线性代数 06.04 综合模拟 02

一、填空题
$1.四阶方阵A的特征值为1、3、4、5,\\ 则a_{11} + a_{22} + a_{33} + a_{44} = \underline{\ \ {\color{blue}{13} } \ \ }; |A| = \underline{\ \ {\color{blue}{60} } \ \ }.$
$解: a_{11} + a_{22} + a_{33} + a_{44} = 1 + 3 + 4 + 5 = 13 \\ |A| = \lambda_1 \lambda_2 \lambda_3 \lambda_4 = 1 \times 3 \times 4 \times 5 = 60$

$2.设 A = \begin{pmatrix} \lambda_1 & & \\ & \lambda_2 & \\ & & \lambda_3 \end{pmatrix}, B = \begin{pmatrix} \lambda_2 & & \\ & \lambda_3 & \\ & & \lambda_1 \end{pmatrix}, \\ 则可存在可逆矩阵P,使P^{-1}AP = B,其中P = \underline{\ \ {\color{blue}{\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix}}} \ \ }.$
$解: \\ A经过第一行与第二行交换,再进行第一列与第二列交换, \\ 然后第二行与第三行交换,再进行第二行与第三列交换即得B. \\ 而每次初等行变换相当于用一个初等矩阵左乘矩阵A,每次\\ 列变换相当于用一个初等矩阵右乘矩阵A,所用的初等矩阵为\\ P_1 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}, P_2 = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix}, \\ 因为P_1^{-1} = P_1, P_2^{-1} = P_2 \\ 所以P_2^{-1}P_1^{-1}AP_1P_2 = B \\ \therefore P = P_1P_2 = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix}$

$3.已知四阶行列式D的第三行元素分别为-1, 3, 2, 0,\\ 第二行元素的余子式依次为5,-2, a, 4,则a = \underline{ \ \ {\color{blue}{-\dfrac{1}{2}} }\ \ }.$
$解: 因为行列式第三行元素与第二行元素对应的代数\\ 余子式乘积之和为零,所以有:\\ -1 \times (-1)^{(2 + 1)} \times 5 + 3 \times (-1)^{(2 + 2)} \times (-2) \\ + 2 \times (-1)^{(2 + 3)} \times a + 0 \times (-1)^{(2 + 4)} \times 4 = 0 \\ 5 - 6 -2a + 0 = 0 \\ a = -\dfrac{1}{2}$

$4.已知A是满秩方阵,且AB = \begin{pmatrix} 1 & 2 &3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \\ \end{pmatrix}, \\ 则B的秩为\underline{ \ \ {\color{blue}{1} } \ \ }$
$解:因为A为满秩矩阵,所以A可以写成有限个初等\\ 矩阵的乘积, 用有限个初等矩阵左乘矩阵B,相当于\\ 对矩阵B进行了有限次初等行变换, 而初等变换不改\\ 变矩阵的秩,所以矩阵B的秩等于AB的秩.而AB的\\ 秩为1,所以B的秩为1.$

$5.设1是实对称矩阵A的一个特征值,\\ 且B = A^3 - A,则|B| = \underline{\ \ {\color{blue}{0} } \ \ }.$
$解: \\ \because 1为实对称矩阵A的一个特征值，\\ \therefore |A - E| = 0 \\ B = A(A^2 - E) = A(A + E)(A - E) \\ |B| = |A(A + E)(A - E)| \\ = |A||A+E||A - E| \\ = |A||A+E| \cdot 0 = 0$

二、选择题
$1.设\alpha_1, \alpha_2, \alpha_3线性无关,则下列向量组线性相关的是(\ \ {\color{blue}{A} }\ \ )$
$A. \alpha_1 - \alpha_2, \alpha_2 - \alpha_3, \alpha_3 - \alpha_1; \\ B. \alpha_1 + \alpha_2, \alpha_2 + \alpha_3, \alpha_3 + \alpha_1; \\ C. \alpha_1, \alpha_1 + \alpha_2, \alpha_1 + \alpha_3; \\ D. \alpha_1, \alpha_2, \alpha_3 - \alpha_1 .$
$解: \\ A中每个向量都可由其它两个向量线性表示. 如:\\ \alpha_1 - \alpha_2 = -[(\alpha_2 - \alpha_3) + (\alpha_3 - \alpha_1)] \\ \\设一组数k_1, k_2, k_3,使k_1(\alpha_1 - \alpha_2) + k_2(\alpha_2 - \alpha_3) + k_3(\alpha_3 - \alpha_1) = 0 \\ (k_1 - k_3) \alpha_1 + (k_2 - k_1)\alpha_2 + (k_3 - k_2) \alpha_3 = 0 \\ \because \alpha_1, \alpha_2, \alpha_3线性无关,所以 \\ k_1 - k_3 = 0 \\ k_2 - k_1 = 0 \\ k_3 - k_2 = 0 \\ 解得k_1 = k_2 = k_3 \\ 令k_1= k_2 = k_3 = 1,则有一组不全为零的数,使 \\ (\alpha_1 - \alpha_2) + (\alpha_2 - \alpha_3) + (\alpha_3 - \alpha_1) = 0 \\ 故选A.$

$2. 设A是n阶矩阵,且A的行列式|A| = 0, 则A中(\ \ {\color{blue}{C} }\ \ ).$
$A. 必有一列元素为零; \\ B. 必有两列元素对应成比例; \\ C. 必有一列向量是其余向量的线性组合; \\ D. 任一列向量是其余列向量的线性组合.$
$解: 由|A| = 0,可知,A的列向量组是线性相关的,\\ 所以其中至少有一个列向量可由其余列向量线性表示,\\ 因此选C.$

$3. 设A,B均是n阶正交阵,若|A| + |B| = 0,\\ 则 A + B必为(\ \ {\color{blue}{D} }\ \ )$
$A. 初等阵; \quad B. 正交阵; \quad C. 对称阵; \quad D. 奇异阵.$
$解: \\ \because A^{-1}(A+B)B^{-1} = B^{-1} + A^{-1} \\ \therefore A^{-1}(A + B) = (B^{-1} + A^{-1})B \\ \therefore |A^{-1}||A + B| = |B^{-1} + A^{-1}||B| = |B||A^{-1} + B^{-1}| \\ \because A^{-1} = A^T, B^{-1} = B^T, |A^T| = |A| \\ \therefore (|A| - |B|)| A + B | = 0 \\ \because |A| + |B| = 0 \\ \therefore |A| - |B| \neq 0 \\ \therefore |A + B| = 0$

$4. 已知 A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \\ \end{pmatrix}, B = \begin{pmatrix} 2 & 4 & 6 \\ 1 & 2 & 3 \\ 4 & 8 & 12 \\ \end{pmatrix}, \\ P_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}, P_2 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}, \\ 则B = (\ \ {\color{blue}{C} } \ \ )$
$A. AP_2P_1; \quad B. AP_1P_2; \quad C. P_2P_1A; \quad D. P_1P_2A.$
$解: \\ 1.先 r_3 + r_1 , 2. r2 \leftrightarrow r_2$

$5. 设\beta能由\alpha_1, \alpha_2, \alpha_3线性表示, 但不能由\alpha_1, \alpha_2线性表示,则(\ \ {\color{blue}{A} } \ \ )$
$A. \alpha_3不能由\alpha_1, \alpha_2线性表示，但能由\beta, \alpha_1, \alpha_2线性表示; \\ B. \alpha_3不能由\alpha_1, \alpha_2线性表示，也不能由\beta, \alpha_1, \alpha_2线性表示; \\ C. \alpha_3能由\alpha_1, \alpha_2线性表示,但不能由\beta,\alpha_1, \alpha_2线性表示; \\ D. \alpha_3能由\alpha_1,\alpha_2线性表示,也能由\beta,\alpha_1, \alpha_2线性表示.$
$解: \\ 若\alpha_3能由\alpha_1,\alpha_2线性表示,因为\beta能由\alpha_1, \alpha_2, \alpha_3线性表示,\\ 则\beta能由\alpha_1,\alpha_2线性表示,与已知矛盾,\\ 所以\alpha_3不能由\alpha_1,\alpha_2线性表示.C、D错误. \\ \because \beta = \lambda_1 \alpha_1 + \lambda_2 \alpha_2 + \lambda_3 \alpha_3 \\ \therefore \lambda_3 \neq 0, [\because \lambda_3 = 0时,将使\beta可由\alpha_1, \alpha_2线性表示.] \\ \therefore \alpha_3 = \dfrac{1}{\lambda_3}\beta - \dfrac{\lambda_1}{\lambda_3} \alpha_1 - \dfrac{\lambda_2}{\lambda_3} \alpha_3 \\ 即\alpha_3能由\beta,\alpha_1,\alpha_2线性表示.\\ 故选A.$

三、计算题
$1.解矩阵方程 \\ \left[ x \begin{pmatrix} 1 & 2 & 0 \\ 2 & 0 & 2 \end{pmatrix} \right] ^ T = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}.$
$解: \begin{pmatrix} 1 & 2 \\ 2 & 0 \\ 0 & 2 \end{pmatrix} x^T = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$
$对增广矩阵B施行初等行变换 \\ B = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 0 & -2 \\ 0 & 2 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 1 \\ 0 & -4 & -4 \\ 0 & 2 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} \\ \therefore x^T = \begin{pmatrix} -1 \\ 1 \end{pmatrix}, \\ \therefore x = \begin{pmatrix} -1 & 1 \end{pmatrix}$

$2.设A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 2 & -3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}, 求A^{-1}.$
$解: 对矩阵(A | E) 作初等行变换 \\ (A | E) = \begin{pmatrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 2 & -3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & -2 & 7 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ A^{-1} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & -2 & 7 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$

$3.验证\\ \alpha_1 = (1, -1, 0 )^T, \alpha_2 = (2, 1, 3)^T, \alpha_3(3, 1, 2)^T, \\ 为R^3的一个基,并求\beta = (5, 0, 7)^T这组基下的坐标.$
$解: 令 A = (\alpha_1, \alpha_2, \alpha_3) \\ \because |A| = \begin{vmatrix} 1 & 2 & 3 \\ -1 & 1 & 1 \\ 0 & 3 & 2 \\ \end{vmatrix} = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 3 & 4 \\ 0 & 3 & 2 \\ \end{vmatrix} = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 3 & 4 \\ 0 & 0 & -2 \\ \end{vmatrix} = -6 \neq 0 \\ \therefore \alpha_1, \alpha_2, \alpha_3为R^3的一个基. \\ 令\beta = k_1 \alpha_1 + k_2 \alpha_2 + k_3 \alpha_3 \\ 求k_1, k_2, k_3就是解方程组\\ (\alpha_1, \alpha_2, \alpha_3) \begin{pmatrix}k_1 \\ k_2 \\ k_3 \end{pmatrix} = \beta \\ 对矩阵(\alpha_1, \alpha_2, \alpha_3, \beta)作初等行变换\\ \begin{pmatrix} 1 & 2 & 3 & 5 \\ -1 & 1 & 1 & 0 \\ 0 & 3 & 2 & 7 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 3 & 5 \\ 0 & 3 & 4 & 5 \\ 0 & 3 & 2 & 7 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 1 & 0 & \frac{1}{3} & \frac{5}{3} \\ 0 & 3 & 4 & 5 \\ 0 & 0 & -2 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & \frac{1}{3} & \frac{5}{3} \\ 0 & 1 & \frac{4}{3} & \frac{5}{3} \\ 0 & 0 & 1 & -1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \\ \end{pmatrix} \\ 所以\beta在这组基下的坐标为2, 3, -1.$

$4. 设A = \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 1 \\ -2 & 1 & 0 & 1 \\ 1 & -2 & 1 & 0 \\ \end{pmatrix} \\ 求矩阵A的秩及A的列向量的极大无关组.$
$解:对A施以行初等变换 \\ A = \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 1 \\ -2 & 1 & 0 & 1 \\ 1 & -2 & 1 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 1 \\ 0 & 1 & 2 & -3 \\ 0 & -2 & 0 & 2 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 4 & -4 \\ 0 & 0 & -4 & 4 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \\ 所以矩阵A的秩为3，第一列、第二列、第三列为A的列向量的一个极大无关组.$

四、证明题
$1.若n阶可逆阵A的任意行元素之和都等于a, \\ 证明:a为矩阵A的一个特征值,且a \neq 0.$
$证:由已知\\ \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \cdots & \cdots & \ddots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} = \begin{pmatrix} a \\ a \\ \vdots \\ a \end{pmatrix} = a \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} \\ 因为a \neq 0,所以a是矩阵A的一个特征值.$

$2.矩阵A满足A^2 + 6A + 8E = 0,且A = A^T,\\ 证明:A + 3E是正交阵.$
$证:由A^2 + 6A + 8E = 0得 \\ (A + 3E)^2 = E \\ \therefore (A + 3E)可逆,且(A+3E)^{-1} = (A+3E) \\ \because A = A^T, \therefore A + 3E = (A + 3E)^T \\ \therefore (A + 3E)^{-1} = (A + 3E)^T \\ 所以 A + 3E是正交阵.$

$3.已知向量组\alpha_1, \alpha_2, \alpha_3是齐次线性方程组AX = 0的基础\\ 解系,且\beta_1 = \alpha_1 + \alpha_2, \beta_2 = \alpha_2 + \alpha_3, \beta_3 = \alpha_3 + \alpha_1, \\ 证明:向量组\beta_1, \beta_2, \beta_3也是AX = 0的一个基础解系.$
$证: \because\alpha_1, \alpha_2, \alpha_3是AX = 0基础解系,\\ 故\beta_1, \beta_2, \beta_3为AX = 0的解,\\ 所以只需证明\beta_1, \beta_2, \beta_3线性无关, \\ 设k_1, k_2, k_3使k_1 \beta_1 + k_2 \beta_2 + k_3 \beta_3 = 0 \\ 即(k_1 + k_3) \alpha_1 + (k_1 + k_2)\alpha_2 + (k_2 + k_3) \alpha_3 = 0 \\ 因为\alpha_1, \alpha_2, \alpha_3线性无关,所以$
$\left \{ \begin{array}{l} k_1 + k_3 = 0 \\ k_1 + k_2 = 0 \\ k_2 + k_3 = 0 \end{array} \right.$
$解得k_1 = k_2 = k_3 = 0 \\ 所以\beta_1,\beta_2,\beta_3线性无关, \\ 所以\beta_1,\beta_2,\beta_3亦是齐次线性方程组的基础解系.$

$五、设A = \begin{pmatrix} 1 & -1 & 1 \\ x & 4 & y \\ -3 & -3 & 5 \\ \end{pmatrix}, \\ 已知A有3个线性无关的特征向量,\\ \lambda = 2是A的二重特征值,求: \\ (1) x,y的值. \\ (2) 求一个可逆矩阵P,使P^{-1}AP = \Lambda.$
$解:\\ (1) 由题意可知, 必有R(A - 2E) = 1, 于是 \\ A - 2E = \begin{pmatrix} -1 & -1 & 1 \\ x & 2 & y \\ -3 & -3 & 3 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & -1 \\ 0 & 2-x & y+x \\ 0 & 0 & 0 \\ \end{pmatrix}$
$必有\left \{ \begin{array}{l} 2 - x = 0 \\ x + y = 0 \end{array} \right. 解得\left \{ \begin{array}{l} x = 2 \\ y = -2 \end{array} \right.$
$(2) 又 \lambda_1 + \lambda_2 + \lambda_3 = 1 + 4 + 5 = 10\\ 2 + 2 + \lambda_3 = 10 \\ \lambda_3 = 6 \\ 由(A - \lambda E)x = 0,解A的特征向量\\ 当\lambda_1 = \lambda_2 = 2时, \\ P_1 = (-1, 1, 0)^T, P_2 = (1, 0, 1)^T, \\ A - 6E = \begin{pmatrix} -5 & -1 & 1 \\ 2 & -2 & -2 \\ -3 & -3 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -\frac{1}{3} \\ 0 & 1 & \frac{2}{3} \\ 0 & 0 & 0 \\ \end{pmatrix} \\ P_3 = (1, -2, 3)^T, \\ 于是有可逆矩阵P=（p_1,p_2, p_3) = \begin{pmatrix} -1 & 1 & 1 \\ 1 & 0 & -2 \\ 0 & 1 & 3 \\ \end{pmatrix},使得 \\ P^{-1}AP = \Lambda = \begin{pmatrix} 2 & & \\ & 2 & \\ & & 6 \\ \end{pmatrix}$

$六、已知 \alpha_1 = \begin{pmatrix} 1 \\ -1 \\ x \\ 0 \end{pmatrix}, \alpha_2 = \begin{pmatrix} 7 \\ 1 \\ 3 \\ 2 \end{pmatrix}, \alpha_3 = \begin{pmatrix} 4 \\ 0 \\ 2 \\ 1 \end{pmatrix}, \beta = \begin{pmatrix} 10 \\ y \\ 4 \\ 3 \end{pmatrix}, \\ (1) x,y取什么值时，\beta能由\alpha_1, \alpha_2,\alpha_3唯一线性表示,并写出表达式; \\ (2) x,y取什么值时,\beta不能由\alpha_1, \alpha_2, \alpha_3线性表示; \\ (3) x,y取什么值时,\beta能由\alpha_1, \alpha_2, \alpha_3多种线性表示,并写出表达式.$
$解:对矩阵B = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 & \beta \end{pmatrix}施以行初等变换 \\ B = \begin{pmatrix} 1 & 7 & 4 & 10 \\ -1& 1 & 0 & y \\ x & 3 & 2 & 4 \\ 0 & 2 & 1 & 3 \\ \end{pmatrix} \\ \stackrel{r_1 - 3r_4, r_3 -r_4}{\sim} \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1& 1 & 0 & y \\ x & 1 & 1 & 1 \\ 0 & 2 & 1 & 3 \\ \end{pmatrix} \\ \stackrel{r_2 + r_1, r_3 -r_1}{\sim} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0& 2 & 1 & y +1 \\ x -1 & 0 & 0 & 0 \\ 0 & 2 & 1 & 3 \\ \end{pmatrix} \\ \stackrel{r_2 - r_4}{\sim} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & y - 2 \\ x -1 & 0 & 0 & 0 \\ 0 & 2 & 1 & 3 \\ \end{pmatrix} \\ \stackrel{r_2 \leftrightarrow r_4}{\sim} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 2 & 1 & 3 \\ x -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & y - 2 \\ \end{pmatrix} \\ (1) 当y = 2时,且x \neq 1时, R(A) = R(B) = 3,\\ \beta能由\alpha_1, \alpha_2, \alpha_3唯一线性表示, \\ 表示式为:\beta = 0\alpha_1 + 2\alpha_2 - \alpha_3 \\ (2) 当y \neq 2时，\beta不能由\alpha_1, \alpha_2, \alpha_3线性表示, \\ (3) 当y = 2, x = 1时,\beta能由\alpha_1, \alpha_2, \alpha_3多种线性表示. \\ B {\sim} \begin{pmatrix} 1 & 0 & \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \\ 表达式为\beta = -\dfrac{c + 1}{2} \alpha_1 -\dfrac{c - 3}{2} \alpha_2 +c \alpha_3, c \in R$

$七、设A是正定阵,试证存在正定阵B,使得A = B^2.$
$证:设\lambda_1,\lambda_2, \cdots, \lambda_n是A的特征值,\\ 因A是正定矩阵,故\lambda_i > 0 (i = 1, 2, \cdots, n), \\ 且存在正交矩阵P,使 \\ P^{-1}AP = diag \lbrace \lambda_1, \lambda_2, \cdots, \lambda \rbrace \\ = (diag \lbrace \sqrt{\lambda_1}, \sqrt{\lambda_2}, \cdots, \sqrt{\lambda_n} \rbrace )^2 \\ \therefore A = P(diag \lbrace \sqrt{\lambda_1}, \sqrt{\lambda_2}, \cdots, \sqrt{\lambda_n} \rbrace )^2 P^{-1} \\ =P \begin{pmatrix} \sqrt{\lambda_1} & & & \\ & \sqrt{\lambda_2} & & \\ & & \ddots & \\ & & & \sqrt{\lambda_n} \\ \end{pmatrix}P^{-1} \\ \times P \begin{pmatrix} \sqrt{\lambda_1} & & & \\ & \sqrt{\lambda_2} & & \\ & & \ddots & \\ & & & \sqrt{\lambda_n} \\ \end{pmatrix}P^{-1} \\ = B^2 \\ 其中 B = P \begin{pmatrix} \sqrt{\lambda_1} & & & \\ & \sqrt{\lambda_2} & & \\ & & \ddots & \\ & & & \sqrt{\lambda_n} \\ \end{pmatrix}P^{-1} \\ 显然B为正定阵,且有A = B^2 .$