22}+a_{23}|}$ and ${|a_{32}+a_{33}|}$ are even numbers ${0}$ or ${2}$.
• 2.1 When ${|a_{22}+a_{23}|=|a_{32}+a_{33}|=2}$,${a_{22}=a_{23}}$ and ${a_{32}=a_{33}}$.In such case,${ \begin{bmatrix} a_{22}&a_{23} \end{bmatrix}, \begin{bmatrix} a_{32}&a_{33} \end{bmatrix} }$are linearly dependent vectors,which means that ${ \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}=0 }$,contradiction.
• 2.2 When ${|a_{22}+a_{23}|=|a_{32}+a_{33}|=0}$,the sign of ${a_{22},a_{23}}$ must be opposite,and the sign of ${a_{32},a_{33}}$ are also opposite.In such case,${ \begin{bmatrix} a_{22}&a_{23} \end{bmatrix}, \begin{bmatrix} a_{32}&a_{33} \end{bmatrix} }$are linearly dependent vectors,which means that ${ \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}=0 }$,contradiction.

The above argument shows that ${\det A\neq 6}$.So the largest possible value of ${\det A}$ is ${4}$.Now we construct an example in which ${\det A}$ can reach ${4}$:

$\displaystyle \det \begin{vmatrix} 1&0&1\\ -1&1&1\\ -1&-1&1 \end{vmatrix} =4.$

$\Box$

Exercise 4.3.23 How many permutations of ${(1,2,3,4)}$ are even and what are they?Extra credit:What are all the possible ${4}$ by ${4}$ determinants of ${I+P_{even}}$?

Solution: The number of even permutations and odd permutations of ${(1,2,3,4)}$ are same,because we can construct a bijection from the set of even permutations and the set of odd permutations:Exchange the last two numbers of an even permutation will get an odd permutation and vise versa.

So there are ${\frac{4!}{2}=12}$ even permutations of ${(1,2,3,4)}$.They are

$\displaystyle (1,2,3,4),(1,3,4,2),(1,4,2,3),(2,3,1,4),(2,4,3,1),(2,1,4,3),$

$\displaystyle (3,1,2,4),(3,2,4,1),(3,4,1,2),(4,2,1,3),(4,1,3,2),(4,2,3,1).$

The corresponding permutation matrices are

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

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

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

The corresponding matrices of the type ${I+P_{even}}$ are

$\displaystyle \begin{bmatrix} 2&0&0&0\\ 0&2&0&0\\ 0&0&2&0\\ 0&0&0&2 \end{bmatrix}, \begin{bmatrix} 2&0&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 0&1&0&1 \end{bmatrix}, \begin{bmatrix} 2&0&0&0\\ 0&1&0&1\\ 0&1&1&0\\ 0&0&1&1 \end{bmatrix}, \begin{bmatrix} 1&1&0&0\\ 0&1&1&0\\ 1&0&1&0\\ 0&0&0&2 \end{bmatrix}, \begin{bmatrix} 1&1&0&0\\ 0&1&0&1\\ 0&0&2&0\\ 1&0&0&1 \end{bmatrix},$

$\displaystyle \begin{bmatrix} 1&1&0&0\\ 1&1&0&0\\ 0&0&1&1\\ 0&0&1&1 \end{bmatrix}, \begin{bmatrix} 1&0&1&0\\ 1&1&0&0\\ 0&1&1&0\\ 0&0&0&2 \end{bmatrix}, \begin{bmatrix} 1&0&1&0\\ 0&2&0&0\\ 0&0&1&1\\ 1&0&0&1 \end{bmatrix}, \begin{bmatrix} 1&0&1&0\\ 0&1&0&1\\ 1&0&1&0\\ 0&1&0&1 \end{bmatrix}, \begin{bmatrix} 1&0&0&1\\ 0&2&0&0\\ 1&0&1&0\\ 0&0&1&1 \end{bmatrix},$

$\displaystyle \begin{bmatrix} 1&0&0&1\\ 1&1&0&0\\ 0&0&2&0\\ 0&1&0&1 \end{bmatrix}, \begin{bmatrix} 1&0&0&1\\ 0&2&0&0\\ 0&0&2&0\\ 1&0&0&1 \end{bmatrix}.$

The determinants of the matrices of the type ${I+P_{even}}$ are

$\displaystyle 16,2,2,0,4,0,4,4,0,4,4,0$

respectively. $\Box$

Exercise 4.3.24 Find cofactors and then transpose.Multiply ${C_A^T}$ and ${C_B^T}$ by ${A}$ and ${B}$!

$\displaystyle A= \begin{bmatrix} 2&1\\ 3&6 \end{bmatrix},B= \begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&0&0 \end{bmatrix}.$

Solution:

$\displaystyle C_A= \begin{bmatrix} 6&-3\\ -1&2 \end{bmatrix},C_A^T= \begin{bmatrix} 6&-1\\ -3&2 \end{bmatrix}.$

$\displaystyle C_B= \begin{bmatrix} 0&42&-35\\ 0&-21&14\\ -3&6&-3 \end{bmatrix},C_B^T= \begin{bmatrix} 0&0&-3\\ 42&-21&6\\ -35&14&-3 \end{bmatrix}.$

$\displaystyle AC_A^T= \begin{bmatrix} 9&0\\ 0&9 \end{bmatrix},BC_B^T= \begin{bmatrix} -21&0&0\\ 0&-21&0\\ 0&0&-21 \end{bmatrix}.$

$\Box$

Exercise 4.3.25 Find the cofactor matrix ${C}$ and compare ${AC^T}$ with ${A^{-1}}$:

$\displaystyle A= \begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{bmatrix},A^{-1}=\frac{1}{4} \begin{bmatrix} 3&2&1\\ 2&4&2\\ 1&2&3 \end{bmatrix}.$

Solution:

$\displaystyle C= \begin{bmatrix} 3&2&1\\ 2&4&2\\ 1&2&3 \end{bmatrix},C^T=C.$

${AC^T=(\det A) I=(\det A)AA^{-1}}$,so ${A^{-1}=\frac{1}{\det A}C^T}$. $\Box$

Exercise 4.3.26 The matrix ${B_n}$ is the ${-1,2,-1}$ matrix ${A_n}$ except that ${b_{11}=1}$ instead of ${a_{11}=2}$.Using cofactors of the last row of ${B_4}$,show that ${|B_4|=2|B_3|-|B_2|=1}$

$\displaystyle B_4= \begin{bmatrix} 1 & -1 & & \\ -1 & 2 &-1 & \\ & -1 &2 &-1\\ & &-1 &2 \end{bmatrix},B_3= \begin{bmatrix} 1&-1& \\ -1&2&-1\\ &-1&2 \end{bmatrix}.$

The recursion ${|B_n|=2|B_{n-1}|-|B_{n-2}|}$ is the same as for the ${A}$‘s.The difference is in the starting value ${1,1,1}$ for ${n=1,2,3}$.What are the pivots?

Solution: In this exercise,we only need to find the pivots(I have checked the recursion relation on the scratch paper).By using elimination steps,the pivots are ${1}$‘s. $\Box$

Exercise 4.3.28 The ${n}$ by ${n}$ determinant ${C_n}$ has ${1}$ s above and below the main diagonal:

$\displaystyle C_1= \begin{vmatrix} 0 \end{vmatrix},C_2= \begin{vmatrix} 0&1\\ 1&0 \end{vmatrix},C_3= \begin{vmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{vmatrix},C_4= \begin{vmatrix} 0&1&0&0\\ 1&0&1&0\\ 0&1&0&1\\ 0&0&1&0 \end{vmatrix}.$

• What are the determinants of ${C_1,C_2,C_3,C_4}$?
• By cofactors find the relation between ${C_n}$ and ${C_{n-1}}$ and ${C_{n-2}}$.Find ${C_{10}}$.

Solution:

• ${\det C_1=0,\det C_2=-1}$,${\det C_3=0}$,${\det C_4=1}$.
• ${C_n=-C_{n-2}}$.So ${C_{10}=C_2=-1}$.

$\Box$

Exercise 4.3.29 Problem 28 has ${1}$s just above and below the main diagonal.Going down the matrix,which order of columns (if any) gives all ${1}$s?Explain why that permutation is even for ${n=4,8,12,\cdots}$ and odd for ${n=2,6,10,\cdots}$

Solution: Going down the matrix,column of the order ${(2,1,4,3,6,5,\cdots,2n,2n-1,\cdots)}$ gives all ${1}$s. $\Box$

Exercise 4.3.30 Explain why this Vandermonde determinant contains ${x^3}$ but not ${x^4}$ or ${x^5}$:

$\displaystyle V_4=\det \begin{bmatrix} 1&a&a^2&a^3\\ 1&b&b^2&b^3\\ 1&c&c^2&c^3\\ 1&x&x^2&x^3 \end{bmatrix}.$

Solution: The determinant is zero at ${x=a,b,}$and ${c}$.The cofactor of ${x^{3}}$ is ${V_3=(b-a)(c-a)(c-b)}$.Then ${V_4=(x-a)(x-b)(x-c)V_3}$. $\Box$

Exercise 4.3.31 Compute the determinants ${S_1,S_2,S_3}$ of these ${1,3,1}$ tridiagonal matrices:

$\displaystyle S_1= \begin{vmatrix} 3 \end{vmatrix},S_2= \begin{vmatrix} 3&1\\ 1&3 \end{vmatrix},S_3= \begin{vmatrix} 3&1&0\\ 1&3&1\\ 0&1&3 \end{vmatrix}.$

Make a Fibonacci guess for ${S_4}$ and verify that you are right.

Solution: ${S_1=3}$,${S_2=8}$,${S_3=21}$.In general,${S_n=3S_{n-1}-S_{n-2}}$,for ${n\geq 3}$.So ${S_4=3S_3-S_2=55}$. $\Box$

Exercise 4.3.32 Cofactors of those ${1,3,1}$ matrices give ${S_n=3S_{n-1}-S_{n-2}}$.Challenge:Show that ${S_n}$ is the Fibonacci number ${F_{2n+2}}$ by proving ${F_{2n+2}=3F_{2n}-F_{2n-2}}$.Keep using Fibonacci’s rule ${F_k=F_{k-1}+F_{k-2}}$.

Solution: ${F_{4}=3,F_{6}=8}$,and

$\displaystyle \begin{array}{rcl} F_{2n+2}&=&F_{2n+1}+F_{2n} \\&=&(F_{2n}+F_{2n-1})+F_{2n} \\&=&2F_{2n}+F_{2n-1} \\&=&2F_{2n}+(F_{2n}-F_{2n-2}) \\&=&3F_{2n}-F_{2n-2} \end{array}$

$\Box$

Exercise 4.3.33 Change ${3}$ to ${2}$ in the upper left corner of the matrices in Problem 32.Why does that subtract ${S_{n-1}}$ from the determinant ${S_n}$?Show that the determinants become the Fibonacci numbers ${2,5,13}$(always ${F_{2n+1}}$).

Solution: This can be easily seen by using the linearity of the determinant in the first row:${ \begin{bmatrix} 2&1&0 \end{bmatrix}= \begin{bmatrix} 3&1&0 \end{bmatrix}- \begin{bmatrix} 1&0&0 \end{bmatrix}. }$Denote the new determinant by ${D_n}$.Then by cofactor expansion in the first row,

$\displaystyle \begin{array}{rcl} D_n&=&2S_{n-1}-S_{n-2} \\&=&2F_{2n}-F_{2n-2} \\&=&2(F_{2n-1}+F_{2n-2})-F_{2n-2} \\&=&2F_{2n-1}+F_{2n-2} \\&=&F_{2n-1}+F_{2n} \\&=&F_{2n+1} \end{array}$

$\Box$

Exercise 4.3.34 With ${2}$ by ${2}$ blocks,you cannot always use block determinants!

$\displaystyle \begin{vmatrix} A&B\\ 0&D \end{vmatrix}= \begin{vmatrix} A \end{vmatrix} \begin{vmatrix} D \end{vmatrix}~but~ \begin{vmatrix} A&B\\ C&D \end{vmatrix}\neq \begin{vmatrix} A \end{vmatrix} \begin{vmatrix} D \end{vmatrix}- \begin{vmatrix} C \end{vmatrix} \begin{vmatrix} B \end{vmatrix}$

•  Why is the first statement true?Somehow ${B}$ doesn’t enter.
• Show by example that equality fails(as shown) when ${C}$ enters.
• Show by example that the answer ${\det (AD-CB)}$ is also wrong.

Solution:

• This can be seen directly from the big formula of the determinant.
• Counterexample:${A=C=D= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix},B= \begin{bmatrix} 1&1\\ 1&1 \end{bmatrix}. }$Then

$\displaystyle \begin{vmatrix} A&B\\ C&D \end{vmatrix}= \begin{vmatrix} 1&0&1&1\\ 0&1&1&1\\ 1&0&1&0\\ 0&1&0&1 \end{vmatrix}=-1,$

but ${|A||D|-|C||B|=1}$.

• Counterexample:Let ${A=C}$ be nonsingular,and ${D=B+I}$.Then ${ \begin{vmatrix} A&B\\ C&D \end{vmatrix}=0, }$but ${\det (AD-CB)=\det A\neq 0}$.

$\Box$

Exercise 4.3.35 With block multiplication,${A=LU}$ has ${A_k=L_kU_k}$ in the upper left corner:

$\displaystyle A= \begin{bmatrix} A_k&*\\ *&* \end{bmatrix}= \begin{bmatrix} L_k&0\\ *&* \end{bmatrix} \begin{bmatrix} U_k&*\\ 0&* \end{bmatrix}.$

• Suppose the first three pivots of ${A}$ are ${2,3,-1}$.What are the determinants of ${L_1,L_2,L_3}$(with diagonal ${1}$s),${U_1,U_2,U_3}$ and ${A_1,A_2,A_3}$?
• If ${A_1,A_2,A_3}$ have determinants ${5,6,7}$,find the three pivots.

Solution:

• ${\det L_1=\det L_2=\det L_3=1}$.${\det U_1=\det A_{1}=2,\det U_2=\det A_{2}=3,\det U_3=\det A_{3}=-1}$.
• The three pivots are ${5,\frac{6}{5},\frac{7}{6}}$.

$\Box$

Exercise 4.3.38 For ${A_4}$ in Problem 6,five of the ${4!=24}$ terms in the big formula (6) are nonzero.Find those five terms to show that ${D_4=-1}$.

Solution:

$\displaystyle A_4= \begin{bmatrix} 1&1&0&0\\ 1&1&1&0\\ 0&1&1&1\\ 0&0&1&1 \end{bmatrix}.$

$\displaystyle \begin{array}{rcl} D_4&=&\delta_{1234}^{1234}\times 1\times 1\times 1\times 1+\delta_{1234}^{1243}\times 1\times 1\times 1\times 1+\delta_{1234}^{1324}\times 1\times 1\times 1\times 1 \\&+&\delta_{1234}^{2134}\times 1\times 1\times 1\times 1+\delta_{1234}^{2143}\times 1\times 1\times 1\times 1=1-1-1-1+1=-1. \end{array}$

(The meaning of the ${\delta}$” notation can be found in Yichao Xu(许 以超)’s Linear Algebra and Matrix Theory,second edition,section 2.1) $\Box$

Exercise 4.3.39 For the ${4}$ by ${4}$ tridiagonal matrix(entries ${-1,2,-1}$),find the five terms in the big formula that give ${\det A=16-4-4-4+1}$

Solution:

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

$\displaystyle \begin{array}{rcl} \det A&=&\delta_{1234}^{1234}2\times 2\times 2\times 2+\delta_{1234}^{1243}2\times 2\times (-1)\times (-1)+\delta_{1234}^{1324}2\times (-1)\times (-1)\times 2 \\&+&\delta_{1234}^{2134}(-1)\times (-1)\times 2\times 2+\delta_{1234}^{2143}(-1)\times (-1)\times (-1)\times (-1) \\&=&16-4-4-4+1=5. \end{array}$

$\Box$

Exercise 4.3.40 Find the determinant of this cyclic ${P}$ by cofactors of row ${1}$.How many exchanges reorder ${4,1,2,3}$ into ${1,2,3,4}$?Is ${|P^2|=+1}$ or ${-1}$?

$\displaystyle P= \begin{bmatrix} 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{bmatrix},P^2= \begin{bmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{bmatrix}= \begin{bmatrix} 0&I\\ I&0 \end{bmatrix}.$

Solution: ${\det P=\delta_{1234}^{4123}=-1}$.${3}$ exchanges reorder ${4,1,2,3}$ into ${1,2,3,4}$.${|P^2|=(\det P)^2=1}$. $\Box$

Exercise 4.3.43 All Pascal matrices have determinant ${1}$.If I subtract ${1}$ from the ${n,n}$ entry,why does the determinant become zero?(Use rule 3 or a cofactor.)

$\displaystyle \det \begin{bmatrix} 1&1&1&1\\ 1&2&3&4\\ 1&3&6&10\\ 1&4&10&20 \end{bmatrix}=1~~~~(known)~~~\det \begin{bmatrix} 1&1&1&1\\ 1&2&3&4\\ 1&3&6&10\\ 1&4&10&19 \end{bmatrix}=0~~~(explain).$

Solution:

$\displaystyle \begin{array}{rcl} \begin{vmatrix} 1&1&1&1\\ 1&2&3&4\\ 1&3&6&10\\ 1&4&10&19 \end{vmatrix}&=& \begin{vmatrix} 1&1&1&1\\ 1&2&3&4\\ 1&3&6&10\\ 1&4&10&20 \end{vmatrix}- \begin{vmatrix} 1&1&1&1\\ 1&2&3&4\\ 1&3&6&10\\ 0&0&0&1 \end{vmatrix} \\&=& \begin{vmatrix} 1&1&1&1\\ 1&2&3&4\\ 1&3&6&10\\ 1&4&10&20 \end{vmatrix}- \begin{vmatrix} 1&1&1\\ 1&2&3\\ 1&3&6\\ \end{vmatrix} \\&=&1-1=0. \end{array}$

$\Box$