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张祖锦2023年数学专业真题分类70天之第59天

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发表于 2023-3-5 13:15:09 | 显示全部楼层 |阅读模式
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## 张祖锦2023年数学专业真题分类70天之第59天 --- 1335、 5、 (20 分) 设 $\displaystyle A\in \mathbb\{R\}^\{n\times n\}$ 是正定矩阵, $\displaystyle \beta\in \mathbb\{R\}^n, c\in\mathbb\{R\}$. 如果存在 $\displaystyle x\in\mathbb\{R\}^n$ 使得 \begin\{aligned\} x^\mathrm\{T\} Ax+2\beta^\mathrm\{T\} x+c=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 证明: $\displaystyle \beta^\mathrm\{T\} A^\{-1\}\beta\geq c$. (南开大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A$ 正定知存在正交矩阵 $\displaystyle P$ 使得 \begin\{aligned\} &P^\mathrm\{T\} AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n),\quad \lambda\_i > 0\\\\ \Rightarrow&P^\mathrm\{T\} A^\{-1\}P=\mathrm\{diag\}(\lambda\_1^\{-1\},\cdots,\lambda\_n^\{-1\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle \beta=P\gamma, x=Py$, 则 \begin\{aligned\} &\beta^\mathrm\{T\} A^\{-1\}\beta+x^\mathrm\{T\} Ax\\\\ =&\gamma^\mathrm\{T\} \mathrm\{diag\}(\lambda\_1^\{-1\},\cdots,\lambda\_n^\{-1\})\gamma +y^\mathrm\{T\} \mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)y\\\\ =&\sum\_\{i=1\}^n \frac\{\gamma\_i^2\}\{\lambda\_i\} +\sum\_\{i=1\}^n \lambda\_iy\_i^2 =\sum\_\{i=1\}^n \left(\frac\{\gamma\_i^2\}\{\lambda\_i\}+\lambda\_iy\_i^2\right)\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\geq\}&\sum\_\{i=1\}^n 2\left|\frac\{\gamma\_i\}\{\sqrt\{\lambda\_i\}\}\cdot \sqrt\{\lambda\_i\}y\_i\right| =2\left|\sum\_\{i=1\}^n \gamma\_iy\_i\right|\\\\ =&2|\gamma^\mathrm\{T\} y| =2|\beta^\mathrm\{T\} P\cdot P^\mathrm\{T\} x| =2|\beta^\mathrm\{T\} x|\geq -2\beta^\mathrm\{T\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 移项后利用题中等式即知结论成立.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1336、 (10)、 设 $\displaystyle f(x)=ax\_1^2+ax\_2^2+(a-1)x\_3^2+2x\_1x\_3-2x\_2x\_3$, 则当 $\displaystyle a=\underline\{\ \ \ \ \ \ \ \ \ \ \}$ 时, $\displaystyle f(x)$ 的规范形为 $\displaystyle y\_1^2+y\_2^2$. [题目有问题, 理由见参考解答.] (厦门大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle 2=\mathrm\{tr\} f=a+a+(a-1)\Rightarrow a=1$. 但代入一算知 $\displaystyle f$ 的规范形为 $\displaystyle y\_1^2+y\_2^2-y\_3^2$. 与题设矛盾.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1337、 9、 (15 分) 设二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3)=5x\_1^2+5x\_2^2-6x\_1x\_3+6x\_1x\_2+6x\_2x\_3+tx\_3^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的秩为 $\displaystyle 2$. (1)、 求 $\displaystyle t$ 的值; (2)、 用正交替换把二次型化为标准形, 并把正交替换写出. (山西大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}5&3&-3\\\\ 3&5&3\\\\ -3&3&t\end\{array\}\right)$. 由题设, $\displaystyle 0=|A|=16(t-9)\Rightarrow t=9$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 11,8,0$. 由 \begin\{aligned\} &11E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\frac\{1\}\{3\}\\\\ 0&1&-\frac\{1\}\{3\}\\\\ 0&0&0 \end\{array\}\right), 8E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&0&1\\\\ 0&0&0 \end\{array\}\right),\\\\ &0E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-\frac\{3\}\{2\}\\\\ 0&1&\frac\{3\}\{2\}\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 11,8,0$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\3 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}3\\\\-3\\\\2 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} -\frac\{1\}\{\sqrt\{11\}\}&\frac\{1\}\{\sqrt\{2\}\}&\frac\{3\}\{\sqrt\{22\}\}\\\\ \frac\{1\}\{\sqrt\{11\}\}&\frac\{1\}\{\sqrt\{2\}\}&-\frac\{3\}\{\sqrt\{22\}\}\\\\ \frac\{3\}\{\sqrt\{11\}\}&0&\frac\{2\}\{\sqrt\{22\}\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(11,8,0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交线性替换 $\displaystyle X=PY$ 后, $\displaystyle f$ 化为了标准形 $\displaystyle 11y\_1^2+8y\_2^2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1338、 5、 (15 分) 求使实二次型 \begin\{aligned\} f(x\_1,\cdots,x\_n)=\sum\_\{i=1\}^n (x\_i+a\_ix\_\{i+1\})^2\left(\mbox\{约定 $\displaystyle x\_\{n+1\}=x\_1$\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 正定的充分必要条件. (陕西师范大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 \begin\{aligned\} y\_1=x\_1+a\_1x\_2, y\_2=x\_2+a\_2x\_3, \cdots, y\_\{n-1\}=x\_\{n-1\}+a\_\{n-1\}x\_n, y\_n=x\_n+a\_nx\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则当且仅当上述线性替换非奇异时, $\displaystyle f=y\_1^2+\cdots+y\_n^2$ 是正定二次型. 算出上述线性替换的矩阵的行列式 \begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}1&0&\cdots&a\_n\\\\ a\_1&1&\cdots&\vdots\\\\ &\ddots&\ddots&\vdots\\\\ &&a\_\{n-1\}&1\end\{array\}\right|=1+a\_n\cdot(-1)^\{1+n\}a\_1\cdots a\_\{n-1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 即知 $\displaystyle f$ 正定 $\displaystyle \Leftrightarrow a\_1\cdots a\_n\neq (-1)^n$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1339、 8、 二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3)=2x\_1^2+5x\_2^2+5x\_3^2+2ax\_1x\_2+2bx\_1x\_3-8x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 可经过正交变换化为标准形 $\displaystyle y\_1^2+y\_2^2+cy\_3^2$. 求 $\displaystyle a,b,c$ 及正交变换矩阵 $\displaystyle P$. (上海财经大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&a&b\\\\ a&5&-4\\\\ b&-4&5\end\{array\}\right)$. 于是 $\displaystyle 12=\mathrm\{tr\} A=1+1+c\Rightarrow c=10$. 故 \begin\{aligned\} A\sim \mathrm\{diag\}(1,1,10)\Rightarrow \mathrm\{rank\}(A-E)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle A-E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&a&b\\\\ 0&4-a^2&-(ab+4)\\\\ a+b&0&0\end\{array\}\right)$ 知 $\displaystyle a=\pm 2, b=\mp 2$. (1)、 当 $\displaystyle a=2, b=-2$ 时, $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&-2\\\\ 2&5&-4\\\\ -2&-4&5\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,10$. 由 \begin\{aligned\} E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&-2\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), 10E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\frac\{1\}\{2\}\\\\ 0&1&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 1,10$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\0\\\\1 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\-2\\\\2 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} -\frac\{2\}\{\sqrt\{5\}\}&\frac\{2\}\{3\sqrt\{5\}\}&-\frac\{1\}\{3\}\\\\ \frac\{1\}\{\sqrt\{5\}\}&\frac\{4\}\{3\sqrt\{5\}\}&-\frac\{2\}\{3\}\\\\ 0&\frac\{\sqrt\{5\}\}\{3\}&\frac\{2\}\{3\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(1,1,10\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交变换 $\displaystyle X+PY$ 后, $\displaystyle f$ 化为了标准形 $\displaystyle y\_1^2+y\_2^2+10y\_3^2$. (2)、 当 $\displaystyle a=-2, b=2$ 时, $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&-2&2\\\\ -2&5&-4\\\\ 2&-4&5\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,10$. 由 \begin\{aligned\} E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&2\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), 10E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-\frac\{1\}\{2\}\\\\ 0&1&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 1,10$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\0\\\\1 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-2\\\\2 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} \frac\{2\}\{\sqrt\{5\}\}&-\frac\{2\}\{3\sqrt\{5\}\}&\frac\{1\}\{3\}\\\\ \frac\{1\}\{\sqrt\{5\}\}&\frac\{4\}\{3\sqrt\{5\}\}&-\frac\{2\}\{3\}\\\\ 0&\frac\{\sqrt\{5\}\}\{3\}&\frac\{2\}\{3\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(1,1,10\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交变换 $\displaystyle X+PY$ 后, $\displaystyle f$ 化为了标准形 $\displaystyle y\_1^2+y\_2^2+10y\_3^2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1340、 (2)、 (15 分) 设二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3)=ax\_1^2-3x\_2^2-2x\_3^2+2x\_1x\_3+2x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 是负定二次型, 且整数 $\displaystyle a$ 为 $\displaystyle 3$ 的倍数, 试确定 $\displaystyle a$ 的最大值, 并在此时用正交变换将此二次型化为标准形 (需写出正交变换及标准形). (上海大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&0&1\\\\ 0&-3&1\\\\ 1&1&-2\end\{array\}\right)$. 由 $\displaystyle A$ 负定知 $\displaystyle a < 0, -3a > 0, |A|=5a+3 < 0\Leftrightarrow a < -\frac\{3\}\{5\}$. 联合 $\displaystyle a$ 是 $\displaystyle 3$ 的倍数知 $\displaystyle a=-3$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle -1,-3,-4$. 由 \begin\{aligned\} &-E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-\frac\{1\}\{2\}\\\\ 0&1&-\frac\{1\}\{2\}\\\\ 0&0&0 \end\{array\}\right), -3 E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 0&0&1\\\\ 0&0&0 \end\{array\}\right),\\\\ &-4 E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle -1,-3,-4$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\2 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\0 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\-1\\\\1 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} \frac\{1\}\{\sqrt\{6\}\}&-\frac\{1\}\{\sqrt\{2\}\}&-\frac\{1\}\{\sqrt\{3\}\}\\\\ \frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{2\}\}&-\frac\{1\}\{\sqrt\{3\}\}\\\\ \frac\{2\}\{\sqrt\{6\}\}&0&\frac\{1\}\{\sqrt\{3\}\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(-1,-3,-4\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交线性替换 $\displaystyle X=PY$ 后, $\displaystyle f$ 化为了标准形 $\displaystyle -y\_1^2-3y\_2^2-4y\_3^2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1341、 10、 试用正交线性替换把二次型 \begin\{aligned\} 7x\_1^2+x\_2^2+x\_3^2-8x\_1x\_2+8x\_1x\_3+16x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 化为标准形. (首都师范大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 题中二次型 $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}7&-4&4\\\\ -4&1&8\\\\ 4&8&1\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 9,9,-9$. 由 \begin\{aligned\} 9E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&2&-2\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), -9E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&\frac\{1\}\{2\}\\\\ 0&1&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 9,-9$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\0\\\\1 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\-2\\\\2 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} -\frac\{2\}\{\sqrt\{5\}\}&\frac\{2\}\{3\sqrt\{5\}\}&-\frac\{1\}\{3\}\\\\ \frac\{1\}\{\sqrt\{5\}\}&\frac\{4\}\{3\sqrt\{5\}\}&-\frac\{2\}\{3\}\\\\ 0&\frac\{\sqrt\{5\}\}\{3\}&\frac\{2\}\{3\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(9,9,-9\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交线性替换 $\displaystyle X=PY$ 后, $\displaystyle f$ 化为标准形 $\displaystyle 9y\_1^2+9y\_2^2-9y\_3^2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1342、 8、 二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3)=x^\mathrm\{T\} Ax =x\_1^2+x\_2^2+x\_3^2+2ax\_1x\_2+2x\_1x\_3+4bx\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 在正交线性替换 $\displaystyle x=Qy$ 下化为标准形 $\displaystyle y\_1^2+2y\_2^2$. (1)、 求 $\displaystyle a,b$ 及 $\displaystyle Q$; (2)、 若 $\displaystyle tE+3A+A^2$ 正定, 确定 $\displaystyle t$ 的范围; (3)、 设 $\displaystyle x,y,z\in \mathbb\{R\}, x^2+y^2+z^2=1$, 证明: $\displaystyle 0\leq x^2+y^2+z^2+2xz\leq 2$. (太原理工大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&a&1\\\\ a&1&2b\\\\ 1&2b&1\end\{array\}\right)$. 由题设, $\displaystyle A$ 的特征值为 $\displaystyle 1,2,0$. 故 \begin\{aligned\} 0=|0E-A|=(-1)^3|A|=(a-2b)^2\Rightarrow a=2b. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 进而算出 $\displaystyle 0=|E-A|=-2a^2\Rightarrow a=0\Rightarrow b=0$. 由 \begin\{aligned\} 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&-1\\\\ 0&1&0\\\\ 0&0&0 \end\{array\}\right), E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&0\\\\ 0&0&1\\\\ 0&0&0 \end\{array\}\right), 0E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&1\\\\ 0&1&0\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 2,1,0$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\1 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\1\\\\0 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\1 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} Q=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} \frac\{1\}\{\sqrt\{2\}\}&0&-\frac\{1\}\{\sqrt\{2\}\}\\\\ 0&1&0\\\\ \frac\{1\}\{\sqrt\{2\}\}&0&\frac\{1\}\{\sqrt\{2\}\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle Q$ 正交, 且 \begin\{aligned\} Q^\mathrm\{T\} AQ=Q^\{-1\}AQ=\mathrm\{diag\}\left(2,1,0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 设 $\displaystyle f(s)=t+3s+s^2$, 则 \begin\{aligned\} Q^\mathrm\{T\} (tE+3A+A^2)Q =\mathrm\{diag\}\left(f(2),f(1),f(0)\right) =\mathrm\{diag\}(10+t, 4+t, t). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle tE+3A+A^2$ 正定 $\displaystyle \Leftrightarrow 10+t > 0, 4+t > 0, t > 0\Leftrightarrow t > 0$. (3)、 设 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right)=Q\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\\\\w\end\{array\}\right)$, 则 \begin\{aligned\} &u^2+v^2+w^2=(u,v,w)\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\\\\w\end\{array\}\right)\\\\ =&(x,y,z)Q Q^\mathrm\{T\}\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right)=x^2+y^2+z^2=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} x^2+y^2+z^2+2xz=f(x,y,z)\overset\{\tiny\mbox\{第1步\}\}\{=\} 2u\_1^2+u\_2^2\in [0,2]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1343、 5、 (10 分) 设 $\displaystyle A\in \mathbb\{R\}^\{n\times n\}$ 为正定矩阵, $\displaystyle \beta$ 与 $\displaystyle x$ 均为 $\displaystyle n$ 维实向量, 试证二次函数 \begin\{aligned\} f(x)=x^\mathrm\{T\} Ax-2\beta^\mathrm\{T\} x+c \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的极小值为 $\displaystyle c-\beta^\mathrm\{T\} A^\{-1\}\beta$. (西安电子科技大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A$ 正定知存在可逆矩阵 $\displaystyle C$, 使得 $\displaystyle A=C^\mathrm\{T\} C$. 设 $\displaystyle y=Cx$, 则 \begin\{aligned\} f(x)=&y^\mathrm\{T\} y-2\beta^\mathrm\{T\} C^\{-1\}y+c\\\\ \stackrel\{\gamma=\beta^\mathrm\{T\} C^\{-1\}=(r\_1,\cdots,r\_n)\}\{=\}&\sum\_\{i=1\}^n y\_i^2-2\sum\_\{i=1\}^n r\_iy\_i+c =\sum\_\{i=1\}^n (y\_i-c\_i)^2+c-\sum\_\{i=1\}^n r\_i^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} \min f=c-\gamma\gamma^\mathrm\{T\} =c-\beta^\mathrm\{T\} C^\{-1\}C^\{-\mathrm\{T\}\}\beta=c-\beta^\mathrm\{T\} A^\{-1\}\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1344、 11、 设实二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3,x\_4)=&k(x\_1^2+x\_2^2+x\_3^2+x\_4^2)\\\\ &+2x\_1x\_2+2x\_1x\_3-2x\_1x\_4-2x\_2x\_3+2x\_2x\_4+2x\_3x\_4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (1)、 若 $\displaystyle f$ 为正定二次型, 求 $\displaystyle k$ 的取值范围; (2)、 若 $\displaystyle f$ 通过正交线性替换化为标准形 $\displaystyle g=3y\_1^2+3y\_2^2+3y\_3^2-y\_4^2$, 求 $\displaystyle k$ 的值. (西北大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}k&1&1&-1\\\\ 1&k&-1&1\\\\ 1&-1&k&1\\\\ -1&1&1&k\end\{array\}\right)$. 而 $\displaystyle f$ 正定 \begin\{aligned\} &\Leftrightarrow k > 0, k^2-1 > 0, (k+1)^2(k-2) > 0, |A|=(k-3)(k+1)^3 > 0\\\\ &\Leftrightarrow k > 3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 由题设, $\displaystyle 4k=3+3+3-1\Rightarrow k=2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1345、 5、 (20 分) 已知二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3)=(1+a)x\_1^2+(1+a)x\_2^2+2x\_3^2+2(a-1)x\_1x\_2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的秩为 $\displaystyle 2$, 求 $\displaystyle a$ 的值, 并将二次型经过正交替换 $\displaystyle X=QY$ 化为标准形, 同时求 $\displaystyle f(x\_1,x\_2,x\_3)=0$ 的解. (西南财经大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1+a&a-1&0\\\\ a-1&1+a&0\\\\ 0&0&2\end\{array\}\right)$. 由题设, $\displaystyle 0=|A|=8a\Rightarrow a=0\Rightarrow A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ -1&1&0\\\\ 0&0&2\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,2,0$. 由 \begin\{aligned\} 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), 0E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&0&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 2,0$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\1 \end\{array\}\right);\quad \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{\sqrt\{2\}\}&0&\frac\{1\}\{\sqrt\{2\}\}\\\\ \frac\{1\}\{\sqrt\{2\}\}&0&\frac\{1\}\{\sqrt\{2\}\}\\\\ 0&1&0 \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(2,2,0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交线性替换 $\displaystyle X=PY$ 后, $\displaystyle f$ 化为了标准形 $\displaystyle 2y\_1^2+2y\_2^2=0$. 进一步, \begin\{aligned\} f=0\Leftrightarrow y\_1=y\_2=0\Leftrightarrow x=P\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\y\_3\end\{array\}\right)=\frac\{y\_3\}\{\sqrt\{2\}\}\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0\end\{array\}\right) =k\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0\end\{array\}\right), \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1346、 7、 (10 分) 设 $\displaystyle \alpha,\beta$ 为实数域上的 $\displaystyle n$ 维非零列向量, 且 $\displaystyle A=\alpha\beta^\mathrm\{T\}+\beta\alpha^\mathrm\{T\}$, 求二次型 $\displaystyle f(x)=x^\mathrm\{T\} Ax$ 的符号差. (西南交通大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 8、 若 $\displaystyle \alpha,\beta$ 线性相关, 则由 $\displaystyle \alpha,\beta$ 均不为零知 \begin\{aligned\} \exists\ k\in\mathbb\{R\}\backslash\left\\{0\right\\},\mathrm\{ s.t.\} \beta=k\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 \begin\{aligned\} f(x)&=x^\mathrm\{T\} (2k\alpha\alpha^\mathrm\{T\})x =2kx^\mathrm\{T\} \alpha \alpha^\mathrm\{T\} x=2ky^\mathrm\{T\} y\left(y=\alpha^\mathrm\{T\} x\right)\\\\ &=\left\\{\begin\{array\}\{llllllllllll\} z^\mathrm\{T\} z\Rightarrow\mbox\{$f$ 的符号差为 $\displaystyle 1$\},&z=\sqrt\{2k\}y, k\geq 0,\\\\ -z^\mathrm\{T\} z\Rightarrow\mbox\{$f$ 的符号差为 $\displaystyle -1$\},&z=\sqrt\{-2k\}y, k < 0. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 9、 若 $\displaystyle \alpha,\beta$ 线性无关, 则 \begin\{aligned\} \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}a\_1&\cdots&a\_n\\\\ b\_1&\cdots&b\_n\end\{array\}\right)=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 不妨设 $\displaystyle \left|\begin\{array\}\{cccccccccc\}a\_1&a\_2\\\\ b\_1&b\_2\end\{array\}\right|\neq 0$, 而非退化线性替换 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}y\_1\\\\\vdots\\\\ y\_n\end\{array\}\right) =\left(\begin\{array\}\{cccccccccccccccccccc\} a\_1&a\_2&a\_3&\cdots&a\_n\\\\ b\_1&b\_2&b\_3&\cdots&b\_n\\\\ 0&0&1&\cdots&0\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 0&0&0&\cdots&1 \end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\\vdots\\\\x\_n\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 后, \begin\{aligned\} f(x)&=x^\mathrm\{T\} \alpha\beta^\mathrm\{T\} x+x^\mathrm\{T\} \beta \alpha^\mathrm\{T\} x =x^\mathrm\{T\} \alpha \beta^\mathrm\{T\} x+(x^\mathrm\{T\} \beta \alpha^\mathrm\{T\} x)^\mathrm\{T\} =2x^\mathrm\{T\} \alpha \beta^\mathrm\{T\} x\\\\ & =2(\alpha^\mathrm\{T\} x)^\mathrm\{T\} (\beta^\mathrm\{T\} x) =2y\_1y\_2 =\frac\{1\}\{2\}\left\[(y\_1+y\_2)^2-(y\_1-y\_2)^2\right\]\\\\ &=z\_1^2-z\_2^2\left(z\_1=\frac\{y\_1+y\_2\}\{\sqrt\{2\}\}, z\_2=\frac\{y\_1-y\_2\}\{\sqrt\{2\}\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 此时, $\displaystyle f$ 的符号差为 $\displaystyle 0$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1347、 7、 (20 分) 证明: 一个实二次型可以分解成两个实系数的一次齐次多项式的乘积的充分必要条件是它的秩为 $\displaystyle 2$ 且符号差为 $\displaystyle 0$, 或者秩为 $\displaystyle 1$. (新疆大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Rightarrow$: 设实二次型 $\displaystyle f(x)$ 的秩为 $\displaystyle 2$, 且符号差为 $\displaystyle 0$, 则存在非退化线性替换 $\displaystyle x=Py$ 使得 \begin\{aligned\} f(x)&=y\_1^2-y\_2^2=(y\_1+y\_2)(y\_1-y\_2)\\\\ &=\left\[(q\_\{11\}+q\_\{21\})x\_1+\cdots+(q\_\{1n\}+q\_\{2n\})x\_n\right\]\\\\ &\quad \cdot \left\[(q\_\{11\}-q\_\{21\})x\_1+\cdots+(q\_\{1n\}-q\_\{2n\})x\_n\right\]\left(y=Qx, Q=P^\{-1\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 若 $\displaystyle f$ 的秩为 $\displaystyle 1$, 则存在非退化线性替换 $\displaystyle x=Py$ 使得 \begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\} y\_1^2=(q\_\{11\}x\_1+\cdots+q\_\{1n\}x\_n)^2,\\\\ -y\_1^2=-(q\_\{11\}x\_1+\cdots+q\_\{1n\}x\_n)^2. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 $\displaystyle \Leftarrow$: 设实二次型 $\displaystyle f$ 可以分解为两个实系数的一次齐次多项式的乘积, 则 \begin\{aligned\} f(x)=(k\_1x\_1+\cdots+k\_nx\_n)(l\_1x\_1+\cdots+l\_nx\_n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 \begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}k\_1&\cdots&k\_n\\\\ l\_1&\cdots&l\_n\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2-1)、 若 $\displaystyle \mathrm\{rank\}(A)=1$, 则不妨设 \begin\{aligned\} k\_1\neq 0, \frac\{l\_1\}\{k\_1\}=\cdots=\frac\{l\_n\}\{k\_n\}=a\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是在可逆线性替换 \begin\{aligned\} y\_1=k\_1x\_1+\cdots+k\_nx\_n, y\_2=x\_2,\cdots, y\_n=x\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 下, \begin\{aligned\} f(x)=y\_1\cdot ay\_1=ay\_1^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 $\displaystyle f$ 的秩为 $\displaystyle 1$. (2-2)、 若 $\displaystyle \mathrm\{rank\}(A)=2$, 则不妨设 \begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}k\_1&k\_2\\\\l\_1&l\_2\end\{array\}\right|\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 在非退化线性替换 \begin\{aligned\} y\_1=k\_1x\_1+\cdots+k\_nx\_n, y\_2=l\_1x\_1+\cdots+l\_nx\_n, y\_3=x\_3,\cdots, y\_n=x\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 下, \begin\{aligned\} f(x)&=y\_1y\_2=\left(\frac\{y\_1+y\_2\}\{2\}\right)^2 -\left(\frac\{y\_1-y\_2\}\{2\}\right)^2=z\_1^2+z\_2^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} $\displaystyle f$ 的秩为 $\displaystyle 2$, 且符号差为 $\displaystyle 0$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1348、 4、 (15 分) 找出所有使三元实二次型 \begin\{aligned\} 2x\_1^2+2x\_2^2+2x\_3^2+2x\_1x\_2+2ax\_1x\_3+2x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 正定的实数 $\displaystyle a$, 并在 $\displaystyle a=-1$ 时用正交的线性替换将其化为平方和. (云南大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 题中二次型 $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&a\\\\ 1&2&1\\\\ a&1&2\end\{array\}\right)$, 而 $\displaystyle f$ 正定 $\displaystyle \Leftrightarrow A$ 的顺序主子式 \begin\{aligned\} 2 > 0, 3 > 0, |A|=-2(a+1)(a-2) > 0\Leftrightarrow -1 < a < 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 当 $\displaystyle a=-1$ 时, $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&-1\\\\ 1&2&1\\\\ -1&1&2\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 3,3,0$. 由 \begin\{aligned\} 3E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), 0E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&-1\\\\ 0&1&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 3,0$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\1 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\\\\1 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} \frac\{1\}\{\sqrt\{2\}\}&-\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{3\}\}\\\\ \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{6\}\}&-\frac\{1\}\{\sqrt\{3\}\}\\\\ 0&\frac\{2\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{3\}\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(3,3,0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交线性替换 $\displaystyle X=PY$ 后, $\displaystyle f$ 化为了平方和 $\displaystyle 3y\_1^2+3y\_2^2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1349、 8、 (15 分) 将二次曲面方程 \begin\{aligned\} 5x^2+5y^2+8z^2-10xy-4xz+4yz-3\sqrt\{2\}x-3\sqrt\{2\}y=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 化为标准方程并指出它表示什么曲面. (长安大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 二次项的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}5&-5&-2\\\\ -5&5&2\\\\ -2&2&8\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 12,6,0$. 由 \begin\{aligned\} 12E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&1\\\\ 0&1&-1\\\\ 0&0&0 \end\{array\}\right), 6E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-\frac\{1\}\{2\}\\\\ 0&1&\frac\{1\}\{2\}\\\\ 0&0&0 \end\{array\}\right), 0E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&0&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 12,6,0$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\1 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\\\\2 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} -\frac\{1\}\{\sqrt\{3\}\}&\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{2\}\}\\\\ \frac\{1\}\{\sqrt\{3\}\}&-\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{2\}\}\\\\ \frac\{1\}\{\sqrt\{3\}\}&\frac\{2\}\{\sqrt\{6\}\}&0\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(12,6,0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过正交线性替换 $\displaystyle X=PY$ 后, 二次项化为了标准形 $\displaystyle 12y\_1^2+6y\_2^2$. 再由 \begin\{aligned\} &(-3\sqrt\{2\},-3\sqrt\{2\},0)\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right) =(-3\sqrt\{2\},-3\sqrt\{2\},0)P\left(\begin\{array\}\{cccccccccccccccccccc\}y\_1\\\\y\_2\\\\y\_3\end\{array\}\right)\\\\ =&(0,0,-6)\left(\begin\{array\}\{cccccccccccccccccccc\}y\_1\\\\y\_2\\\\y\_3\end\{array\}\right)=-6y\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知原二次曲面方程化为了 $\displaystyle 12y\_1^2+6y\_2^2-6y\_3=0\Leftrightarrow 2y\_1^2+y\_2^2-y\_3=0$. 这是一张椭圆抛物面.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1350、 (3)、 设二次型 \begin\{aligned\} f(x)=2x\_1^2+4x\_2^2+x\_3^2-2x\_1x\_2-2x\_1x\_3+2\alpha x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 经过正交变换 $\displaystyle X=QY$ 化为标准形 $\displaystyle 2y\_1^2+2y\_2^2+\beta y\_3^2$. (3-1)、 求 $\displaystyle \alpha,\beta$ 及正交矩阵 $\displaystyle Q$; (3-2)、 当 $\displaystyle X^\mathrm\{T\} X=3$ 时, 求 $\displaystyle f$ 的最大值. [题目有问题, 跟锦数学微信公众号没法做哦.] (郑州大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目有问题, 跟锦数学微信公众号没法做哦.]跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1351、 (6)、 二次型 \begin\{aligned\} (x-y)^2+2(y-z)^2+(z-x)^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的正惯性指数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国科学技术大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 二次型的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&-1&-1\\\\ -1&3&-2\\\\ -1&-2&3\end\{array\}\right)$, 特征值为 $\displaystyle 5,3,0$, 而正惯性指数为 $\displaystyle 2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1352、 4、 设有 $\displaystyle n$ 元实二次型 \begin\{aligned\} f(x\_1,\cdots,x\_n)=&(x\_1+a\_1x\_2)^2+(x\_2+a\_2x\_3)^2 +\cdots\\\\ &+(x\_\{n-1\}+a\_\{n-1\}x\_n)^2+(x\_n+a\_nx\_1)^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle a\_i\ (i=1,\cdots,n)$ 为实数. 试问: 当 $\displaystyle a\_1,\cdots,a\_n$ 满足何种条件时, 二次型为正定二次型? (中国科学院大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 \begin\{aligned\} &y\_1=x\_1+a\_1x\_2, y\_2=x\_2+a\_2x\_3, \cdots,\\\\ &y\_\{n-1\}=x\_\{n-1\}+a\_\{n-1\}x\_n, y\_n=x\_n+a\_nx\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则当且仅当上述线性替换非奇异时, $\displaystyle f=y\_1^2+\cdots+y\_n^2$ 是正定二次型. 算出上述线性替换的矩阵的行列式 \begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}1&0&\cdots&a\_n\\\\ a\_1&1&\cdots&\vdots\\\\ &\ddots&\ddots&\vdots\\\\ &&a\_\{n-1\}&1\end\{array\}\right|=1+a\_n\cdot(-1)^\{1+n\}a\_1\cdots a\_\{n-1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 即知当 $\displaystyle a\_1\cdots a\_n\neq (-1)^n$ 时, 二次型为正定二次型.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1353、 7、 设实二次型 $\displaystyle f(x\_1,\cdots,x\_n)=\sum\_\{i=1\}^n (a\_\{i1\}x\_1+\cdots+a\_\{in\}x\_n)^2$. (1)、 求 $\displaystyle f$ 的矩阵 $\displaystyle B$; (2)、 说明 $\displaystyle \mathrm\{rank\} B=\mathrm\{rank\} A$, 其中 $\displaystyle A=(a\_\{ij\})\_\{n\times n\}$. (中国矿业大学(北京)2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 \begin\{aligned\} f&=\sum\_i (x\_1,\cdots,x\_n)\left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{i1\}\\\\\vdots\\\\a\_\{in\}\end\{array\}\right) (a\_\{i1\},\cdots,a\_\{in\})\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\\vdots\\\\x\_n\end\{array\}\right)\\\\ &=(x\_1,\cdots,x\_n)\left(\begin\{array\}\{cccccccccccccccccccc\} \sum\_ia\_\{i1\}^2&\cdots&\sum\_ia\_\{i1\}a\_\{in\}\\\\ \vdots&\ddots&\vdots\\\\ \sum\_ia\_\{in\}a\_\{i1\}&\cdots&\sum\_ia\_\{in\}^2 \end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\\vdots\\\\x\_n\end\{array\}\right)\\\\ &=X^\mathrm\{T\} (A^\mathrm\{T\} A)X. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle f$ 的矩阵 $\displaystyle B=A^\mathrm\{T\} A$. (2)、 证明方程组 $\displaystyle AX=0$ 与 $\displaystyle A^\mathrm\{T\} AX=0$ 同解. 显然 $\displaystyle AX=0\Rightarrow A^\mathrm\{T\} AX=0$. 反之, \begin\{aligned\} A^\mathrm\{T\} AX=0&\Rightarrow 0=X^\mathrm\{T\} A^\mathrm\{T\} AX=(AX)^\mathrm\{T\} (AX)\Rightarrow AX=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (3)、 由第 2 步知 \begin\{aligned\} &\left\\{x\in\mathbb\{R\}^n; AX=0\right\\}=\left\\{x\in\mathbb\{R\}^n; A^\mathrm\{T\} Ax=0\right\\}\\\\ \Rightarrow&\dim\left\\{x\in\mathbb\{R\}^n; AX=0\right\\}=\dim\left\\{x\in\mathbb\{R\}^n; A^\mathrm\{T\} Ax=0\right\\}\\\\ \Rightarrow&n-\mathrm\{rank\} A=n-\mathrm\{rank\}(A^\mathrm\{T\} A)\\\\ \Rightarrow& \mathrm\{rank\} A=\mathrm\{rank\}(A^\mathrm\{T\} A)=\mathrm\{rank\} B. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1354、 (5)、 二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3)=\lambda(x\_1^2+x\_2^2+x\_3^2)-4x\_1x\_2-4x\_1x\_3+4x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 是负定的, 则 $\displaystyle \lambda$ 的取值范围是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国矿业大学(徐州)2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f$ 的矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda&-2&-2\\\\ -2&\lambda&2\\\\ -2&2&\lambda\end\{array\}\right)$. 故 $\displaystyle f$ 负定等价于 $\displaystyle A$ 的顺序主子式 \begin\{aligned\} \lambda < 0, \lambda^2-4 > 0, |A|=(\lambda-2)^2(\lambda+4) < 0\Leftrightarrow \lambda < -4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1355、 (6)、 若 $\displaystyle t$ 满足条件 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$, 则二次型 \begin\{aligned\} x\_1^2+x\_2^2+5x\_3^2+2tx\_1x\_2-2x\_1x\_3+4x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 是正定的. (中国人民大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 二次型的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&t&-1\\\\ t&1&2\\\\ -1&2&5\end\{array\}\right)$, 而二次型正定 \begin\{aligned\} \Leftrightarrow 1 > 0, 1-t^2 > 0, |A|= -4t-5t^2 > 0\Leftrightarrow -\frac\{4\}\{5\} < t < 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1356、 4、 (15 分) 设 $\displaystyle \mathbb\{R\}^4$ 中多项式 \begin\{aligned\} Q=x\_1^2+x\_2^2+x\_3^2-2x\_4^2+6x\_1x\_2+4x\_1x\_4+4x\_2x\_3+6x\_2x\_4-2x\_3x\_4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 求 $\displaystyle Q$ 的正负惯性指数. (中山大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle Q$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&3&0&2\\\\ 3&1&2&3\\\\ 0&2&1&-1\\\\ 2&3&-1&-2\end\{array\}\right)$. 由 \begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-3&0&-2\\\\ 0&1&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\end\{array\}\right)\Rightarrow P\_1^\mathrm\{T\} AP\_1=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &-8&2&-3\\\\ &2&1&-1\\\\ &-3&-1&-6\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &1&\frac\{1\}\{4\}&-\frac\{3\}\{8\}\\\\ &0&1&0\\\\ &0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2=A\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &-8&&\\\\ &&\frac\{3\}\{2\}&-\frac\{7\}\{4\}\\\\ &&-\frac\{7\}\{4\}&-\frac\{39\}\{8\}\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} P\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &1&&\\\\ &&1&\frac\{7\}\{6\}\\\\ &&&1\end\{array\}\right)\Rightarrow P\_3^\mathrm\{T\} A\_2P\_3=\mathrm\{diag\}\left(1,-8,\frac\{3\}\{2\},-\frac\{83\}\{12\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle f$ 的正负惯性指数都是 $\displaystyle 2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1357、 6、 (12 分) 已知二次型 (有一项的系数为 $\displaystyle 2a$) \begin\{aligned\} f(x\_1,x\_2,x\_3)=\cdots, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 问 $\displaystyle a$ 为何值时, 存在正交变换可将 $\displaystyle f$ 化为 $\displaystyle y\_1^2+2y\_2^2+5y\_3^2$. [题目不全, 跟锦数学微信公众号没法做哦.] (重庆师范大学2023年高等代数考研试题) [二次型 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目不全, 跟锦数学微信公众号没法做哦.]跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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