Linear Algebra and Its Applications_Strang

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下面我们解几道关于离散动力系统的应用题,以及其它题目.这些题目来自Gilbert Strang的《线性代数及其应用》第三版第5.3节.

(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
$$
A=
\begin{bmatrix}
0&0&6\\
\frac{1}{2}&0&0\\
0&\frac{1}{3}&0
\end{bmatrix}.
$$
证明$A^3=I$.若开始时每组有3000个甲虫,求6年的甲虫分布.

解:
设矩阵$A$的特征值为$\lambda$.则
$$
\begin{vmatrix}
-\lambda&0&6\\
\frac{1}{2}&-\lambda&0\\
0&\frac{1}{3}&-\lambda
\end{vmatrix}=0,
$$
解得$\lambda_{1}=1,\lambda_2=e^{\frac{-\pi}{3}i},\lambda_3=e^{\frac{2\pi}{3}i}$.可得$\lambda_{1}^3=\lambda_{2}^{3}=\lambda_{3}^{3}=1$.因此$A^3=I$.若开始时每组有$3000$只甲虫,则$6$年后每组还是有$3000$只甲虫.


(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.

解:设在第$n$个月,波士顿有$x_n$辆卡车,洛杉矶有$y_n$辆卡车,芝加哥有$z_n$辆卡车.则
$$
\begin{bmatrix}
x_{n+1}\\
y_{n+1}\\
z_{n+1}
\end{bmatrix}=
\begin{bmatrix}
\frac{1}{2}&0&\frac{1}{2}\\
0&\frac{1}{2}&\frac{1}{2}\\
\frac{1}{2}&\frac{1}{2}&0
\end{bmatrix}
\begin{bmatrix}
x_n\\
y_n\\
z_n
\end{bmatrix}
$$
所以转移矩阵为
$$
A=\begin{bmatrix}
\frac{1}{2}&0&\frac{1}{2}\\
0&\frac{1}{2}&\frac{1}{2}\\
\frac{1}{2}&\frac{1}{2}&0
\end{bmatrix},
$$
可求得矩阵$A$的特征值为$-\frac{1}{2},\frac{1}{2},1$.对应于特征值$1$的某个特征向量为$
\begin{bmatrix}
1\\
1\\
1\\
\end{bmatrix}.
$所以对应于特征值$1$的稳定状态是:三个城市的货车一样多.


(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.
$$
\begin{bmatrix}
d_{k+1}\\
s_{k+1}\\
w_{k+1}
\end{bmatrix}=
\begin{bmatrix}
1&\frac{1}{4}&0\\
0&\frac{3}{4}&\frac{1}{2}\\
0&0&\frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
d_k\\
s_k\\
w_k
\end{bmatrix}.
$$


解:显然,转移矩阵的特征值为$1,\frac{3}{4},\frac{1}{2}$.特征值$1$对应的某个特征向量为$
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}
$.即人全死光为稳定状态.

(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?



$$
\begin{pmatrix}
V_{n+1}\\
W_{n+1}
\end{pmatrix}=
\begin{pmatrix}
\alpha&\alpha\\
\alpha&\alpha
\end{pmatrix}
\begin{pmatrix}
V_n\\
W_n
\end{pmatrix}.
$$
矩阵
$$
A=
\begin{pmatrix}
\alpha&\alpha\\
\alpha&\alpha
\end{pmatrix}
$$
的特征值为$0,2\alpha$.当$\alpha>\frac{1}{2}$时,产生不稳定性.


Tags:

我曾经在2017.12.3和2017.12.4日用两种方法推导过正规方程.现在把它们整理到一起,发布在此.

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.
设$n$个$m$维向量$\mathbf{v}_{1},\mathbf{v}_{2},\cdots,\mathbf{v}_{n}$张成子空间$W$.设$m$维向量$\mathbf{b}$在子空间$W$中的射影为$\mathbf{b}’$.则$\mathbf{b}-\mathbf{b}’$正交于子空间$W$,也即与$\mathbf{v}_{1},\mathbf{v}_{2},\cdots,\mathbf{v}_{n}$都正交.设
$$\mathbf{b}’=x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n},$$
则$\mathbf{b}-\mathbf{b}’=\mathbf{b}-(x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots+x_{n}\mathbf{v}_{n})$.我们有

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

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

$$\begin{pmatrix}
\mathbf{v}_{1}\cdot\mathbf{v}_{1}&\mathbf{v}_{2}\cdot
  \mathbf{v}_{1}&\cdots&\mathbf{v}_{n}\cdot\mathbf{v}_{1}\\
\mathbf{v}_{1}\cdot \mathbf{v}_{2}&\mathbf{v}_{2}\cdot
\mathbf{v}_{2}&\cdots& \mathbf{v}_{n}\cdot \mathbf{v}_{2}\\
\vdots&\vdots&\cdots&\vdots\\
\mathbf{v}_{1}\cdot \mathbf{v}_{n}&\mathbf{v}_{2}\cdot
\mathbf{v}_{n}&\cdots& \mathbf{v}_{n}\cdots \mathbf{v}_{n}
\end{pmatrix}
\begin{pmatrix}
  x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{pmatrix}=
\begin{pmatrix}
  \mathbf{b}\cdot \mathbf{v}_{1}\\
\mathbf{b}\cdot \mathbf{v}_{2}\\
\vdots\\
\mathbf{b}\cdot \mathbf{v}_{n}
\end{pmatrix}.$$
事实上,这就是正规方程.如果更详细地记$\mathbf{v}_{i}=
\begin{pmatrix}
  a_{1i}\\
a_{2i}\\
\vdots\\
a_{mi}
\end{pmatrix}(1\leq i\leq n),
$则上式可以改写成如下形式:
$$A^{T}Ax=A^{T}b,$$
其中
$$A=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\cdots&\vdots\\
  a_{mn}&a_{m2}&\cdots&a_{mn}
\end{pmatrix},x=
\begin{pmatrix}
  x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{pmatrix}.$$
两种形式的正规方程其实是一样的.

2.用微分法推导正规方程

我们考虑,当$x$满足什么条件时,$||Ax-b||$取极小值,其中$x$是$n$维向量,$A$是$m\times n$矩阵,$b$是$m$维向量.设
\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*}
令$y_{1}=(Ax)^{T}Ax$,$y_{2}=(Ax)^{T}b,y_{3}=b^{T}Ax,y_{4}=b^{T}b$,则$y=y_1-y_2-y_3+y_4$,则
$$
dy=dy_1-dy_2-dy_3+dy_4,
$$
而$dy_2=[A(dx)]^{T}b=(dx)^TA^{T}b$,$dy_3=b^TAdx,dy_4=0$.
\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
\\&=[x^TA^{T}Ax+x^TA^TAdx+(dx)^{T}A^{T}Ax]-x^TA^TAx
\\&=x^TA^{T}Adx+(dx)^TA^{T}Ax
\end{align*}
\begin{align*}
dy&=x^TA^TAdx+(dx)^TA^TAx-(dx)^TA^Tb-b^TAdx
\\&=(x^{T}A^{T}A-b^{T}A)dx+(dx)^T(A^{T}Ax-A^{T}b)
\\&=M^{T}+M,
\end{align*}
其中$M=(dx)^T(A^{T}Ax-A^{T}b)$.当$||Ax-b||$取极小值时,$y$也取极小值,因此$dy=0$.即
$$
M^{T}+M=0
$$
因此只能有$M=0$,即$M=(dx)^T(A^{T}Ax-A^{T}b)=0$.因此$A^TAx-A^Tb=0$,即
\begin{equation}\label{eq:1}
A^TAx=A^Tb
\end{equation}
即当$x$满足式\eqref{eq:1}时,$||Ax-b||$取极小值.而式\eqref{eq:1}就是正规方程.

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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 $AB=BA$.

个人觉得这个结论的叙述有它不清楚的地方.反而是该书的第四版对同一个结论表述得好一些.我给Strang教授发了一封邮件,内容如下:

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.

 

不过在书中并没有给出该结论的完整证明.下面我们将完整证明如下结论:

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

证明:对于$n$阶方阵 $A$ 和 $B$,若 $A,B$ 拥有共同的特征向量矩阵$S$,则存在对角阵$D_1,D_2$,使得$A=SD_1S^{-1}$,$B=SD_2S^{-1}$,则
$$
AB=SD_1D_2S^{-1}=SD_2D_1S^{-1}=BA.
$$
反之,若$AB=BA$,设$\mathbf{x}$是$B$的任意一个特征向量,且设$B$的特征向量$\mathbf{x}$对应特征值$\lambda$,则
$$
(AB)\mathbf{x}=A(B\mathbf{x})=A(\lambda \mathbf{x})=\lambda A\mathbf{x}=(BA)\mathbf{x}=B(A\mathbf{x}),
$$
即$B(A\mathbf{x})=\lambda A\mathbf{x}$.对于矩阵$B$来说,设特征值$\lambda$对应于特征空间$W_{\lambda}$,则由$B(A\mathbf{x})=\lambda A\mathbf{x}$可得$A\mathbf{x}\in W_{\lambda}$.可见,$B$的某个特征空间中的向量在矩阵$A$的作用下仍然在该特征空间中.

设矩阵$A$是线性变换$T_A$在基$\alpha$下的矩阵,因为矩阵$A$是可对角化矩阵,所以当线性变换$T_A$限制在$W_{\lambda}$上时(注意在上面我们已经证明$W_{\lambda}$在$T_A$的作用下成为自己的子空间),必然也成为可对角化的线性变换.鉴于$W_{\lambda}$是$B$的任意一个特征空间,因此矩阵$A$和$B$可同时对角化.

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

Tags: , ,

 

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

Tags: ,

 

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:

  • No need to answer.
  • \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:

\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

Tags:

在这里我发布Gilbert Strang所著的教材Linear Algebra and Its Applications 第4版习题4.3的习题解答.

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

See also 2018年北京大学硕士研究生入学考试《高等代数与解析几何》题1.

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

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