## 张祖锦2023年数学专业真题分类70天之第58天
---
1312、 (3)、 (20 分) 设 $\displaystyle A$ 为 $\displaystyle n$ 阶实对称矩阵, $\displaystyle n\geq 2$,
\begin\{aligned\} W=\left\\{X\in\mathbb\{R\}^n; X^\mathrm\{T\} AX=0\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \mathbb\{R\}^n$ 为 $\displaystyle n$ 维实列向量空间. 证明: $\displaystyle W$ 为 $\displaystyle \mathbb\{R\}^n$ 的子空间当且仅当 $\displaystyle A$ 为半正定阵或半负定阵. (安徽大学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) / $\displaystyle W$ 为 $\displaystyle \mathbb\{R\}^n$ 的一个子空间 $\displaystyle \Leftrightarrow A$ 半正定或半负定, 且 $\displaystyle \dim W=n-\mathrm\{rank\} A$. (3-1)、 $\displaystyle \Leftarrow$: 如有必要, 用 $\displaystyle -W$ 代替 $\displaystyle W$, 而可设 $\displaystyle W$ 半正定, 存在可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}(E\_r,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
\begin\{aligned\} W&=\left\\{X\in\mathbb\{R\}^n; X^\mathrm\{T\} AX=0\right\\}\\\\ &=\left\\{PY\in\mathbb\{R\}^n; Y^\mathrm\{T\} \mathrm\{diag\}(E\_r,0)Y=0, Y\in\mathbb\{R\}^n\right\\}\\\\ &=\left\\{PY; y\_1^2+\cdots+y\_r^2=0\right\\}\\\\ &=\left\\{P(\underbrace\{0,\cdots,0\}\_r,y\_\{r+1\},\cdots,y\_n)^\mathrm\{T\}; y\_i\in\mathbb\{R\}, r+1\leq i\leq n\right\\}\\\\ &=L(Pe\_\{r+1\},\cdots, Pe\_n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据此即知 $\displaystyle W$ 是 $\displaystyle \mathbb\{R\}^n$ 的子空间, 且 $\displaystyle \dim W=n-r=n-\mathrm\{rank\} A$. (3-2)、 $\displaystyle \Rightarrow$: 用反证法. 若 $\displaystyle A$ 既不是半正定, 也不是半负定, 则存在可逆阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}(E\_r,-E\_s,0), r\geq 1, s\geq 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\} W&=\left\\{X\in\mathbb\{R\}^n; X^\mathrm\{T\} AX=0\right\\}\\\\ &=\left\\{PY; Y^\mathrm\{T\} \mathrm\{diag\}(E\_r,-E\_s,0)Y=0, Y\in\mathbb\{R\}^n\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} &(e\_1\pm e\_\{p+1\})^\mathrm\{T\} \mathrm\{diag\}(E\_r,-E\_s,0)(e\_1\pm e\_\{p+1\}) =(\pm 1)^2-(\pm 1)^2=0,\\\\ &(2e\_1)^\mathrm\{T\} \mathrm\{diag\}(E\_r,-E\_s,)(2e\_1)=4\neq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} P(e\_1\pm e\_\{p+1\})\in W, P(2e\_1)=P(e\_1+e\_\{p+1\})+P(e\_1-e\_\{p+1\})\not\in W. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle W$ 不是 $\displaystyle \mathbb\{R\}^n$ 的子空间. 故有结论.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)/
---
1313、 4、 (15 分) 设 $\displaystyle f(x\_1,x\_2,x\_3,x\_4)=X^\mathrm\{T\} AX$ 为实系数二次型, 实对称矩阵 $\displaystyle A$ 的特征值为 $\displaystyle \lambda\_1=1$ (二重), $\displaystyle \lambda\_2=-1$ (二重), 且
\begin\{aligned\} \varepsilon\_1=(1,1,0,0)^\mathrm\{T\}, \varepsilon\_2=(1,1,0,1)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为属于特征值 $\displaystyle \lambda\_1=1$ 的特征向量. 求二次型 $\displaystyle f(x\_1,x\_2,x\_3,x\_4)$ 的表达式. (北京科技大学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) / 设 $\displaystyle x$ 是 $\displaystyle A$ 的属于特征值 $\displaystyle \lambda\_2=-1$ 的特征向量, 则
\begin\{aligned\} \varepsilon\_i^\mathrm\{T\} x=0, i=1,2\Leftrightarrow Bx=0, B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&0\\\\ 1&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\}
由 $\displaystyle B\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&0\\\\ 0&0&0&1\end\{array\}\right)$ 知取 $\displaystyle x\_2,x\_3$ 为自由变量后, $\displaystyle Bx=0$ 的基础解系为
\begin\{aligned\} \varepsilon\_3=(-1,1,0,0)^\mathrm\{T\}, \varepsilon\_4=(0,0,1,0)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle P=(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3,\varepsilon\_4)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&-1&0\\\\ 1&1&1&0\\\\ 0&0&0&1\\\\ 0&1&0&0\end\{array\}\right)$, 则
\begin\{aligned\} &P^\{-1\}AP=\mathrm\{diag\}(1,1,-1,-1)\Rightarrow A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&-1&0\\\\ 0&0&0&1\end\{array\}\right)\\\\ \Rightarrow& f(X)=2x\_1x\_2-x\_3^2+x\_4^2. \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)/
---
1314、 3、 已知 $\displaystyle A$ 而二阶实对称矩阵, 且 $\displaystyle |A|=36, \mathrm\{tr\} A=13$. (1)、 证明: $\displaystyle A$ 为正定矩阵; (2)、 判断二次曲线 $\displaystyle X^\mathrm\{T\} AX=4$ 的形状, 其中 $\displaystyle X=\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right)$. (北京理工大学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) / 设 $\displaystyle A$ 的特征值为 $\displaystyle \lambda\_1,\lambda\_2\in\mathbb\{R\}$, 则由题设,
\begin\{aligned\} \lambda\_1\lambda\_2=36, \lambda\_1+\lambda\_2=13\Rightarrow \lambda\_1=4, \lambda\_2=9. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 正定. 再者, 存在正交矩阵 $\displaystyle P$, 使得 $\displaystyle P^\mathrm\{T\} AP=\mathrm\{diag\}(4,9)$. 故经过正交线性替换 $\displaystyle X=PY$ 后, 二次曲线
\begin\{aligned\} 4=X^\mathrm\{T\} AX=4y\_1^2+9y\_2^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个椭圆.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)/
---
1315、 2、 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}4&4&4\\\\ 0&4&4\\\\ 0&0&4\end\{array\}\right)$. 用正交线性替换将二次型 $\displaystyle f(x)=x^\mathrm\{T\} Ax$ 化为标准形. (北京师范大学2023年高等代数与解析几何考研试题) [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 $\displaystyle B=\frac\{A^\mathrm\{T\}+A\}\{2\}=\left(\begin\{array\}\{cccccccccccccccccccc\}4&2&2\\\\ 2&4&2\\\\ 2&2&4\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 8,2,2$. 由
\begin\{aligned\} 8E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&-1\\\\ 0&1&-1\\\\ 0&0&0 \end\{array\}\right), 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 1&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 $\displaystyle 8,2$ 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\\\\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\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{\sqrt\{3\}\}&-\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\}\}\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\}(8,2,2). \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 $\displaystyle 8y\_1^2+2y\_2^2+2y\_3^2$.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)/
---
1316、 (2)、 用正交线性替换化二次型
\begin\{aligned\} f(x\_1,x\_2,x\_3)=2x\_1^2+3x\_2^2-x\_3^2+4x\_1x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为标准形, 写出所作正交线性替换以及标准形. (大连理工大学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) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&0&2\\\\ 0&3&0\\\\ 2&0&-1\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 3,3,-2$. 由
\begin\{aligned\} 3E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&-2\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), -2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\frac\{1\}\{2\}\\\\ 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 $\displaystyle 3,-2$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\0\\\\1 \end\{array\}\right); \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\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\} 0&\frac\{2\}\{\sqrt\{5\}\}&-\frac\{1\}\{\sqrt\{5\}\}\\\\ 1&0&0\\\\ 0&\frac\{1\}\{\sqrt\{5\}\}&\frac\{2\}\{\sqrt\{5\}\}\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,-2\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 $\displaystyle 3y\_1^2+3y\_2^2-2y\_3^2$.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)/
---
1317、 3、 混合题. (1)、 已知 $\displaystyle A=(a\_\{ij\})$ 为一个 $\displaystyle n$ 阶实矩阵, 对角元 $\displaystyle a\_\{ii\}=a > 0\ (i=1,\cdots,n)$, 且对任意的 $\displaystyle 1\leq i\leq n$, 有
\begin\{aligned\} \sum\_\{j=1\}^n (|a\_\{ij\}|+|a\_\{ji\}|) < 4a. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设二次型 $\displaystyle f=X^\mathrm\{T\} AX$, 其中 $\displaystyle X$ 为实 $\displaystyle n$ 维列向量. (1-1)、 求二次型 $\displaystyle f$ 的矩阵 $\displaystyle B=(b\_\{ij\})$; (1-2)、 证明: $\displaystyle b\_\{ii\} > \sum\_\{j\neq i\}|b\_\{ij\}|, i=1,\cdots,n$; (1-3)、 证明: $\displaystyle f$ 是正定二次型. (大连理工大学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) / 由
\begin\{aligned\} f&=\sum\_\{i,j=1\}^n a\_\{ij\}x\_ix\_j =\sum\_\{j,i=1\}^n a\_\{ji\}x\_jx\_i\\\\ & =\frac\{1\}\{2\}\sum\_\{i,j=1\}^n (a\_\{ij\}+a\_\{ji\})x\_ix\_j \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 的矩阵为 $\displaystyle B=\frac\{A+A^\mathrm\{T\}\}\{2\}$. 由题意,
\begin\{aligned\} &b\_\{11\}=\cdots=b\_\{nn\}=a > 0; 1\leq i\leq n\\\\ \Rightarrow& \sum\_\{j=1\}^n |b\_\{ij\}| \leq \frac\{1\}\{2\}\sum\_\{j=1\}^n (|a\_\{ij\}|+|a\_\{ji\}|) < 2a\\\\ \Rightarrow& \sum\_\{j=1,j\neq i\}^n |b\_\{ij\}| < a=|b\_\{ii\}|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \lambda$ 为 $\displaystyle B$ 的特征值, $\displaystyle \alpha=(a\_1,\cdots,a\_n)^T$ 为其对应的特征向量, 记 $\displaystyle |a\_i|=\max\_\{1\leq j\leq n\}|a\_j|$, 则
\begin\{aligned\} &B\alpha =\lambda \alpha =\sum\_\{j=1\}^n b\_\{ij\}a\_j=\lambda a\_i\\\\ \Rightarrow& \lambda =\frac\{1\}\{a\_i\} \sum\_\{j=1\}^n b\_\{ij\}a\_j \geq b\_\{ii\}-\sum\_\{j=1,j\neq i\} |b\_\{ij\}|\left|\frac\{a\_j\}\{a\_i\}\right| \geq b\_\{ii\}-\sum\_\{j=1,j\neq i\} |b\_\{ij\}| > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了实对称矩阵 $\displaystyle B$ 的特征值均大于零, 而 $\displaystyle B$ 正定, $\displaystyle f$ 的规范形为
\begin\{aligned\} y\_1^2+\cdots+y\_n^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)/
---
1318、 (5)、 设实二次型
\begin\{aligned\} f(x\_1,x\_2,x\_3)=x\_1^2-x\_2^2+2ax\_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\}
的负惯性指数是 $\displaystyle 1$, 则 $\displaystyle a$ 的取值范围是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (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) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&a\\\\ 0&-1&2\\\\ a&2&0\end\{array\}\right)$. 由
\begin\{aligned\} &P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-a\\\\ 0&1&0\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_1^\mathrm\{T\} AP\_1=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&-1&2\\\\ 0&2&-a^2\end\{array\}\right),\\\\ &P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&2\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2=\mathrm\{diag\}(1,-1,4-a^2) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle 4-a^2\geq 0\Leftrightarrow -2\leq a\leq 2$.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)/
---
1319、 5、 设 $\displaystyle n\ ( > 1)$ 元实二次型
\begin\{aligned\} f(x\_1,\cdots,x\_n)=(x\_1,\cdots,x\_n)A\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\}
其中 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\cdots&0&0\\\\ 2a&1&\cdots&0&0\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 2a&2a&\cdots&1&0\\\\ 2a&2a&\cdots&2a&1\end\{array\}\right)$, $\displaystyle a\in \mathbb\{R\}$. 6、 求 $\displaystyle f$ 在正交线性替换下的标准形 (不用写出正交线性替换). 7、 若实二次型 $\displaystyle f$ 正定, 求 $\displaystyle a$ 的取值范围. (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) / $\displaystyle f$ 的矩阵为 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&a&\cdots&a\\\\ a&1&\cdots&a\\\\ \vdots&\vdots&&\vdots\\\\ a&a&\cdots&1\end\{array\}\right)$. 设 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则
\begin\{aligned\} &|\lambda E-B|=|(\lambda-1+a)E-aee^\mathrm\{T\}|\\\\ =&|(\lambda-1+a)E|\cdot |E-(\lambda-1+a)^\{-1\}aee^\mathrm\{T\}|\\\\ =&(\lambda-1+a)^n [1-(\lambda-1+a)^\{-1\}ae^\mathrm\{T\} e]\\\\ =&(\lambda-1+a)^n [\lambda-1+a-na]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle B$ 的特征值为 $\displaystyle 1-a$ ($n-1$ 重), $\displaystyle (n-1)a+1$ (单重). $\displaystyle f$ 在正交线性替换下的标准形为
\begin\{aligned\} (1-a)(y\_1^2+\cdots+y\_\{n-1\}^2)+[(n-1)a+1]y\_n^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle f$ 正定 $\displaystyle \Leftrightarrow 1-a > 0, (n-1)a+1 > 0\Leftrightarrow -\frac\{1\}\{n-1\} < a < 1$.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)/
---
1320、 (4)、 确定二次型
\begin\{aligned\} f(x\_1,x\_2,x\_3)=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\}
的正负惯性指数. (福州大学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) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&2&2\\\\ 2&0&2\\\\ 2&2&0\end\{array\}\right)$, 易知 $\displaystyle A$ 的特征值为 $\displaystyle 4,-2,-2$, 而 $\displaystyle f$ 的正负惯性指数分别为 $\displaystyle 1,2$.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)/
---
1321、 (6)、 (12 分) 将二次型
\begin\{aligned\} f(x\_1,x\_2,x\_3)=3x\_1^2+3x\_3^2+4x\_1x\_2+8x\_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\}
利用正交线性替换化为标准形, 并写出所用的正交线性替换和标准型. (福州大学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) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}3&2&4\\\\ 2&0&2\\\\ 4&2&3\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 8,-1,-1$. 由
\begin\{aligned\} 8E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&-\frac\{1\}\{2\}\\\\ 0&0&0 \end\{array\}\right), -E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&\frac\{1\}\{2\}&1\\\\ 0&0&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 $\displaystyle 8,-1$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\1\\\\2 \end\{array\}\right);\quad \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\2\\\\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\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{2\}\{3\}&-\frac\{1\}\{\sqrt\{5\}\}&-\frac\{4\}\{3\sqrt\{5\}\}\\\\ \frac\{1\}\{3\}&\frac\{2\}\{\sqrt\{5\}\}&-\frac\{2\}\{3\sqrt\{5\}\}\\\\ \frac\{2\}\{3\}&0&\frac\{\sqrt\{5\}\}\{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(8,-1,-1\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 $\displaystyle 8y\_1^2-y\_2^2-y\_3^2$.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)/
---
1322、 (7)、 (14 分) 已知 $\displaystyle A=(a\_\{ij\})\_\{n\times n\}$ 为正定矩阵. 记
\begin\{aligned\} f(x\_1,\cdots,x\_n)=\left|\begin\{array\}\{cccccccccc\}0&x\_1&\cdots&x\_n\\\\ x\_1&a\_\{11\}&\cdots&a\_\{1n\}\\\\ \vdots&\vdots&&\vdots\\\\ x\_n&a\_\{n1\}&\cdots&a\_\{nn\}\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\}
证明: $\displaystyle f(x\_1,\cdots,x\_n)$ 是一个负定二次型. (福州大学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) / 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&-x^\mathrm\{T\} A^\{-1\}\\\\ 0&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}0&x^\mathrm\{T\}\\\\ x&A\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}-x^\mathrm\{T\} A^\{-1\}x&0\\\\ x&A\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 $\displaystyle f(x)=-|A| x^\mathrm\{T\} A^\{-1\}x$. 再由 $\displaystyle A$ 正定知 $\displaystyle A^\{-1\}$ 正定, $\displaystyle -|A|A^\{-1\}$ 负定, 从而 $\displaystyle f$ 负定.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)/
---
1323、 (5)、 设
\begin\{aligned\} f(x\_1,x\_2,x\_3)=5x\_1^2+5x\_2^2+cx\_3^2-2x\_1x\_2+6x\_1x\_3-6x\_2x\_3 \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 $\displaystyle 2$, 则 $\displaystyle c=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (广西大学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 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}5&-1&3\\\\ -1&5&-3\\\\ 3&-3&c\end\{array\}\right)$. 由题设, $\displaystyle 0=|A|=24(c-3)\Rightarrow c=3$.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)/
---
1324、 8、 用正交变换将二次型
\begin\{aligned\} f(x\_1,x\_2,x\_3)=2x\_1^2+5x\_2^2+5x\_3^2+4x\_1x\_2-4x\_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\}
化为标准形, 并写出所用的正交变换. (哈尔滨工程大学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
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&-2\\\\ 2&5&-4\\\\ -2&-4&5\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\}
易知 $\displaystyle A$ 的特征值为 $\displaystyle 10,1,1$. 由
\begin\{aligned\} 10E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}2&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 $\displaystyle 10$ 的特征向量为
\begin\{aligned\} \xi\_1=\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\}
由
\begin\{aligned\} 1E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&-2\\\\ 0&0&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 $\displaystyle 1$ 的特征向量为
\begin\{aligned\} \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\1\\\\0\end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\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 $\displaystyle \eta\_1,\eta\_2,\eta\_3$, 并设
\begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{3\}&-\frac\{2\}\{\sqrt\{5\}\}&\frac\{2\}\{3\sqrt\{5\}\}\\\\ -\frac\{2\}\{3\}&\frac\{1\}\{\sqrt\{5\}\}&\frac\{4\}\{3\sqrt\{5\}\}\\\\ \frac\{2\}\{3\}&0&\frac\{\sqrt\{5\}\}\{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=\mathrm\{diag\}(10,1,1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故经过正交替换 $\displaystyle x=Py$ 后, $\displaystyle f$ 化为标准形 $\displaystyle 10y\_1^2+y\_2^2+y\_3^2$.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)/
---
1325、 7、 若实对称矩阵 $\displaystyle A\_\{n\times n\}$ 为二次型 $\displaystyle f(x\_1,\cdots,x\_n)$ 的矩阵. 证明: $\displaystyle f(x\_1,\cdots,x\_n)$ 半正定的充要条件为存在实对称矩阵 $\displaystyle B$, 使得 $\displaystyle A=B^2$. (哈尔滨工业大学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)、 $\displaystyle \Leftarrow$: 由 $\displaystyle A=B^\mathrm\{T\} B$ 知
\begin\{aligned\} \forall\ x\in\mathbb\{R\}^n, x^\mathrm\{T\} Ax\stackrel\{y=Bx\}\{=\}y^\mathrm\{T\} y\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ (而 $\displaystyle f$) 半正定. (2)、 $\displaystyle \Rightarrow$: 由 $\displaystyle f$ 半正定知 $\displaystyle A$ 半正定, 而存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} A=P\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)P^\mathrm\{T\}, \quad \lambda\_i\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle B=P\mathrm\{diag\}(\sqrt\{\lambda\_1\},\cdots,\sqrt\{\lambda\_n\})P^\mathrm\{T\}$, 则 $\displaystyle B$ 实对称, 且 $\displaystyle A=B^2$.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)/
---
1326、 3、 (1)、 设实二次型
\begin\{aligned\} f(x\_1,x\_2,x\_3)=x\_1^2+2x\_2^2+6x\_3^2-2x\_1x\_2+2x\_1x\_3-6x\_2x\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
用可逆线性变换将 $\displaystyle f$ 化为规范形, 并求出所用的可逆线性替换, 并说明该二次型对应的矩阵 $\displaystyle A$ 是正定矩阵. (2)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ -1&2&-3\\\\ 1&-3&6\end\{array\}\right)$, 求可逆矩阵 $\displaystyle D$ 使得 $\displaystyle A=D^\mathrm\{T\} D$. (合肥工业大学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 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ -1&2&-3\\\\ 1&-3&6\end\{array\}\right)$. 由
\begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&-1\\\\ 0&1&0\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_1^\mathrm\{T\} AP\_1=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&-2\\\\ 0&-2&5\end\{array\}\right),\\\\ P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&2\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2=E\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知经过可逆线性替换
\begin\{aligned\} X=PY, P=P\_1P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 0&1&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\}
后, $\displaystyle f$ 化为了规范形 $\displaystyle y\_1^2+y\_2^2+y\_3^2$. 故 $\displaystyle f$ 正定. (2)、 由第 1 步知 $\displaystyle P^\mathrm\{T\} AP=E\_3\Rightarrow A=P^\{-\mathrm\{T\}\}P^\{-1\}$. 故取 $\displaystyle D=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ 0&1&-2\\\\ 0&0&1\end\{array\}\right)$, 则 $\displaystyle A=D^\mathrm\{T\} D$.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)/
---
1327、 8、 设 $\displaystyle A$ 为实可逆矩阵, (1)、 $\displaystyle f(x)=\left|\begin\{array\}\{cccccccccc\}0&-x^\mathrm\{T\}\\\\ x&A\end\{array\}\right|$, 求 $\displaystyle f$ 的矩阵; (2)、 $\displaystyle A$ 为实对称矩阵, 求 $\displaystyle A$ 与 $\displaystyle f$ 的正负惯性指数之间的关系. (河北工业大学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)、 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&-x^\mathrm\{T\} A^\{-1\}\\\\ 0&E\_n\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}0&-x^\mathrm\{T\}\\\\ x&A\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}-x^\mathrm\{T\} A^\{-1\}x&0\\\\ x&A\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
\begin\{aligned\} f(x)=-|A|\cdot x^\mathrm\{T\} A^\{-1\}x=x^\mathrm\{T\} A^\star x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 的矩阵为 $\displaystyle \frac\{A^\star+(A^\star)^\mathrm\{T\}\}\{2\}$. (2)、 由 $\displaystyle A$ 实对称知存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} &P^\mathrm\{T\} AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i\left\\{\begin\{array\}\{llllllllllll\} > 0,&1\leq i\leq r,\\\\ < 0,&r+1\leq i\leq n\end\{array\}\right.\\\\ \Rightarrow&P^\mathrm\{T\} A^\star P=\mathrm\{diag\}\left(\frac\{|A|\}\{\lambda\_1\},\cdots,\frac\{|A|\}\{\lambda\_n\}\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 $\displaystyle |A| > 0$, 则 $\displaystyle A$ 与 $\displaystyle A^\star (f)$ 的正负惯性指数相同, 分别为 $\displaystyle r,n-r$. (2-2)、 若 $\displaystyle |A| < 0$, 则 $\displaystyle A$ 的正惯性指数 $\displaystyle =r=A^\star (f)$ 的负惯性指数, $\displaystyle A$ 的负惯性指数 $\displaystyle =n-r=A^\star (f)$ 的正惯性指数.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)/
---
1328、
(10)、 使得二次型
\begin\{aligned\} q(x,y,z)=x^2+y^2+z^2+yz+xz+txy\left(t\in\mathbb\{R\}\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 $\displaystyle t$ 的取值范围为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学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) / $\displaystyle q$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&\frac\{t\}\{2\}&\frac\{1\}\{2\}\\\\ \frac\{t\}\{2\}&1&\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}&\frac\{1\}\{2\}&1\end\{array\}\right)$. 而 $\displaystyle q$ 正定
\begin\{aligned\} \Leftrightarrow 1 > 0, 1-\frac\{t^2\}\{4\} > 0, |A|=-\frac\{1\}\{4\}(t+1)(t-2) > 0\Leftrightarrow -1 < t < 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)/
---
1329、 4、 $\displaystyle A$ 为 $\displaystyle n$ 阶正定矩阵, $\displaystyle B$ 为 $\displaystyle n$ 阶实可逆矩阵, 二次型 $\displaystyle f(X)$ 的矩阵为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&B^\mathrm\{T\}\\\\ B&0\end\{array\}\right)$. (1)、 证明: $\displaystyle B^\mathrm\{T\} A^\{-1\}B$ 是正定矩阵; (2)、 求 $\displaystyle f$ 的正负惯性指数. (华南理工大学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)、 由 $\displaystyle A$ 正定知存在可逆阵 $\displaystyle P$, 使得 $\displaystyle P^\mathrm\{T\} AP=E$, 而 $\displaystyle P^\mathrm\{T\} B$ 可逆, $\displaystyle P^\mathrm\{T\} Bx=0$ 只有零解. 对 $\displaystyle \forall\ 0\neq \alpha\in\mathbb\{R\}^n$, $\displaystyle \beta=P^\mathrm\{T\} B\alpha\neq 0$, 而
\begin\{aligned\} \alpha^\mathrm\{T\}(B^\mathrm\{T\} A^\{-1\}B)\alpha=\alpha^\mathrm\{T\} B^\mathrm\{T\} PP^\mathrm\{T\} B\alpha=\beta^\mathrm\{T\} \beta > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle B^\mathrm\{T\} A^\{-1\}B$ 正定. (2)、 第 1 问中用 $\displaystyle B^\mathrm\{T\}$ 代替 $\displaystyle B$ 得到 $\displaystyle BA^\{-1\}B^\mathrm\{T\}$ 正定. 再由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -BA^\{-1\}&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A&B^\mathrm\{T\}\\\\ B&0\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A^\{-1\}B^\mathrm\{T\}\\\\ 0&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A&0\\\\ 0&-BA^\{-1\}B^\mathrm\{T\}\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 $\displaystyle f$ 的正负惯性指数都为 $\displaystyle n$.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)/
---
1330、 5、 设二次型 $\displaystyle f(x\_1,x\_2,x\_3,x\_4)=\mu(x\_1^2+x\_2^2+x\_3^2)+2x\_1x\_2-2x\_1x\_3-2x\_2x\_3+x\_4^2$. (1)、 写出 $\displaystyle f(x\_1,x\_2,x\_3,x\_4)$ 的矩阵 $\displaystyle A$; (2)、 依据参数 $\displaystyle \mu$ 的值, 讨论 $\displaystyle f(x\_1,x\_2,x\_3,x\_4)$ 的秩, 正惯性指数和符号差. (华南师范大学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) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\mu&1&-1&0\\\\ 1&\mu&-1&0\\\\ -1&-1&\mu&0\\\\ 0&0&0&1\end\{array\}\right)$. $\displaystyle A$ 的特征值为 $\displaystyle \mu+2, \mu-1, \mu-1, 1$. 故
\begin\{aligned\} \begin\{array\}\{cccccc\} \mu&\mathrm\{rank\} f&\mbox\{正惯性指数\}&\mbox\{符号差\}\\\\ \mu < -2&4&1&-2\\\\ \mu=-2&3&1&-1\\\\ -2 < \mu < 1&4&2&0\\\\ \mu=1&2&2&2\\\\ \mu > 1&4&4&4 \end\{array\} \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)/
---
1331、 5、 二次型 $\displaystyle f(x,y,z)=\alpha x^2+2y^2-2z^2+2\beta xz\ (\beta > 0)$ 的矩阵为 $\displaystyle A$. 若 $\displaystyle A$ 的特征值之积为 $\displaystyle -12$, 和为 $\displaystyle 1$. 求 $\displaystyle \alpha,\beta$, 并求正交变换使之为标准形, 写出对应的正交矩阵. (暨南大学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) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha&0&\beta\\\\ 0&2&0\\\\ \beta&0&-2\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,2,-3$. 由
\begin\{aligned\} 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-2\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), -3E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\frac\{1\}\{2\}\\\\ 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 $\displaystyle 2,-3$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\2 \end\{array\}\right); \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\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\} 0&\frac\{2\}\{\sqrt\{5\}\}&-\frac\{1\}\{\sqrt\{5\}\}\\\\ 1&0&0\\\\ 0&\frac\{1\}\{\sqrt\{5\}\}&\frac\{2\}\{\sqrt\{5\}\}\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,-3\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 $\displaystyle 2y\_1^2+2y\_2^2-3y\_3^2$.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)/
---
1332、 6、 (15 分) 求 $\displaystyle n$ 元实二次型 $\displaystyle \sum\_\{1\leq i < j\leq n\}x\_ix\_j$ 的负惯性指数. (南京大学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 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&\frac\{1\}\{2\}&\cdots&\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}&0&\cdots&\frac\{1\}\{2\}\\\\ \vdots&\vdots&&\vdots\\\\ \frac\{1\}\{2\}&\frac\{1\}\{2\}&\cdots&0\end\{array\}\right)$. 设 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则 $\displaystyle A=-\frac\{1\}\{2\}I\_n+\frac\{1\}\{2\}ee^\mathrm\{T\}$,
\begin\{aligned\} &|\lambda I\_n-A|=\left|\left(\lambda+\frac\{1\}\{2\}\right)I\_n-\frac\{1\}\{2\}ee^\mathrm\{T\}\right|\\\\ =&\left|\left(\lambda+\frac\{1\}\{2\}\right)I\_n\right|\cdot \left|I\_n-\left(\lambda+\frac\{1\}\{2\}\right)^\{-1\}\frac\{1\}\{2\}ee^\mathrm\{T\}\right|\\\\ =&\left(\lambda+\frac\{1\}\{2\}\right)^n \left\[1-\left(\lambda+\frac\{1\}\{2\}\right)^\{-1\}\frac\{1\}\{2\}e^\mathrm\{T\} e\right\]\\\\ =&\left(\lambda+\frac\{1\}\{2\}\right)^\{n-1\}\left(\lambda+\frac\{1\}\{2\}-\frac\{n\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 的特征值为 $\displaystyle -\frac\{1\}\{2\}$ ($n-1$ 重), $\displaystyle \frac\{n-1\}\{2\}$ (单重). 进而二次型的负惯性指数为 $\displaystyle n-1$.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)/
---
1333、 5、 设三阶实矩阵 $\displaystyle A$ 的 $\displaystyle 3$ 个列向量 $\displaystyle \alpha,\beta,\gamma$ 线性无关, 二次型
\begin\{aligned\} f(x)=(\alpha^\mathrm\{T\} x)^2+(\beta^\mathrm\{T\} x)^2+(\gamma^\mathrm\{T\} x)^2, x=(x\_1,x\_2,x\_3)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求此二次型的矩阵 $\displaystyle B$; (2)、 问: 此二次型是否正定? 并写出此二次型的规范形; (3)、 是否存在正定矩阵 $\displaystyle S$, 使得 $\displaystyle B=S^3$? 并说明理由. (南京航空航天大学大学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)、 由
\begin\{aligned\} f(x)=&x^\mathrm\{T\} \alpha\alpha^\mathrm\{T\} x+x^\mathrm\{T\} \beta\beta^\mathrm\{T\} x+x^\mathrm\{T\} \gamma\gamma^\mathrm\{T\} x\\\\ =&x^\mathrm\{T\}(\alpha\alpha^\mathrm\{T\}+\beta\beta^\mathrm\{T\}+\gamma\gamma^\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\}
知
\begin\{aligned\} B=\alpha\alpha^\mathrm\{T\}+\beta\beta^\mathrm\{T\}+\gamma\gamma^\mathrm\{T\}=(\alpha,\beta,\gamma)\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha^\mathrm\{T\}\\\\\beta^\mathrm\{T\}\\\\\gamma^\mathrm\{T\}\end\{array\}\right) =AA^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle A$ 可逆知 $\displaystyle A^\mathrm\{T\}$ 可逆, $\displaystyle A^\mathrm\{T\} x=0$ 只有零解, 而
\begin\{aligned\} x\neq 0\Rightarrow y=A^\mathrm\{T\} x\neq 0\Rightarrow f(x)=x^\mathrm\{T\} AA^\mathrm\{T\} x=y^\mathrm\{T\} y > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 正定, 规范形为 $\displaystyle z\_1^2+z\_2^2+z\_3^2$. (3)、 存在. 由第 2 步知 $\displaystyle B$ 正定, 而存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} B=P\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)P^\mathrm\{T\}, 0 < \lambda\_i\in\mathbb\{R\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle S=P\mathrm\{diag\}\left(\sqrt[3]\{\lambda\_1\},\cdots,\sqrt[3]\{\lambda\_n\}\right)P^\mathrm\{T\}$, 则 $\displaystyle S$ 正定, 且 $\displaystyle S^3=B$.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)/
---
1334、 (6)、 $\displaystyle t$ 满足条件 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ 时, 二次型
\begin\{aligned\} f(x\_1,x\_2,x\_3)=x\_1^2+4x\_2^2+2x\_3^2+2tx\_1x\_2-2x\_1x\_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. (南京理工大学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) / $\displaystyle f$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&t&-1\\\\ t&4&0\\\\ -1&0&2\end\{array\}\right)$, 而 $\displaystyle f$ 正定 $\displaystyle \Leftrightarrow A$ 的顺序主子式
\begin\{aligned\} 1 > 0, 4-t^2 > 0, |A|=4-2t^2 > 0\Leftrightarrow -\sqrt\{2\} < t < \sqrt\{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)/
发表于 2023-3-5 13:14:38
