八月 2018

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数学交流群(QQ号415243762)里山西大同罗锦伟问了一道题:

题目:已知正方形$ABCD-A_1B_1C_1D_1$棱长为$1$,$M$是$AB$的中点,点$Q$是面$CBB_1C_1$所在平面内的动点,且满足$\angle DQC=\angle MQB$,则三棱锥$Q-ABC$的体积的最大值是_______.

有人说点Q的轨迹可以利用阿波罗尼斯圆来做.我不管他,只管建立空间直角坐标系用解析几何方法按部就班来做,照样能得到答案.

如图1所示,以点$C_1$为原点$O$,$\ov{C_1D_1}$方向为$x$轴正方向,$\ov{C_{1}B_{1}}$方向为$y$轴正方向,$\ov{C_1C}$方向为$z$轴正方向,建立空间直角坐标系.

因为点$Q$在$yz$平面上,设点$Q$坐标为$(0,y,z)$.易得
$$
\ov{QC}=\ov{OC}-\ov{OQ}=(0,0,1)-(0,y,z)=(0,-y,1-z),
$$
$$
\ov{QD}=\ov{OD}-\ov{OQ}=(1,0,1)-(0,y,z)=(1,-y,1-z),
$$
$$
\ov{QB}=\ov{OB}-\ov{OQ}=(0,1,1)-(0,y,z)=(0,1-y,1-z),
$$
$$
\ov{QM}=\ov{OM}-\ov{OQ}=\left(\frac{1}{2},1,1\right)-(0,y,z)=\left(\frac{1}{2},1-y,1-z\right).
$$
$\angle DQC=\angle MQB$当且仅当$\cos \angle DQC=\cos\angle MQB$,而
\begin{align*}
\cos \angle
DQC&=\frac{\ov{QC}\cdot\ov{QD}}{|\ov{QC}|\cdot|\ov{QD}|}\\&=\frac{0\times 1+(-y)^2+(1-z)^2}{\sqrt{0^{2}+(-y)^2+(1-z)^2}\sqrt{1^{2}+(-y)^2+(1-z)^2}}\\&=\sqrt{\frac{y^2+(1-z)^2}{1+y^2+(1-z)^{2}}}
\end{align*}
\begin{align*}
\cos
MQB&=\frac{\ov{QB}\cdot\ov{QM}}{|\ov{QB}|\cdot|\ov{QM}|}\\&=\frac{0\times \frac{1}{2}+(1-y)^2+(1-z)^2}{\sqrt{0^2+(1-y)^2+(1-z)^2}\sqrt{\left(\frac{1}{2}\right)^2+(1-y)^{2}+(1-z)^2}}\\&=\sqrt{\frac{(1-y)^2+(1-z)^2}{\frac{1}{4}+(1-y)^2+(1-z)^2}},
\end{align*}
因此
$$
\sqrt{\frac{y^2+(1-z)^2}{1+y^2+(1-z)^{2}}}=\sqrt{\frac{(1-y)^2+(1-z)^2}{\frac{1}{4}+(1-y)^2+(1-z)^2}},
$$
化简并整理可得
$$
(1-z)^2=\frac{4}{9}-\left(y-\frac{4}{3}\right)^2\leq \frac{4}{9},
$$
因此$|1-z|\leq \frac{2}{3}$.因此三棱锥$Q-ABC$以$ABC$为底面的高的最大值是$\frac{2}{3}$.因此三棱锥$Q-ABC$的体积的最大值是
$$\frac{1}{3}S_{\triangle ABC}\times |1-z|_{max}=\frac{1}{3}\times
\frac{1}{2}\times \frac{2}{3}=\frac{1}{9}.$$
体积最大值当且仅当点$Q$位于$\left(0,\frac{4}{3},\frac{1}{3}\right)$或者$\left(0,\frac{4}{3},\frac{5}{3}\right)$时达到.

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.

Tags: ,

设$\mathbf{x}_i=
\begin{bmatrix}
a_{i1}\\
a_{i2}\\
\vdots\\
a_{im}
\end{bmatrix}
$,其中$a_{ij}\in \mathbf{R}$,$1\leq i\leq n,1\leq j\leq m$.则
$$
\left(\sum_{k=1}^{n}p_{k}\mathbf{x}_{k}\right)^T\left(\sum_{k=1}^{n}p_{k}\mathbf{x}_{k}\right)\leq \left(\sum_{k=1}^np_k^2\right)\left(\sum_{k=1}^{n}\mathbf{x}_{k}^T\mathbf{x}_{k}\right),
$$
其中$\forall 1\leq k\leq n$,$p_k\in \mathbf{R}$.上面的式子,换一句话说,是
$$
\left|\left|\sum_{k=1}^{n}p_k\mathbf{x}_{k}\right|\right|^2\leq \left(\sum_{k=1}^{n}p_{k}^{2}\right)\left(\sum_{k=1}^{n}||\mathbf{x}_{k}||^{2}\right).
$$

如果令矩阵$A=
\begin{bmatrix}
|&|&|&|\\
\mathbf{x}_1&\mathbf{x}_2&\cdots&\mathbf{x}_n\\
|&|&|&|
\end{bmatrix}
$,向量$\mathbf{p}=
\begin{bmatrix}
p_1\\
p_2\\
\vdots\\
p_n
\end{bmatrix}
$,则上面的式子可以进一步改写成如下更为紧凑的形式:
$$
||A(\mathbf{p})||\leq ||\mathbf{p}||\cdot||A||,
$$
其中$||A||$是矩阵$A$的Frobenius范数.

上面的不等式,等号成立的条件是矩阵$A$的每个行向量都与向量$\mathbf{p}$线性相关.特别地,如果矩阵$A$是$1\times n$矩阵,则上面的不等式就是我们熟悉的Cauchy不等式.

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继续在QQ群过着荒诞不经的生活.中学教研网(QQ号712018203)的六盘水―高中数学―苏世贤问了一道题,

题目:已知$\triangle ABC$的外接圆圆心为$O$,且$\angle A=60^{\circ}$.若$\overrightarrow{AO}=\alpha\overrightarrow{AB}+\beta\overrightarrow{AC}(\alpha,\beta\in \mathbf{R})$,则$\alpha+\beta$的最大值是 _______.

有人说这题用奔驰定理做.话是没错.可是有多少学生掌握了这个所谓的“奔驰定理”呢?事实上,即便不用“奔驰定理”,老老实实地用纯向量方法也能方便地解决这道题.解答如下,连画图都免了.

:由题意,
$$
\overrightarrow{AO}^2=\overrightarrow{BO}^2=\overrightarrow{CO}^2,
$$
结合$\overrightarrow{AO}=\alpha\overrightarrow{AB}+\beta\overrightarrow{AC}$,$\overrightarrow{BO}=\overrightarrow{AO}-\overrightarrow{AB}=(\alpha-1)\overrightarrow{AB}+\beta\overrightarrow{AC}$,$\overrightarrow{CO}=\overrightarrow{AO}-\overrightarrow{AC}=\alpha\overrightarrow{AB}+(\beta-1)\overrightarrow{AC}$,
可得
$$
(\alpha\overrightarrow{AB}+\beta\overrightarrow{AC})^2=[(\alpha-1)\overrightarrow{AB}+\beta\overrightarrow{AC}]^2=[\alpha\overrightarrow{AB}+(\beta-1)\overrightarrow{AC}]^{2}.
$$
化简$(\alpha\overrightarrow{AB}+\beta\overrightarrow{AC})^2=[(\alpha-1)\overrightarrow{AB}+\beta\overrightarrow{AC}]^2$可

\begin{equation}
\label{eq:1}
(2\alpha-1)|\overrightarrow{AB}|+\beta|\overrightarrow{AC}|=0,
\end{equation}
同理,化简$(\alpha\overrightarrow{AB}+\beta\overrightarrow{AC})^2=[\alpha\overrightarrow{AB}+(\beta-1)\overrightarrow{AC}]^{2}$可得
\begin{equation}
\label{eq:2}
\alpha|\overrightarrow{AB}|+(2\beta-1)|\overrightarrow{AC}|=0.
\end{equation}
由\eqref{eq:1}和\eqref{eq:2}可得
$$
\beta=\frac{2}{3}-\frac{|\overrightarrow{AB}|}{3|\overrightarrow{AC}|},\alpha=\frac{2}{3}-\frac{|\overrightarrow{AC}|}{3|\overrightarrow{AB}|},
$$
因此
$$
\alpha+\beta=\frac{4}{3}-\frac{1}{3}\left(\frac{|\overrightarrow{AB}|}{|\overrightarrow{AC}|}+\frac{|\overrightarrow{AC}|}{|\overrightarrow{AB}|}\right)\leq
\frac{4}{3}-\frac{1}{3}\cdot 2 \sqrt{\frac{|\overrightarrow{AB}|}{|\overrightarrow{AC}|}\cdot\frac{|\overrightarrow{AC}|}{|\overrightarrow{AB}|}}=\frac{2}{3}.
$$
最大值当且仅当$\triangle ABC$是等边三角形时取到.

Tags: ,

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:

暑假无聊,关注了两个中学数学QQ群。今天看到中学数学教研网(QQ号712018203)里浙江衢州高中数学毛敏问了一个向量题:

2018绍兴柯桥二模题17 已知向量$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$满足$|\overrightarrow{b}|=|\overrightarrow{c}|=2|\overrightarrow{a}|=1$,求$(\overrightarrow{c}-\overrightarrow{a})\cdot(\overrightarrow{c}-\overrightarrow{b})$的最大值和最小值.

兹解答如下:

解:\begin{eqnarray*}
(\overrightarrow{c}-\overrightarrow{a})\cdot
(\overrightarrow{c}-\overrightarrow{b})&=&\overrightarrow{c}^2-\overrightarrow{b}\cdot\overrightarrow{c}-\overrightarrow{c}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}
\\&=&\overrightarrow{c}^{2}+\frac{(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\overrightarrow{a}^2-\overrightarrow{b}^2-\overrightarrow{c}^2}{2}
\\&=&\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\frac{1}{8}.
\end{eqnarray*}
当且仅当$\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$时,$(\overrightarrow{c}-\overrightarrow{a})\cdot
(\overrightarrow{c}-\overrightarrow{b})$取到最小值$-\frac{1}{8}$.


$$
\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\frac{1}{8}\leq \frac{1}{2}(|\overrightarrow{a}|+|\overrightarrow{b}|+|\overrightarrow{c}|)^2-\frac{1}{8}=3,
$$
当且仅当$\overrightarrow{a},\overrightarrow{b}$都与向量$\overrightarrow{c}$反向时取到最大值.$\Box$

 

收获一堆赞誉,虚荣心得到了小小的满足

再来几道类似题.我们同样用配方法和向量绝对值不等式处理它.

题目2:平面向量$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$的模为$2$,且$\overrightarrow{a}\perp\overrightarrow{b}$,求$(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{c})$的最小值.

解:
\begin{align*}
(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{c})&=\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{c}-\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}^2
\\&=\frac{(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\overrightarrow{a}^{2}-\overrightarrow{b}^{2}-\overrightarrow{c}^2}{2}+\overrightarrow{c}^2
\\&=\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-2
\\&\geq \frac{1}{2}(|\overrightarrow{a}+\overrightarrow{b}|-|\overrightarrow{c}|)^2-2
\\&=4-4 \sqrt{2}.
\end{align*}
当且仅当$\overrightarrow{c}$与$\overrightarrow{a+b}$反向时取到最小值.

题目3:已知$\overrightarrow{a},\overrightarrow{b}$是平面内两个互相垂直的单位向量,若向量$\overrightarrow{c}$满足$(\overrightarrow{a}-\overrightarrow{c})\cdot (\overrightarrow{b}-\overrightarrow{c})=\overrightarrow{0}$,则$|\overrightarrow{c}|$的最大值是______.

:\begin{align*}
(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{c})=\overrightarrow{0}&\iff
\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{c}-\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}^{2}=0
\\&\iff
\frac{(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\overrightarrow{a}^2-\overrightarrow{b}^2-\overrightarrow{c}^{2}}{2}+\overrightarrow{c}^{2}=0
\\&\iff \overrightarrow{c}^2=2-(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2
\end{align*}
$$
|\overrightarrow{c}|^2=2-(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2\leq
2-(|\overrightarrow{a}+\overrightarrow{b}|-|\overrightarrow{c}|)^2=2 \sqrt{2}|\overrightarrow{c}|-|\overrightarrow{c}|^2,
$$
即$2|\overrightarrow{c}|^2\leq 2 \sqrt{2}|\overrightarrow{c}|$,因此$|\overrightarrow{c}|\leq \sqrt{2}$.当且仅当$\overrightarrow{a}+\overrightarrow{b}$与$\overrightarrow{c}$反向时,$|\overrightarrow{c}|$取到最大值$\sqrt{2}$.

 

题目4(2009年全国I卷理科第6题):已知$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$为单位向量,且$\overrightarrow{a}\cdot\overrightarrow{b}=0$,则$(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{c})$的最小值为_____.

证明:\begin{align*}
(\overrightarrow{a}-\overrightarrow{c})\cdot (\overrightarrow{b}-\overrightarrow{c})&=\overrightarrow{a}\cdot
\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{c}-\overrightarrow{c}\cdot\overrightarrow{b}+\overrightarrow{c}^2\\&=\frac{(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\overrightarrow{a}^2-\overrightarrow{b}^{2}-\overrightarrow{c}^2}{2}+\overrightarrow{c}^2
\\&=\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\frac{1}{2}
\\&\geq \frac{1}{2}(|\overrightarrow{a}+\overrightarrow{b}|-|\overrightarrow{c}|)^2-\frac{1}{2}
\\&=1-\sqrt{2}.
\end{align*}
当且仅当$\overrightarrow{c}$与$\overrightarrow{a}+\overrightarrow{b}$反向时,取到最小值.

题目5(2011年辽宁卷第10题):已知$\overrightarrow{a},\overrightarrow{b}$是共面的单位向量,且$\overrightarrow{a}\cdot\overrightarrow{b}=0$,$(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{c})\leq 0$,则$|\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}|$的最大值是_______.

:
\begin{align*}
(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{c})\leq 0&\iff
\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{c}-\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}^2\leq
0
\\&\iff
\frac{(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2-\overrightarrow{a}^2-\overrightarrow{b}^2-\overrightarrow{c}^2}{2}+\overrightarrow{c}^2\leq
0
\\&\iff (\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})^2\leq 1.
\end{align*}
因此$|\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}|\leq 1$.当且仅当$(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{c})=0$时,最大值$1$取到.

 

题目6:$\overrightarrow{e_1},\overrightarrow{e_2}$均为单位向量,且它们的夹角为$45^{\circ}$,设$\overrightarrow{a},\overrightarrow{b}$满足$|\overrightarrow{a}+\overrightarrow{e_2}|=\frac{\sqrt{2}}{4}$,$\overrightarrow{b}=\overrightarrow{e_1}+k\overrightarrow{e_2}(k\in
\mathbf{R})$,则$|\overrightarrow{a}-\overrightarrow{b}|$的最小值为_________.

:\begin{align*}
|\overrightarrow{a}-\overrightarrow{b}|&=|(\overrightarrow{a}+\overrightarrow{e_2})-(\overrightarrow{b}+\overrightarrow{e_2})|
\\&=|(\overrightarrow{a}+\overrightarrow{e_2})-[\overrightarrow{e_1}+(1+k)\overrightarrow{e_2}]|
\\&\geq \bigg||\overrightarrow{a}+\overrightarrow{e_2}|-|\overrightarrow{e_1}+(1+k)\overrightarrow{e}_2|\bigg|
\\&=\bigg||\overrightarrow{e_{1}}+(1+k)\overrightarrow{e_{2}}|-\frac{\sqrt{2}}{4}\bigg|.
\end{align*}

\begin{align*}
|\overrightarrow{e_1}+(1+k)\overrightarrow{e_2}|^2&=\overrightarrow{e_1}^2+(1+k)^2\overrightarrow{e_2}^2+2(1+k)\overrightarrow{e_1}\cdot\overrightarrow{e_2}
\\&=(1+k)^2+\sqrt{2}(1+k)+1
\\&=(k+1+\frac{\sqrt{2}}{2})^2+\frac{1}{2}
\\&\geq \frac{1}{2},
\end{align*}
所以$|\overrightarrow{e_1}+(1+k)\overrightarrow{e_2}|\geq \frac{\sqrt{2}}{2}$,等号当且仅当$k=-1-\frac{\sqrt{2}}{2}$时取到.因此,
$$
\bigg||\overrightarrow{e_{1}}+(1+k)\overrightarrow{e_{2}}|-\frac{\sqrt{2}}{4}\bigg|\geq
|\overrightarrow{e_1}+(1+k)\overrightarrow{e_2}|-\frac{\sqrt{2}}{4}\geq \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}=\frac{\sqrt{2}}{4}.
$$
最小值$\frac{\sqrt{2}}{4}$当且仅当$\overrightarrow{a}+\overrightarrow{e_2}$与$\overrightarrow{b}+\overrightarrow{e_2}$反向,且$k=-1-\frac{\sqrt{2}}{2}$时取到.

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 Chapter 2.7:Classification of Conic Sections

(Exercise 8) Classify and sketch the curve {2xy-y^2=1}.

Solution:

\displaystyle 2xy-y^2=1\iff \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} 0&1\\ 1&-1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}=1.

\displaystyle \begin{bmatrix} 0&1\\ 1&-1 \end{bmatrix}= \begin{bmatrix} -\frac{2}{\sqrt{10+2 \sqrt{5}}}&\frac{1+\sqrt{5}}{\sqrt{10+2 \sqrt{5}}}\\ -\frac{2}{\sqrt{10-2 \sqrt{5}}}&\frac{1-\sqrt{5}}{\sqrt{10-2 \sqrt{5}}} \end{bmatrix}^{T} \begin{bmatrix} \frac{-1-\sqrt{5}}{2}&0\\ 0&\frac{-1+\sqrt{5}}{2} \end{bmatrix} \begin{bmatrix} -\frac{2}{\sqrt{10+2 \sqrt{5}}}&\frac{1+\sqrt{5}}{\sqrt{10+2 \sqrt{5}}}\\ -\frac{2}{\sqrt{10-2 \sqrt{5}}}&\frac{1-\sqrt{5}}{\sqrt{10-2 \sqrt{5}}} \end{bmatrix}

let

\displaystyle \begin{bmatrix} u\\ v \end{bmatrix}= \begin{bmatrix} -\frac{2}{\sqrt{10+2 \sqrt{5}}}&\frac{1+\sqrt{5}}{\sqrt{10+2 \sqrt{5}}}\\ -\frac{2}{\sqrt{10-2 \sqrt{5}}}&\frac{1-\sqrt{5}}{\sqrt{10-2 \sqrt{5}}} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix},

we have

\displaystyle \begin{bmatrix} u&v \end{bmatrix} \begin{bmatrix} \frac{-1-\sqrt{5}}{2}&0\\ 0&\frac{-1+\sqrt{5}}{2} \end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix}=1\iff -\frac{1+\sqrt{5}}{2}u^2+\frac{\sqrt{5}-1}{2}v^2=1.

So {2xy-y^2=1} is a hyperbola.The curve is plotted in red color as below(The curve plotted in black color is {-\frac{1+\sqrt{5}}{2}x^2+\frac{\sqrt{5}-1}{2}y^2=1}.):

\Box

(Exercise 9) Classify and sketch the curve {4x^2+2 \sqrt{2}xy+3y^2=1}.

Solution:

\displaystyle 4x^2+2 \sqrt{2}xy+3y^2=1\iff \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} 4&\sqrt{2}\\ \sqrt{2}&3 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}=1.

\displaystyle \begin{bmatrix} 4&\sqrt{2}\\ \sqrt{2}&3 \end{bmatrix}= \begin{bmatrix} \frac{\sqrt{6}}{3}&\frac{\sqrt{3}}{3}\\ \frac{\sqrt{3}}{3}&-\frac{\sqrt{6}}{3} \end{bmatrix}^T \begin{bmatrix} 5&0\\ 0&2 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{6}}{3}&\frac{\sqrt{3}}{3}\\ \frac{\sqrt{3}}{3}&-\frac{\sqrt{6}}{3} \end{bmatrix}.

Let

\displaystyle \begin{bmatrix} u\\ v \end{bmatrix}= \begin{bmatrix} \frac{\sqrt{6}}{3}&\frac{\sqrt{3}}{3}\\ \frac{\sqrt{3}}{3}&-\frac{\sqrt{6}}{3} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix},

then

\displaystyle \begin{bmatrix} u&v \end{bmatrix} \begin{bmatrix} 5&0\\ 0&2 \end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix}=1\iff 5u^2+2v^2=1.

So {4x^2+2 \sqrt{2}xy+3y^2=1} is an ellipse.Its curve is plotted in red as below(The curve plotted in black is {5x^2+2y^2=1}.)

\Box

(Exercise 10) Classify and sketch the curve {x^2-2xy+y^2=1}.

Solution:

\displaystyle x^2-2xy+y^2=1\iff \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}=1.

\displaystyle \begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix}= \begin{bmatrix} \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2} \end{bmatrix}^T \begin{bmatrix} 0&0\\ 0&2 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2} \end{bmatrix}.

Let

\displaystyle \begin{bmatrix} u\\ v \end{bmatrix}= \begin{bmatrix} \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix},

then

\displaystyle \begin{bmatrix} u&v \end{bmatrix} \begin{bmatrix} 0&0\\ 0&2 \end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix}=1\iff v^2=\frac{1}{2}.

So {x^2-2xy+y^2=1} is two parallel lines.Its curve is plottted in red as below(The curveg plotted in black is {y^2=\frac{1}{2}}.)

\Box

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下面我们求矩阵

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

的Schur分解.设线性变换{T:\mathbf{C}^3\rightarrow \mathbf{C}^3}{\alpha_0=(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)}下的矩阵 {[T]_{\alpha_0}^{\alpha_0}=A}.

易求得{A}的特征值是{-3,-3,6}.{A}的特征值{-3}对应于某个单位 特征向量{\mathbf{v}_1^{(1)}= \begin{bmatrix} \frac{-2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\\ 0 \end{bmatrix}_{\alpha_{0}} }.然后再在{\mathbf{C}^3}中选取两个向量{\mathbf{v}_2^{(1)},\mathbf{v}_3^{(1)}},使得 {\alpha_{1}=(\mathbf{v}_1^{(1)},\mathbf{v}_2^{(1)},\mathbf{v}_3^{(1)})}构成 {\mathbf{C}^3}的一组单位正交基底.不妨令

\displaystyle \mathbf{v}_2^{(1)}= \begin{bmatrix} \frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\\ 0 \end{bmatrix}_{\alpha_{0}},\mathbf{v}_3^{(1)}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}_{\alpha_{0}}.

则线性变换{T}在基{\alpha_1}下的矩阵为

\displaystyle A_1= \begin{bmatrix} -3&0&0\\ 0&2&\frac{10}{\sqrt{5}}\\ 0&\frac{10}{\sqrt{5}}&1 \end{bmatrix}.

然后将矩阵{A_1}的第{1}列和第{1}行去掉,得到其子矩阵

\displaystyle B_1= \begin{bmatrix} 2&\frac{10}{\sqrt{5}}\\ \frac{10}{\sqrt{5}}&1 \end{bmatrix}.

事实上,{\mbox{span}\{\mathbf{v}_2^{(1)},\mathbf{v}_{3}^{(1)}\}}形成{\mathbf{C}^3} 的一个二维子空间{W^2}.且{\beta_{1}=(\mathbf{v}_2^{(1)},\mathbf{v}_3^{(1)})}形成{W^2}的一组基 底.存在线性变换{G_1:W^2\rightarrow W^2},使得{G_1}在基{\beta_1}下的矩阵 {[G_1]_{\beta_1}^{\beta_1}}就是矩阵{B_1}.矩阵{B_1}的特征值为{-3,6}.特 征值{-3}对应于某个单位特征向量{ \begin{bmatrix} -\frac{2}{3}\\ \frac{\sqrt{5}}{3} \end{bmatrix}_{\beta_1} }. 因此矩阵{A}的特征值{-3}还对应于某个单位特征向量{\mathbf{v}_2^{(2)}= \begin{bmatrix} 0\\ -\frac{2}{3}\\ \frac{\sqrt{5}}{3} \end{bmatrix}_{\alpha_1} }.

{\alpha_2=(\mathbf{v}_1^{(1)},\mathbf{v}_2^{(2)},\mathbf{v}_3^{(2)})} 构成{\mathbf{C}^3}的一组单位正交基底,不妨设{\mathbf{v}_3^{(2)}= \begin{bmatrix} 0\\ \frac{\sqrt{5}}{3}\\ \frac{2}{3} \end{bmatrix}_{\alpha_1} }.线性变换{T}在基{\alpha_2}下的矩阵为

\displaystyle [T]_{\alpha_2}^{\alpha_2}= \begin{bmatrix} -3&0&0\\ 0&-3&0\\ 0&0&6 \end{bmatrix}.

\displaystyle \begin{array}{rcl} \alpha_2&=&\left( \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}_{\alpha_1}, \begin{bmatrix} 0\\ -\frac{2}{3}\\ \frac{\sqrt{5}}{3} \end{bmatrix}_{\alpha_1}, \begin{bmatrix} 0\\ \frac{\sqrt{5}}{3}\\ \frac{2}{3} \end{bmatrix}_{\alpha_1} \right) \\&=&\left( \begin{bmatrix} -\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\\ 0 \end{bmatrix}_{\alpha_0}, \begin{bmatrix} \frac{-2}{3 \sqrt{5}}\\ \frac{-4}{3 \sqrt{5}}\\ \frac{\sqrt{5}}{3} \end{bmatrix}_{\alpha_0}, \begin{bmatrix} \frac{1}{3}\\ \frac{2}{3}\\ \frac{2}{3} \end{bmatrix}_{\alpha_{0}} \right) \end{array}

因此矩阵

\displaystyle A=[I]_{\alpha_{2}}^{\alpha_{0}}[T]_{\alpha_2}^{\alpha_{2}}[I]_{\alpha_0}^{\alpha_2}= \begin{bmatrix} -\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}&0\\ -\frac{2}{3 \sqrt{5}}&-\frac{4}{3 \sqrt{5}}&\frac{\sqrt{5}}{3}\\ \frac{1}{3}&\frac{2}{3}&\frac{2}{3} \end{bmatrix}^{T} \begin{bmatrix} -3&0&0\\ 0&-3&0\\ 0&0&6 \end{bmatrix}\begin{bmatrix} -\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}&0\\ -\frac{2}{3 \sqrt{5}}&-\frac{4}{3 \sqrt{5}}&\frac{\sqrt{5}}{3}\\ \frac{1}{3}&\frac{2}{3}&\frac{2}{3} \end{bmatrix}

这就是实对称矩阵{A}的Schur分解.

事实上,对于任意的Hermite矩阵{A}来说,由 Schur分解定理,都存在酉矩阵{O}和上三角矩阵{U},使得{A=\overline{O}^{T}UO},因此 {\overline{A}^{T}=\overline{O}^{T}\overline{U}^{T}O},由于 {\overline{A}^T=A},因此{U=\overline{U}^T},表明上三角矩阵{U}只能是实对 角矩阵{D}.可见,Hermite矩阵{A}的特征值都是实数.

特别地,当{A}是实对称矩阵时,由实对称矩阵{A}进行Schur分解的过程可以断 定,{A}必定存在一组互相正交的单位特征向量.

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题目 1{\max\{a,b\}}{a,b}两数的最大值,当正数{x,y(x>y)}变化时,{t=\max \left\{x^2,\cfrac{25}{y(x-y)}\right\}}的最小值为{\underline{~~~~~~~~~~}}.

解:

\displaystyle \begin{cases} t\geq x^2\\ t\geq \cfrac{25}{y(x-y)}\geq \cfrac{25}{\left[\cfrac{y+(x-y)}{2}\right]^{2}}=\cfrac{100}{x^2}. \end{cases}

所以{t^2\geq x^2\cdot \cfrac{100}{x^2}=100},因此{t\geq 10}.当且仅当 {x=\sqrt{10},y=\cfrac{\sqrt{10}}{2}}时,{t}取到最小值.答案是{10}. \Box

题目 2 试求{u(x,y)=\max\{ |1+x+y|,|4+2x+y|,|9+3x+y|\}}的最小值.

解:

\displaystyle \begin{cases} u(x,y)\geq |1+x+y|\\ u(x,y)\geq |4+2x+y|\\ u(x,y)\geq |9+3x+y| \end{cases},

如图1所示,设{xOy}平面直角坐标系中有直线

\displaystyle l_1:1+x+y=0,l_2:4+2x+y=0,l_3:9+3x+y=0.

记直线{l_1}{l_2}的交点为{A(-3,2)},{l_2}{l_3}的交点为 {B(-5,6)},{l_3}{l_1}的交点为{C(-4,3)}.则

\displaystyle |AB|=2 \sqrt{5},|BC|=\sqrt{10},|CA|=\sqrt{2}.

{P(x,y)}{xOy}平面直角 坐标系中的任意一点.则由点到直线的距离公式以及三角形的面积公式,可得

\displaystyle S_{\triangle PAB}=2 \sqrt{5}\frac{|4+2x+y|}{2\sqrt{2^2+1^2}}=|4+2x+y|,

\displaystyle S_{\triangle PBC}=\sqrt{10}\frac{|9+3x+y|}{2\sqrt{3^2+1^2}}=\frac{|9+3x+y|}{2},

\displaystyle S_{\triangle PCA}=\sqrt{2} \frac{|1+x+y|}{2 \sqrt{1^2+1^2}}=\frac{|1+x+y|}{2}.

\displaystyle S_{\triangle PAB}+S_{\triangle PBC}+S_{\triangle PCA}\geq S_{\triangle ABC}=1,

\displaystyle |4+2x+y|+\frac{|9+3x+y|}{2}+\frac{|1+x+y|}{2}\geq 1, \ \ \ \ \ (1)

等号当且仅当点{P}位于三角形{ABC}区域内部或边界上时成立.事实上,直接使用 绝对值不等式也可以得到不等式(1),具体如下:

\displaystyle \begin{array}{rcl} |4+2x+y|+\frac{|9+3x+y|}{2}+\frac{|1+x+y|}{2}&=&|-4-2x-y|+|\frac{9}{2}+\frac{3}{2}x+\frac{y}{2}|+|\frac{1}{2}+\frac{x}{2}+\frac{y}{2}| \\&\geq &|(-4-2x-y)+(\frac{9}{2}+\frac{3}{2}x+\frac{y}{2})+(\frac{1}{2}+\frac{x}{2}+\frac{y}{2})| \\&=&1. \end{array}

由不等式(1)可得 {u(x,y)+\frac{1}{2}u(x,y)+\frac{1}{2}u(x,y)\geq 1},因此{u(x,y)\geq \frac{1}{2}}.当且仅当{x=-4,y=\frac{11}{3}}时,{u(x,y)}取得最小值 {\frac{1}{2}}.我们发现{(-4,\frac{11}{3})}正好是{\triangle ABC}的重心, 这里面是否有玄机呢? \Box