一些数学基本概念的梳理(高代篇)

\begin{aligned}
1.&由于f(\lambda)=|\lambda E-A|， \\
&f(\lambda) = \lambda^n-(a_{11}+a_{22}+\dots+a_{nn})\lambda^{n-1}+\dots+(-1)^n|A|.\\
&其中a_{11}+a_{22}+\dots+a_{nn}记作tr(A).\\
&这个结论的\lambda^n和\lambda^{n-1}项前面的系数用行列式主对角线展开得到，最后一项令\lambda=0即得.\\
2.&若A的特征值是\lambda，则f(A)的特征值是f(\lambda).
\end{aligned}

\begin{aligned}
&矩阵A的特征多项式为f(\lambda)，则f(A) = O.\\
&\begin{split}
证明：\\
&设 B(\lambda)为\lambda E-A的伴随矩阵. 则B(\lambda)\cdot(\lambda E-A) = f(\lambda)\cdot E.\\
&因为B(\lambda)中全部元素都是\lambda E-A的代数余子式，故其中\lambda的次数不大于n-1.\\
&\begin{split}
故B(\lambda)&可以写作矩阵多项式：\\
&B(\lambda)=\lambda^{n-1}B_0+\lambda^{n-2}B_1+\dots +\lambda B_{n-1}+B_n.
\end{split}\\
&\begin{split}
而f(\lambda)&可以写作：\\
&f(\lambda)=a_0\lambda^n+a_1\lambda^{n-1}+\dots+a_n\lambda+a_{n+1}.\\
\end{split}\\
&\begin{split}
考察B&(\lambda)\cdot(\lambda E-A)=\\
&\lambda^nB_0+\lambda^{n-1}(B_1-B_0\cdot A)+\dots+\lambda(B_{n-1}-B_n\cdot A)+B_n.\\
\end{split}\\
&\begin{split}
而f(\lambda)&\cdot E=\\
&\lambda^{n}a_0E+\lambda^{n-1}a_1E+...+\lambda a_nE+a_{n+1}E.\\
\end{split}\\
&\left\{
\begin{array} {lll}
B_0 &= a_0E,\\
B_1-B_0\cdot A &= a_1E,\\
\dots\\
B_n &= a_{n+1}E.
\end{array}\\
\right.\\
&在上面各式中分别右乘A^n, A^{n-1}, \dots, A, E.\\
&\left\{
\begin{array} {lll}
B_0\cdot A^n &= a_0A^n,\\
B_1\cdot A^{n-1}-B_0\cdot A^n &= a_1A^{n-1},\\
\dots\\
B_n &= a_{n+1}E.
\end{array}\\
\right.\\
&连加即得f(A)=O.
\end{split}
\end{aligned}

\begin{aligned}
&W是V的子空间，对于V上的线性变换\sigma，若\sigma W\in W，则称W是\sigma-不变子空间.\\
&\sigma-不变子空间的性质:\\
&1) 平凡子空间(\text{Trivial Subspace})：\{\textbf 0\} 和V. \\
&(1-1)\because \sigma\mathbf 0 = \mathbf 0 \in \{\mathbf 0\}. \\
&\therefore \{\mathbf0\} 是 \sigma-不变子空间.\\
&(1-2)\forall \alpha \in V, \because\sigma\alpha\in V,\therefore V是\sigma-不变子空间.\\
&2)\sigma V和 \ker \sigma为\sigma-不变子空间.\\
&(2-1) \forall \alpha=\sigma\beta \in \sigma V, \sigma\alpha=\sigma(\sigma\beta)\in\sigma V.\sigma V在\sigma下的像仍在\sigma V下.\therefore \sigma V是\sigma-不变子空间.\\
&(2-2)\forall \alpha \in \ker \sigma, \sigma\alpha = \mathbf 0.而\sigma \mathbf 0=\mathbf 0，即\sigma(\sigma\alpha)=\mathbf 0.即\sigma\alpha \in \ker \sigma. \therefore \sigma V是\sigma-不变子空间.\\
&3)若\sigma和\phi是可交换的，那么\sigma V和\ker \sigma 是\phi-子空间.(反过来也成立)\\
&(3-1)\forall \alpha=\sigma\beta \in \sigma V, \phi\alpha=\phi(\sigma\beta)=(\phi\sigma)\beta=(\sigma\phi)\beta=\sigma(\phi\beta)\in\sigma V. \therefore \sigma V是\sigma-不变子空间.\\
&(3-2)\forall \alpha\in\ker\sigma,\sigma\alpha=\mathbf 0,那么\phi\sigma\alpha=\mathbf 0.即\sigma(\phi\alpha)=0.那么\phi\alpha \in \ker \sigma. \therefore\sigma V是\sigma-不变子空间.\\
&4)任意子空间是数乘变换的不变子空间.\\
&(4)\forall \alpha \in W, \kappa\alpha=k\alpha\in W. (线性空间对数乘封闭)\\
&5)\sigma属于\lambda_i的全部特征向量\xi张成的线性空间是V_{\lambda_i}子空间. \\
&容易知，\sigma在特征子空间上为数乘变换.\\
&6)不变子空间的交和并仍然是不变子空间.\\
&7)W=L(\alpha_1,\alpha_2,\dots,\alpha_s)，W是\sigma-不变子空间的充要条件是\sigma\alpha_1, \sigma\alpha_2, \dots, \sigma\alpha_n 全属于W.
\end{aligned}

\begin{aligned}
&\lambda_1,\lambda_2,\dots,\lambda_s是线性变换A的全部特征值. \\
&f(\lambda)=\prod_{i=1}^s (\lambda-\lambda_i)^{r_i}.\\
&\begin{split}
试证：\\
&V=V_1\oplus V_2\oplus...\oplus V_s.\\
&其中V_i=\{\left.\xi \right| (A-\lambda_iE)^{r_i}\xi=0,\xi\in V \}.\\
&i=1,2,\dots,s.
\end{split}
\end{aligned}

\begin{split}
证明：\\
&(由\text{Hamilton-Cayley}定理，f(A)=O.)\\
&设f_i(\lambda)=f(\lambda)/(\lambda-\lambda_i)^{r_i}.\\
&显然，V_i=f_i(A)V.\\
&(A-\lambda_iE)^{r_i}V_i = f(A)V=\{\bf0\}.\\
&下面先证明，V=V_1+V_2+\dots+V_s.\\
&显然，(f_1(\lambda), f_2(\lambda), \dots,f_s(\lambda)) = 1.\\
&故而\exists u_1(\lambda), u_2(\lambda), \dots, u_s(\lambda),\\
&\text{s.t.} u_1(\lambda)f_1(\lambda)+u_2(\lambda)f_2(\lambda)+\dots+u_s(\lambda)f_s(\lambda)=1.\\
&令\lambda=A，那么：\\
&u_1(A)f_1(A)+u_2(A)f_2(A)+\dots+u_s(A)f_s(A)=E.\\
&取\alpha\in V.上式两边右乘\alpha.\\
&u_1(A)f_1(A)\alpha+u_2(A)f_2(A)\alpha+\dots+u_s(A)f_s(A)\alpha=\alpha.\\
&由于u_i(A), f_i(A)可交换，故f_i(A)V是u_i(A)-不变子空间.\\
&u_i(A)(f_i(A)V) \in f_i(A)V. i = 1,2,\dots, s.\\
&那么上式中的u_i(A)f_i(A)\alpha 分别属于 V_1, V_2, \dots, V_s.\\
&这就证明了 V = V_1 + V_2 + \dots + V_s.\\
&下面接着证明 V=V_1\oplus V_2\oplus...\oplus V_s：\\
&而这只需证明零元素的表出方式唯一.\\
&取一组\beta_1, \beta_2, \dots, \beta_s 分别属于 V_1, V_2, \dots, V_s，它们的和为零元素.\\
&由于它们的和是V中元素，所以这样的一组向量总能找到.\\
&即\beta_1 + \beta_2 + \dots + \beta_s = \bf 0 \cdots (\#) \\
&对于 i \ne j\in\{1,2,\dots,s\}，有(\lambda-\lambda_j)^{r_j} | f_i(\lambda).\\
&又\because (A-\lambda_jE)^{r_j}\beta_j = 0, \therefore f_i(A)\beta_j = \bf 0.\\
&对(\#)式两边左乘f_i(A),可得f_i(A)\beta_i=\bf 0.\\
&我们下一步要得到\beta_i为\bf 0.\\
&注意到(f_i(\lambda), (\lambda - \lambda_i)^{r_i}) = 1.\\
&那么\exists u(\lambda), v(\lambda). \text{s.t.} u(\lambda)f_i(\lambda) + v(\lambda)(\lambda - \lambda_i)^{r_i}=1.\\
&同样用\lambda = A代入，并且将其作用于\beta_i上.\\
&\begin{split}
\beta_i 
&= u(\lambda)f_i(\lambda)\beta_i + v(\lambda)(\lambda - \lambda_i)^{r_i}\beta_i\\
&= \bf {0}. \\(i=1,2,\dots,s)
\end{split}\\
&故零元素的表法唯一.即得证.
\end{split}

\left( \vec a, \vec b \right)^2 \le \left( \vec a, \vec a \right)  \left( \vec b, \vec b\right), 当且仅当\vec a和\vec b线性相关时取“=”。

\begin{aligned}
&a = x_1\epsilon_1+ x_2\epsilon_2 + \dots + x_n\epsilon_n,\\
&b = y_1\epsilon_1+ y_2\epsilon_2 + \dots + y_n\epsilon_n.\\
&\\
&记X=(x_1, x_2, \dots, x_n), Y = (y_1, y_2, \dots, y_n).\\
& A = \left(
\begin{array}{ccc}
(\epsilon_1, \epsilon_1) &(\epsilon_1, \epsilon_2) & \dots &(\epsilon_1, \epsilon_n)\\
(\epsilon_2, \epsilon_1) &(\epsilon_2, \epsilon_2) & \dots &(\epsilon_2, \epsilon_n)\\
\vdots &\vdots &\ddots &\vdots\\
(\epsilon_n, \epsilon_1) &(\epsilon_n, \epsilon_2) &\dots &(\epsilon_n, \epsilon_n)
\end{array}
\right).\\
&\\
&a \cdot b = X^TAY.
\end{aligned}