$color{blue}{第五章 相似矩阵及二次型}$

$color{blue}{S 第五章 第一节 向量的内积}$

$color{blue}{一、向量的内积}$

$color{blue}{1.内积的概念}$

$定义1.设有n维向量\ x = begin{pmatrix} x_1 \ x_2 \ vdots \ x_n end{pmatrix}, y = begin{pmatrix} y_1 \ y_2 \ vdots \ y_n end{pmatrix} \ 令[x, y] = x_1y_1 + x_2y_2 + cdots + x_ny_n,\ 称[x, y]为向量x与y的内积.$
$1)内积是一个数(或一个多项式)$
$2)内积是向量的一种运算,可用矩阵的运算.$
$列向量: [x, y] = x^T y$
$行向量: [x, y] = xy^T$

$color{blue}{2.内积的性质}$

$1)对称性:[x, y] = [y, x];$
$2)齐次性:[lambda x, y] = lambda[x, y];$
$3)线性性:[x+y, z] = [x, z] + [y, z].$

$color{blue}{二、向量的范数与夹角}$

$color{blue}{1.向量的范数(长度)}$

$定义2.令 Vert x Vert = sqrt{[x, x]} = sqrt{x_1^2 + x_2^2 + cdots + x_n^2}\ 称Vert x Vert 为n维向量x的范数.$

$color{blue}{2.范数的性质}$

$1) 非负性:当x neq 0时,Vert x Vert > 0; 当x = 0时, Vert x Vert = 0;$
$2) 齐次性:Vert lambda x Vert = lambda Vert x Vert;$
$3) 三角不定式: Vert x+y Vert leq Vert x Vert + Vert y Vert .$

$color{blue}{3.单位向量}$

$称Vert x Vert = 1时的向量x为单位向量.任意alpha neq 0, e = dfrac{alpha}{Vert alpha Vert}为单位向量.$

$color{blue}{4.许瓦兹不定式}$

$[x, y]^2 leq [x, x][y, y], \ 即left| dfrac{[x, y]}{Vert x Vert Vert y Vert } right| leq 1 (当Vert x Vert neq 0, Vert y Vert neq 0 时)$

$color{blue}{5.向量的夹角}$

$theta = arccos dfrac{[x, y]}{Vert x Vert Vert y Vert } (当Vert x Vert neq 0, Vert y Vert neq 0 时)$

$color{blue}{三.向量组的正交性}$

$color{blue}{1.正交}$

$设x、y为n维向量,当[x, y] = 0时,称x与y正交的.$
$若x = 0,则x与任何向量都正交.$

$color{blue}{2.正交向量组}$

$正交向量组：指一组两两正交的非零向量组.$

$定理1.若n维向量alpha_1, alpha_2, cdots, alpha_r是一组两两正交的非零向量,\ 则alpha_1, alpha_2, cdots, alpha_r线性无关.$
$证:设有lambda_1, lambda_2, cdots, lambda_r,使\ lambda_1 alpha_1 + lambda_2 alpha_2 + cdots + lambda_r alpha_r = 0, \ 取alpha_i(i = 1, 2, cdots, r)在上式两端作内积. \ [lambda_1 alpha_1 + lambda_2 alpha_2 + cdots + lambda_r alpha_r, a_i] = [0, alpha_i] \ [lambda_i alpha_i, alpha_i] = 0 \ 亦即 lambda_i[alpha_i, alpha_i] = 0 \ 因a_i neq 0,故[alpha_i, alpha_i] = Vert a_i Vert ^2 neq 0,\ 从而必有lambda_i = 0(i = 1, 2, cdots, r),\ 于是向量组alpha_1, alpha_2, cdots, alpha_a 线性无关.$

$color{blue}{3.正交基}$

$用正交向量组作向量空间的基,称为向量空间的正交基.$
$例如:n个两两正交的n维非零向量,\ 可构成向量空间R^n的一个正交基.$

$例1.已知3维向量空间R^3中的两个向量\ alpha_1 = begin{pmatrix}1 \ 1 \ 1 end{pmatrix}, alpha_2 = begin{pmatrix}1 \ -2 \ 1 end{pmatrix} \ 正交,式求一个非零向量alpha_3,使alpha_1, alpha_2, alpha_3两两正交.$
$解:设所求的向量alpha_3 = (x_1, x_2, x_3)^T \ 依题意可得:[alpha_1, alpha_3] = 0, [alpha_2, alpha_3] = 0,即$
$left { begin{array}{l} x_1 + x_2 + x_3 = 0 \ x_1 -2x_2 + x_3 = 0 end{array} right.$
$由于A = begin{pmatrix} 1 & 1 & 1 \ 1 & -2 & 1 end{pmatrix} sim begin{pmatrix} 1 & 1 & 1 \ 0 & -3 & 0 end{pmatrix} \ sim begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 end{pmatrix}$
$得left { begin{array}{l} x_1 = -x_3 \ x_2 = 0 end{array} right.$
$从而有基础解系begin{pmatrix} -1 \ 0 \ 1 end{pmatrix}, 取alpha_3 = begin{pmatrix} -1 \ 0 \ 1 end{pmatrix} ,即为所求.$

$color{blue}{四.规范正交基(标准正交基)}$

$color{blue}{1.规范正交基的概念}$

$定义3.设有n维向量e_1, e_2, cdots, e_r是向量空间V(V subset R^n)的一个基,\ 如果e_1, e_2, cdots, e_r是两两正交的单位向量,\ 则称e_1, e_2, cdots, e_r是向量空间V的一个规范正交基. \ 显然,若e_1, e_2, cdots, e_r是V的一个规范正交基.则$
$[e_i, e_j] = left { begin{array}{l} 1 quad i = j \ 0 quad i neq j end{array} right.$
$例如:\ e_1 = begin{pmatrix} dfrac{1}{sqrt{2}} \ dfrac{1}{sqrt{2}} \ 0 \ 0 end{pmatrix}, e_2 = begin{pmatrix} dfrac{1}{sqrt{2}} \ -dfrac{1}{sqrt{2}} \ 0 \ 0 end{pmatrix}, e_3 = begin{pmatrix} 0 \ 0 \ dfrac{1}{sqrt{2}} \ dfrac{1}{sqrt{2}} end{pmatrix}, e_4 = begin{pmatrix} 0 \ 0 \ dfrac{1}{sqrt{2}} \ -dfrac{1}{sqrt{2}} end{pmatrix},$
$由于 [e_i, e_j] = left { begin{array}{l} 1 quad i = j \ 0 quad i neq j end{array} right.（i, j = 1, 2, 3, 4)$
$所以e_1, e_2, e_3, e_4是一个规范正交基.$

$color{blue}{2.向量的坐标}$

$设e_1, e_2, cdots, e_n是V的一个规范正交基,\ 那么V中任何一向量alpha = (x_1, x_2, cdots, x_n)\ 应能由e_1, e_2, cdots, e_n 线性表示,表示法为:\ alpha = x_1e_1 + x_2e_2 + cdots + x_ne_n \ 为求表示法中的系数x_i,可用e_i与alpha作内积(i = 1, 2, cdots, n), \ [alpha, e_i] = [x_1e_1 + x_2e_2 + cdots + x_ne_n, e_i] \ = x_i[e_i, e_i] = x_i \ 称x_1, x_2, cdots, x_n是alpha在基e_1, e_2, cdots, e_n下的坐标.$

$color{blue}{3.施密特标准正交化}$

$设alpha_1, alpha_2, cdots, alpha_r是向量空间V的一个基, \ 令b_1 = alpha_1, quad e_1 = dfrac{b_1}{Vert b_1 Vert} \ b_2 = alpha_2 - [alpha_2, e_1]e_1, quad e_2 = dfrac{b_2}{Vert b_2 Vert } \ b_3= alpha_3 - [alpha_3, e_1]e_1 - [alpha_3, e_2]e_2, quad e_3= dfrac{b_3}{Vert b_3 Vert } \ cdots qquad cdots qquad cdots \ b_r = alpha_r - [alpha_r, e_1]e_1 - cdots - [alpha_r, e_{r-1}]e_{r-1}, quad e_r = dfrac{b_r}{Vert b_r Vert } \ 可以证明,e_1, e_2, cdots, e_r是两两正交的单位向量,\ 故e_1, e_2, cdots, e_r是V的一个规范正交基.$

$例2.设\ alpha_1 = begin{pmatrix}1 \ 2 \ -1 end{pmatrix}, alpha_2 = begin{pmatrix}-1 \ 3 \ 1 end{pmatrix}, alpha_3 = begin{pmatrix}4 \ -1 \ 0 end{pmatrix}, \ 试用施密特正交化过程把这组向量规范正交化.$
$解:取 b_1 = alpha_1, e_1 = dfrac{b_1}{Vert b_1 Vert} = dfrac{1}{sqrt{6}}begin{pmatrix}1 \ 2 \ -1 end{pmatrix}, \ b_2 = alpha_2 - [alpha_2, e_1]e_1 \ = alpha_2 - left[ alpha_2, dfrac{b_1}{Vert b_1 Vert } right] cdot dfrac{b_1}{Vert b_1 Vert } \ = alpha_2 - dfrac{[alpha_2, b_1] b_1}{Vert b_1 Vert ^2} \ = begin{pmatrix}-1 \ 3 \ 1 end{pmatrix} - dfrac{4}{6}begin{pmatrix}1 \ 2 \ -1 end{pmatrix} \ = dfrac{5}{3}begin{pmatrix}-1 \ 1 \ 1 end{pmatrix} \ e_2 = dfrac{b_2}{Vert b_2 Vert } = dfrac{1}{sqrt{3}} begin{pmatrix}-1 \ 1 \ 1 end{pmatrix} \ b_3 = alpha_3 - dfrac{[alpha_3, b_1]}{Vert b_1 Vert ^2} b_1 - dfrac{[alpha_3, b_2]}{Vert b_2 Vert ^2} b_2 \ = begin{pmatrix}4 \ -1 \ 0 end{pmatrix} -dfrac{1}{3}begin{pmatrix}1 \ 2 \ -1 end{pmatrix} + dfrac{5}{3}begin{pmatrix}-1 \ 1 \ 1 end{pmatrix} \ = 2begin{pmatrix}1 \ 0 \ 1 end{pmatrix} \ e_3 = dfrac{b_3}{Vert b_3 Vert } = dfrac{1}{sqrt{2}}begin{pmatrix}1 \ 0 \ 1 end{pmatrix} \ 故e_1, e_2, e_3即合所求.$

$例3.已知alpha_3 = (1, 1, 1)^T \ 求非零向量alpha_1, alpha_2,使alpha_3与alpha_1, alpha_2正交,\ 并把它们化成R^3的规范正交基.$
$解:alpha_1, alpha_2应满足alpha_3^Tx = 0的非零解,即\ x_1 + x_2 + x_3 = 0 \ 它的基础解系为xi_1 = begin{pmatrix}1 \ 0 \ -1 end{pmatrix}, xi_2 = begin{pmatrix}0 \ 1 \ -1 end{pmatrix}, \ 令alpha_1 = xi_1, alpha_2 = xi_2, \ 则alpha_3 与 alpha_1, alpha_2正交,显然alpha_1与alpha_2线性无关,\ 因此可用施密特标准正交化. \ 取 b_1 = alpha_1, 则e_1 = dfrac{b_1}{Vert b_1 Vert } = dfrac{1}{sqrt{2}} begin{pmatrix}1 \ 0 \ -1 end{pmatrix}, \ 取 b_2 = alpha_2 - [alpha_2, e_1]e_1 = begin{pmatrix}-dfrac{1}{2} \ 0 \ -dfrac{1}{2} end{pmatrix} \ 则e_2 = dfrac{b_2}{Vert b_2 Vert } = dfrac{1}{sqrt{2}}begin{pmatrix}-1 \ 2 \ -1 end{pmatrix} \ 再把alpha_3单位化,得e_3 = dfrac{alpha_3}{Vert alpha_3 Vert } = dfrac{1}{sqrt{3}}begin{pmatrix}1 \ 1 \ 1 end{pmatrix}, \ 故e_1, e_2, e_3即为R^3的一个规范正交基.$

$color{blue}{五.正交矩阵与正交变换}$

$color{blue}{1.正交矩阵}$

$定义4.如果n阶矩阵A满足\ A^TA = E(即A^{-1} = A^T),那么称A为正交矩阵.$
$将矩阵A按列分块,即A = (alpha_1, alpha_2, cdots, alpha_n), \ 则begin{pmatrix} alpha_1^T \ alpha_2^t \ vdots \ alpha_n^T end{pmatrix}begin{pmatrix}alpha_1, alpha_2, cdots, alpha_n end{pmatrix} = E.$
$亦即(alpha_i^T, alpha_j) = delta_{ij} = left { begin{array}{l} 1 quad i = j \ 0 quad i neq j end{array} right.$
$这说明方阵A为正交矩阵的充分必要条件是\ A的列向量都是两两正交的单位向量. \ 同理可得方阵A为正交矩阵的充分必要条件是\ A的行向量都是两两正交的单位向量.$

$例4.验证\ P = begin{pmatrix} dfrac{1}{2} & -dfrac{1}{2} & dfrac{1}{2} & -dfrac{1}{2} \ dfrac{1}{2} & -dfrac{1}{2} & -dfrac{1}{2} & dfrac{1}{2} \ dfrac{1}{sqrt{2}} & dfrac{1}{sqrt{2}} & 0 & 0 \ 0 & 0 & dfrac{1}{sqrt{2}} & dfrac{1}{sqrt{2}} \ end{pmatrix}是正交矩阵.$
$解:显然P的每个列向量是两两正交的单位向量.所以P为正交矩阵.$

$例5.设e_1, e_2, cdots, e_n是R^n的一个规范正交基.A为正交矩阵. \ 试证:Ae_1, Ae_2, cdots, Ae_n也是R^n的一个规范正交基.$
$证:由于\ [Ae_i, Ae_j] = (Ae_i)^TAe_j = e_i^TA^TAe_j \ = e_i^Te_j = left lbrace begin{array}{l} 1 quad i = j \ 0 quad i neq j end{array} right.$

$color{blue}{2.正交变换}$

$定义5.设P为正交矩阵,则线性变换y = Px称为正交变换.$
$设y = Px为正交变换,则有 \ Vert y Vert = sqrt{y^T y} \ = sqrt{x^TP^TPx} \ = sqrt{x^Tx} \ = Vert x Vert \ 按Vert x Vert表示向量长度,Vert x Vert = Vert y Vert 说明经过正交变换, \ 向量的长度保持不变,这是正交变换的优良特性.$

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