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$\bf命题2：$设实对称阵$A$的最大特征值等于$x'Ax$的最大值，其中$x$取${R^n}$中的单位向量

方法一：由$A$实对称知，存在正交阵$Q$，使得
\[A = Q'diag\left( {{\lambda _1}, \cdots ,{\lambda _n}} \right)Q\]
其中${{\lambda _1}, \cdots ,{\lambda _n}}$为$A$的特征值，且${{\lambda _1} \ge \cdots \ge {\lambda _n}}$，于是
\[x'Ax = {\left( {Qx} \right)^\prime }diag\left( {{\lambda _1}, \cdots ,{\lambda _n}} \right)Qx\]
令\[Qx = {\left( {{y_1}, \cdots ,{y_n}} \right)^\prime }\]
则\[x'Ax = {\lambda _1}{y_1}^2 + \cdots + {\lambda _n}{y_n}^2 \le {\lambda _1}\sum\limits_{i = 1}^n {{y_i}^2} \]
而\[\sum\limits_{i = 1}^n {{y_i}^2} = {\left| {Qx} \right|^2} = {\left| x \right|^2} = 1\]
所以$x'Ax \le {\lambda _1}$，下面我们设${x_0}$为${\lambda _1}$对应的单位特征向量，则
\[A{x_0} = {\lambda _1}{x_0},\left| {{x_0}} \right| = 1\]
于是\[{x_0}^\prime A{x_0} = {x_0}^\prime {\lambda _1}{x_0} = {\lambda _1}{\left| {{x_0}} \right|^2} = {\lambda _1}\]
故${\lambda _1}{\rm{ = }}\mathop {max}\limits_{\left| x \right| = 1} x'Ax{\rm{ }}$