中国科学院真题 第20146题

3．解 记 $A=\left(\begin{array}{cc}1 & \frac{\alpha}{n} \\ \frac{\alpha}{n} & 1\end{array}\right)$ ， 当 $\alpha=0$ ，显然 $$ \lim _{n \rightarrow+\infty}\left(\begin{array}{cc} 1 & \frac{\alpha}{n} \\ \frac{\alpha}{n} & 1 \end{array}\right)^{n}=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) $$ 当 $\alpha \neq 0$ 时， $$ |\lambda E-A|=(\lambda-1)^{2}-\frac{\alpha^{2}}{n^{2}}=0 $$ 从而 $A$ 的特征值为 $\lambda_{1,2}=1 \pm \frac{\alpha}{n}$ ． 对应于特征值 $\lambda_{1}=1+\frac{\alpha}{n}$ 的特征向量为 $$ \alpha_{1}=\binom{-1}{1} $$ 对应于特征值 $\lambda_{1}=1+\frac{\alpha}{n}$ 的特征向量为 $$ \alpha_{2}=\binom{1}{1} $$ 单位正交化， $$ \beta_{1}=\frac{\alpha_{1}}{\sqrt{2}}, \beta_{2}=\frac{\alpha_{2}}{\sqrt{2}} $$ 从而， $$ P^{-1} A P=\left(\begin{array}{cc} 1+\frac{\alpha}{n} & \\ & 1-\frac{\alpha}{n} \end{array}\right), P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} -1 & 1 \\ 1 & 1 \end{array}\right) \text {, } $$ 从而 $A^{n}=P\left(\begin{array}{cc}\left(1+\frac{\alpha}{n}\right)^{n} & \\ & \left(1-\frac{\alpha}{n}\right)^{n}\end{array} P^{-1}\right.$ ， $$ \begin{aligned} \lim _{n \rightarrow+\infty} A^{n} & =\lim _{n \rightarrow+\infty} P\left(\begin{array}{rl} \left(1+\frac{\alpha}{n}\right)^{n} & \left(1-\frac{\alpha}{n}\right)^{n} \end{array}\right) P^{-1}=\left(\begin{array}{cc} -1 & 1 \\ 1 & 1 \end{array}\right)\left(\begin{array}{ll} e^{\alpha} & \\ & e^{-\alpha} \end{array}\right) \cdot \frac{1}{2}\left(\begin{array}{cc} -1 & 1 \\ 1 & 1 \end{array}\right) \\ & =\frac{1}{2}\left(\begin{array}{cc} e^{\alpha}+e^{-\alpha} & -e^{\alpha}+e^{-\alpha} \\ -e^{\alpha}+e^{-\alpha} & e^{\alpha}+e^{-\alpha} \end{array}\right) \end{aligned} $$ 先用数学归纳法，证 $A^{n}=\left(\begin{array}{cccc}a^{n} & C_{n}^{1} a^{n-1} & \cdots & C_{n}^{100-1} a^{m-100+1} \\ & a^{n} & \cdots & C_{n}^{100-2} a^{m-100+2} \\ & & \ddots & \vdots \\ & & & a^{n}\end{array}\right)$ ． 当 $n=2$ 时， $$ A^{2}=\left(\begin{array}{cccc} a^{2} & 2 a & \cdots & 0 \\ & a^{2} & \cdots & 0 \\ & & \ddots & \vdots \\ & & & a^{2} \end{array}\right), \text { 满足结论. } $$ 假定 $n=k-1$ 时，结论成立．则当 $n=k$ 时， $$ \begin{aligned} A^{k}=A A^{k-1} & =\left(\begin{array}{cccc} a & 1 & & \\ & a & \ddots & \\ & & \ddots & 1 \\ & & & a \end{array}\right)\left(\begin{array}{cccc} a^{k-1} & C_{k-1}^{1} a^{k-2} & \cdots & C_{k-1}^{100-1} a^{k-1-100+1} \\ & a^{k-1} & \cdots & C_{k-1}^{100-2} a^{k-1-100+2} \\ & & \ddots & \vdots \\ & & & a^{k-1} \end{array}\right), \\ & =\left(\begin{array}{cccc} a^{k} & C_{k}^{1} a^{k-1} & \cdots & C_{k}^{100-1} a^{k-100+1} \\ & a^{k} & \cdots & C_{k}^{100-2} a^{k-100+2} \\ & & \ddots & \vdots \\ & & & a^{k} \end{array}\right) \end{aligned} $$ 结论对 $n=k$ 成立． 从而，$A^{50}$ 的第一行元素之和为 $a^{50}+C_{50}^{1} a^{49}+\cdots+C_{50}^{49} a+1=(a+1)^{50}$ ． 假定它们线性相关，即存在不全为零的数 $k_{1}, \cdots, k_{n}$ ，使得 $$ k_{1}\left(\alpha_{1}+\alpha_{2}\right)+\cdots+k_{n-1}\left(\alpha_{n-1}+\alpha_{n}\right)+k_{n}\left(\alpha_{n}+\alpha_{1}\right)=0, $$ 由 $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ 线性无关，知 $$ \begin{aligned} & k_{1}+k_{2}=0 \\ & k_{2}+k_{3}=0 \\ & k_{3}+k_{4}=0 \\ & \vdots \\ & k_{1}+k_{n}=0 \end{aligned} $$ 也即 $$ \left(\begin{array}{ccccc} 1 & 1 & 0 & \cdots & 0 \\ 0 & 1 & 1 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & 1 \end{array}\right)\left(\begin{array}{c} k_{1} \\ \vdots \\ k_{n} \end{array}\right)=0 . $$ 而系数矩阵 $$ A=\left(\begin{array}{ccccc} 1 & 1 & 0 & \cdots & 0 \\ 0 & 1 & 1 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & 1 \end{array}\right), $$ 当 $n$ 为偶数时，$|A|=1+1=2 \neq 0$ ，从而 $r(A)=n+1$ ，即方程组只有零解，产生矛盾．从而当 $n$ 为偶数时向量组 $\alpha_{1}+\alpha_{2}, \alpha_{2}+\alpha_{3}, \cdots, \alpha_{n-1}+\alpha_{n}, \alpha_{n}+\alpha_{1}$ 线性无关。 当 $n$ 为奇数时，$|A|=1-1=0$ ，即方程组有非零解，可解得一组解为 $\left(-1,1, \cdots,(-1)^{k}, \cdots, 1\right)$ ．从而当 $n$ 为奇数时向量组 $\alpha_{1}+\alpha_{2}, \alpha_{2}+\alpha_{3}, \cdots, \alpha_{n-1}+\alpha_{n}, \alpha_{n}+\alpha_{1}$ 线性相关。