二月 2018

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{\mathbf{R}^n}中,有一个{n}维的锥体 {P_{n}}.{P_{n}}{n+1}个顶点,其中一个顶点是 {(0,0,\cdots,0)},剩下的{n}个顶点的坐标是单位矩阵{I_n}的各个行向量(当{n=2}时,{P_{2}}是三角形;当{n=3}{P_{3}}是四面体).可 以用积分的方法来计算{P_{n}}的体积,不过下面我们利用高维祖暅原理和割补法来计算{P_{n}}的体积.

首先,我们来确定{n}维单位方块 {C_{n}}的顶点和组成它的{n-1}维面的个 数.{C_{n}}是二维的单位正 方形和三维的单位正方体的概念的高维推广,且{C_{n}}的每条边都是单位长 度.组成它的{n-1}维面是{n-1}维 的单位方块.计数使用的方法 是递推法,递推依据的是高维方块的形成过程:每个{n+1}维的方块都可以看作是 一个{n}维方块在新的维度上平行拖动一定距离而形成的图形.

{C_{n}}{f(n)}个顶点,则由高维方块的形成过程,可得递推关系{f(n+1)=2f(n)},且 {f(1)=2},因此{f(n)=2^n}.即{n}维方块 {C_{n}}{2^n}个顶点.设{C_{n}}{n-1}维 面有 {g(n)}个.则由高维方块 的形成过程,可得递推关系{g(n+1)=g(n)+2},且{g(1)=2}.因此{g(n)=2n}.即 {C_{n}}{n-1}维面有{2n}个.比如:{2}维的正方形有 {4}条边,{3}维的立方体有{6}个正方形面.

{C_{n}}的某个{n-1}维面上有{2^{n-1}}个顶点,将这些顶点和不在该{n-1}维 面上的某个{C_n}的顶点{D_{i}}连接,形成一个{n}维的锥体 {Q_n}.{C_n}共能分割为{n}个形 状和{Q_n}相同的几何体.这是因为,{C_n}共有{2n}{n-1}维面,其中经过顶点 {D_i}{n-1}维面有{n}个(根据{C_n}的形成过程),因此不包含顶点{D_i}{n-1}维面有{2n-n=n}个,{D_i}与这{2n-n=n}个面中的每一个面都能组成一个和 {Q_n}形状相同的几何体,所有这些几何体恰好能填满{C_{n}}.因此{Q_n}的体积 是{\frac{1}{n}}.

上面这段论证结合高维祖暅原理,可得,一个{n}维椎体的体积,是同底等高的 {n}维柱体体积的{\frac{1}{n}}.

{P_n}的体积是{V_n}.则可得递推关系:

\displaystyle V_{n+1}=\frac{1}{n}V_n,V_1=1.

因此{V_n=\frac{1}{n!}}.

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:

习题2.1.1: 试通过一系列对换,将排列 {13852476} 变为排列 {72453816}.

解:{72453816=a_1a_2a_3a_4a_5a_6a_7a_8}.则排列{13852476=a_7a_5a_6a_4a_2a_3a_1a_8}.现在进行对换操作:

\displaystyle a_7a_5a_6a_4a_2a_3a_1a_8\rightarrow a_5a_7a_6a_4a_2a_3a_1a_8\rightarrow a_5a_6a_7a_4a_2a_3a_1a_8\rightarrow a_5a_6a_4a_7a_2a_3a_1a_8\rightarrow a_5a_6a_4a_2a_7a_3a_1a_8\rightarrow a_5a_6a_4a_2a_3a_7a_1a_8\rightarrow a_5a_6a_4a_2a_3a_1a_7a_8\rightarrow a_5a_4a_6a_2a_3a_1a_7a_8\rightarrow a_5a_4a_2a_6a_3a_1a_7a_8\rightarrow a_5a_4a_2a_3a_6a_1a_7a_8\rightarrow a_5a_4a_2a_3a_1a_6a_7a_8\rightarrow a_4a_5a_2a_3a_1a_6a_7a_8\rightarrow a_4a_2a_5a_3a_1a_6a_7a_8\rightarrow a_4a_2a_3a_5a_1a_6a_7a_8\rightarrow a_4a_2a_3a_1a_5a_6a_7a_8\rightarrow a_2a_4a_3a_1a_5a_6a_7a_8\rightarrow a_2a_3a_4a_1a_5a_6a_7a_8\rightarrow a_2a_3a_1a_4a_5a_6a_7a_8\rightarrow a_2a_1a_3a_4a_5a_6a_7a_8\rightarrow a_1a_2a_3a_4a_5a_6a_7a_8. \Box

习题2.1.2: 设 {1274j56k9}{1j25k4897}{9}个正整数{1,2,3,4,5,6,}{7,8,9}的排列.试决定整数{j}{k},使得前者是奇排列,后者是偶排列.

勘误: 该题目叙述有问题.在第1个排列中,{jk}只可能是{3,8}的一个排列.在第2个排列中,{jk}只可能是{3,6}的一个排列.所以题目有问题. \Box

 

习题2.1.3: 试求下列排列的奇偶性,即计算

  • {\delta_{n(n-1)\cdots ~2~1}^{1~2~\cdots (n-1)n}}.
  • {\delta_{1357\cdots (2n-1)~2~4~\cdots (2n)}^{1234\cdots ~n~(n+1)(n+2)\cdots (2n)}}.
  • {\delta_{246\cdots (2n)1~3\cdots (2n-1)}^{123\cdots n(n+1)(n+2)\cdots (2n)}}.
  • {\delta_{2143\cdots (2n)~(2n-1)}^{1234\cdots (2n-1)~(2n)}}.
  • {\delta_{369(12)\cdots (3n)1~4~7~\cdots (3n-2)~2~5~8\cdots (3n-1)}^{1234\cdots n(n+1)(n+2)(n+3)\cdots ~(2n)(2n+1)(2n+2)(2n+3)\cdots ~(3n)}}.

解:

  • {\delta_{n(n-1)\cdots ~2~1}^{1~2~\cdots (n-1)n}=(-1)^{(n-1)+(n-2)+\cdots+2+1}=(-1)^{\frac{n(n-1)}{2}}}.
  • {\delta_{1357\cdots (2n-1)~2~4~\cdots (2n)}^{1234\cdots ~n~(n+1)(n+2)\cdots (2n)}=(-1)^{1+2+\cdots+(n-1)}=(-1)^{\frac{n(n-1)}{2}}}.
  • {\delta_{246\cdots (2n)1~3\cdots (2n-1)}^{123\cdots n(n+1)(n+2)\cdots (2n)}=(-1)^{1+2+\cdots+n}=(-1)^{\frac{(n+1)n}{2}}}.
  • {\delta_{2143\cdots (2n)~(2n-1)}^{1234\cdots (2n-1)~(2n)}=(-1)^n}.
  • {\delta_{369(12)\cdots (3n)1~4~7~\cdots (3n-2)~2~5~8\cdots (3n-1)}^{1234\cdots n(n+1)(n+2)(n+3)\cdots ~(2n)(2n+1)(2n+2)(2n+3)\cdots ~(3n)}}{=(-1)^{(1+2+\cdots+n)+[1+2+\cdots+(n-1)]}=(-1)^{n^2}}.

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先介绍对换的概念,该概念引自许以超《线性代数与矩阵论》(第二版)第2.1节:

定义 1 (对换) 任取一个排列{i_1i_2\cdots i_n},将其中两个相邻的正整数{i_{j-1},i_j}对换 一下,便造出一个新的排列{i_1i_2\cdots i_{j-2}i_ji_{j-1}i_{j+1}\cdots i_n},称为原来排列的对换排列.这样一种步骤称为对换.

再介绍逆序数的概念:

定义 2 (逆序数) 在一个排列中,如果一对数的前后位置与大小顺序相反,即前面的数大于后面的数,那么它们就称为一个逆序.一个排列中逆序的总数就称为这个排列的逆序数.

该书同一节的定理2.1.1叙述如下:

定理 3 如果有一种对换方式,经过作用偶(奇)数次对换,将排列{i_1i_2\cdots i_n}变为标准排列{12\cdots n},那末,不管用哪种对换方式将排列{i_1i_2\cdots i_n}变为标准排列{12\cdots n},都需要作用偶(奇)数次对换.换句话说,对换次数的奇偶性和对换方式无关.

这个定理是奇排列和偶排列概念的基石.倘若没有这个定理,奇偶排列的定义便是失败的(ill-defined).

许以超是用数学归纳法证明这个定理的.我没有耐心去看他的证明,便自己构思了一个证明.这个证明是我在2010年11月25日构思出来的,发布在自己的QQ空间里.现在重新整理在此.

证明: 首先,进行一次对换,会使排列的逆序数{+1}{-1}:进行交换的两个数,如果原来前面的数小于后面的数,则对换后排列的逆序数加{1};如果前面的数大于后面的数,则对换后逆序数{-1}.

比如排列{312}的逆序数为{2},要经过一系列对换变成排列123,可以有以下对换步骤

\displaystyle 312\rightarrow 132\rightarrow 123.

两次对换的逆序数都是{-1}的,直到变成标准排列{123},标准排列的逆序数为{0}.因此这个过程从逆序数增减的角度来看,即{2-1-1=0}.排列{312}也可以通过对换步骤

\displaystyle 312\rightarrow 321\rightarrow 231\rightarrow 213\rightarrow 123.

变成标准排列.这个过程从逆序数增减的角度来看,可以翻译为{2+1-1-1-1=0}.

一般地,设某个排列的逆序数为{p}.该排列经过{m}次对换变为标准排列,从逆序数增减的角度来看,即

\displaystyle p\cdots+1\cdots-1\cdots-1\cdots+1\cdots=0.

其中{-1}的数目比{+1}的数目多{p}个,且{+1}{-1}{m}个.无论在上式的随便一些位置加上{k}{+1},为了保持等式成立,都要在另外一些 地方地方加上{k}{-1}.同样,无论在哪些位置加上{k}{-1},都要在另外一 些地方加上相同数目的{+1}.原本只用对换{m}次就可以达到标准排列,现在要对 换{m+2k}次才可以达到标准排列.而{m}{m+2k}的奇偶性是相同的.于是定理得 证. \Box

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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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题目1 (北京大学{2018}年硕士研究生招生考试《高等代数与解析几何》试题1) 试确定实数域上所有的{3}{(0,1)}行列式(即所有元素只能是{0,1}的行列式)的最大值,给出证明及取到最大值的一个构造.

解:{3}阶方阵

\displaystyle A= \begin{bmatrix} a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33} \end{bmatrix},

其中{a_{ij}(1\leq i,j\leq 3)}取值{0}{1}.则由行列式的展开可得

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

其中{C_{1j}}{a_{1j}(1\leq j\leq 3)}的代数余子式.由于二阶行列式 {C_{1j}}的四个元素只可能取{0}{1}.因此{C_{1j}}的值只可能是{0,1,-1}.因 此{\det A}不大于{3}.下面证明,{\det A}不可能等于{3}.

如果{\det A=3},则{a_{11},a_{12},a_{13}}都等于{1}.且{C_{11}=C_{12}=C_{13}=1}.即

\displaystyle \begin{vmatrix} a_{22}& a_{23}\\ a_{32}& a_{33} \end{vmatrix}=1, \ \ \ \ \ (1)

\displaystyle \begin{vmatrix} a_{21}& a_{23}\\ a_{31}& a_{33} \end{vmatrix}=-1, \ \ \ \ \ (2)

\displaystyle \begin{vmatrix} a_{21}& a_{22}\\ a_{31}& a_{32} \end{vmatrix}=1. \ \ \ \ \ (3)

由方程(1)可得{a_{22}=a_{33}=1}.由方程(3)可得{a_{21}=a_{32}=1}.由方程(2)可得{a_{31}=a_{23}=1}.因此

\displaystyle \begin{vmatrix} a_{22}& a_{23}\\ a_{32}& a_{33} \end{vmatrix}= \begin{vmatrix} 1&1\\ 1&1 \end{vmatrix}=0.

这与方程(1)成立矛盾.

因此{\det A}最大可能的值是{2}.下面我们给出一个构造:

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

\Box

这道题和来自Linear Algebra and Its Applications 习题4.3.22的题是完全类似的.而且它们的命题背景都是Hadamard’s maximal determinant problem.

 

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