中国科学院真题 第20171题

1．设 $A$ 是 $n$ 阶可逆矩阵，$\alpha, \beta$ 是 $n$ 维列向量，且 $1+\beta^{\prime} A^{-1} \alpha \neq 0$ ．证明：$A+\alpha \beta^{\prime}$ 可逆，并求其逆。 证明 构造 $n+1$ 阶矩阵 $\left(\begin{array}{cc}1 & -\beta^{\prime} \\ 0 & A+\alpha \beta^{\prime}\end{array}\right)$ ，对分块矩阵 $\left(\begin{array}{cc:cc}1 & -\beta^{\prime} & 1 & 0 \\ 0 & A+\alpha \beta^{\prime} & 0 & E_{n}\end{array}\right)$ 作广义初等行变换 $$ \begin{aligned} & \left(\begin{array}{cc:cc} 1 & -\beta^{\prime} & 1 & 0 \\ 0 & A+\alpha \beta^{\prime} & 0 & E_{n} \end{array}\right) \rightarrow\left(\begin{array}{cc:cc} 1 & -\beta^{\prime} & 1 & 0 \\ \alpha & A & \alpha & E_{n} \end{array}\right) \rightarrow\left(\begin{array}{cccc} 1+\beta^{\prime} A^{-1} \alpha & 0 & 1+\beta^{\prime} A^{-1} \alpha & \beta^{\prime} A^{-1} \\ \alpha & A & \alpha & E_{n} \end{array}\right) \\ & \rightarrow\left(\begin{array}{cccc} 1+\beta^{\prime} A^{-1} \alpha & 0 & 1+\beta^{\prime} A^{-1} \alpha & \beta^{\prime} A^{-1} \\ 0 & A & 0 & E_{n}-\frac{\alpha \beta^{\prime} A^{-1}}{1+\beta^{\prime} A^{-1} \alpha} \end{array}\right) \rightarrow\left(\begin{array}{cc:cc} 1 & 0 & 1 & \frac{\beta^{\prime} A^{-1}}{1+\beta^{\prime} A^{-1} \alpha} \\ 0 & E_{n} & 0 & A^{-1}-\frac{A^{-1} \alpha \beta^{\prime} A^{-1}}{1+\beta^{\prime} A^{-1} \alpha} \end{array}\right) \end{aligned} $$ 可见 $\left(\begin{array}{cc}1 & -\beta^{\prime} \\ 0 & A+\alpha \beta^{\prime}\end{array}\right)$ 可以经过广义初等行变换化为单位矩阵 $\left(\begin{array}{cc}1 & 0 \\ 0 & E_{n}\end{array}\right)$ ，所以 $\left(\begin{array}{cc}1 & -\beta^{\prime} \\ 0 & A+\alpha \beta^{\prime}\end{array}\right)$ 可逆，从而 $A+\alpha \beta^{\prime}$ 可逆．且 $\left(\begin{array}{cc}1 & -\beta^{\prime} \\ 0 & A+\alpha \beta^{\prime}\end{array}\right)^{-1}=\left(\begin{array}{cc}1 & \frac{\beta^{\prime} A^{-1}}{1+\beta^{\prime} A^{-1} \alpha} \\ 0 & A^{-1}-\frac{A^{-1} \alpha \beta^{\prime} A^{-1}}{1+\beta^{\prime} A^{-1} \alpha}\end{array}\right)=\left(\begin{array}{cc}M & R \\ N & Q\end{array}\right)$ ，由 $$ \left(\begin{array}{cc} 1 & -\beta^{\prime} \\ 0 & A+\alpha \beta^{\prime} \end{array}\right)\left(\begin{array}{cc} M & R \\ N & Q \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & E_{n} \end{array}\right) $$ 得 $\left(A+\alpha \beta^{\prime}\right) Q=E_{n}$ ，所以 $\quad\left(A+\alpha \beta^{\prime}\right)^{-1}=Q=A^{-1}-\frac{A^{-1} \alpha \beta^{\prime} A^{-1}}{1+\beta^{\prime} A^{-1} \alpha}$ ． 如何升阶的问题应视原矩阵的形状以及所要解决的问题而定，一般来说比较复杂。上例中将矩阵 $A+\alpha \beta^{\prime}$ 升阶为 $\left(\begin{array}{cc}1 & -\beta^{\prime} \\ 0 & A+\alpha \beta^{\prime}\end{array}\right)$ 的做法是比较容易想到的，因为我们希望把 $A+\alpha \beta^{\prime}$ 的第二项消掉，以充分利用 $A$ 的可逆性。