中国科学院真题 第20172题

一、 $\left(10+10=20\right.$ 分）设 $A, B$ 分别是 $n \times m$ 和 $m \times n$ 矩阵，$I_{k}$ 是 $k$ 阶单位矩阵。 （1）求证：$\left|I_{n}-A B\right|=\left|I_{m}-B A\right|$ ； （2）计算行列式 $$ D_{n}=\left|\begin{array}{cccc} 1+a_{1}+x_{1} & a_{1}+x_{2} & \cdots & a_{1}+x_{n} \\ a_{2}+x_{1} & 1+a_{2}+x_{2} & \cdots & a_{2}+x_{n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n}+x_{1} & a_{n}+x_{2} & \cdots & 1+a_{n}+x_{n} \end{array}\right| . $$ 解：（1）证 $$ \begin{aligned} & \because\left(\begin{array}{cc} 0 & I_{m} \\ I_{n} & 0 \end{array}\right)^{-1}\left(\begin{array}{cc} I_{m} & B \\ A & I_{n} \end{array}\right)\left(\begin{array}{cc} 0 & I_{m} \\ I_{n} & 0 \end{array}\right)=\left(\begin{array}{cc} 0 & I_{n} \\ I_{m} & 0 \end{array}\right)\left(\begin{array}{cc} I_{m} & B \\ A & I_{n} \end{array}\right)\left(\begin{array}{cc} 0 & I_{m} \\ I_{n} & 0 \end{array}\right)=\left(\begin{array}{cc} I_{n} & A \\ B & I_{m} \end{array}\right) \\ & \therefore\left|\begin{array}{cc} I_{m} & B \\ A & I_{n} \end{array}\right|=\left|\begin{array}{cc} I_{n} & A \\ B & I_{m} \end{array}\right| \\ & \text { 又 } \because\left|\begin{array}{cc} I_{m} & B \\ A & I_{n} \end{array}\right|=\left|\begin{array}{cc} I_{m} & 0 \\ -A & I_{n} \end{array}\right| \cdot\left|\begin{array}{cc} I_{m} & B \\ A & I_{n} \end{array}\right|=\left|\left(\begin{array}{cc} I_{m} & 0 \\ -A & I_{n} \end{array}\right) \cdot\left(\begin{array}{cc} I_{m} & B \\ A & I_{n} \end{array}\right)\right|=\left|\begin{array}{cc} I_{m} & B \\ 0 & I_{n}-A B \end{array}\right| \\ & \left|\begin{array}{cc} I_{n} & A \\ B & I_{m} \end{array}\right|=\left|\begin{array}{cc} I_{n} & 0 \\ -B & I_{m} \end{array}\right| \cdot\left|\begin{array}{cc} I_{n} & A \\ B & I_{m} \end{array}\right|=\left|\left(\begin{array}{cc} I_{n} & 0 \\ -B & I_{m} \end{array}\right) \cdot\left(\begin{array}{cc} I_{n} & A \\ B & I_{m} \end{array}\right)\right|=\left|\begin{array}{cc} I_{n} & A \\ 0 & I_{m}-B A \end{array}\right| \\ & \therefore\left|\begin{array}{cc} I_{m} & B \\ 0 & I_{n}-A B \end{array}\right|=\left|\begin{array}{cc} I_{n} & A \\ 0 & I_{m}-B A \end{array}\right|, \text { 即 }\left|I_{n}-A B\right|=\left|I_{m}-B A\right| . \\ & \text { (2) 令 } \alpha=\left(a_{1}, a_{2}, \cdots, a_{n}\right)^{T}, X=\left(x_{1}, x_{2}, \cdots, x_{n}\right)^{T}, e=(1,1, \cdots, 1)^{T} \in \mathbb{C}^{n}, \text { 则 } \\ & D_{n}=\left|I_{n}+\alpha e^{T}+e X^{T}\right|=\left|I_{n}-(-\alpha,-e)\binom{e^{T}}{X^{T}}\right| \end{aligned} $$ 利用（1）的结论可得 $$ \begin{aligned} D_{n} & =\left|I_{2}-\binom{e^{T}}{X^{T}}(-\alpha,-e)\right|=\left|\begin{array}{cc} 1+e^{T} \alpha & e^{T} e \\ X^{T} \alpha & 1+X^{T} e \end{array}\right|=\left|\begin{array}{cc} 1+\sum_{i=1}^{n} a_{i} & n \\ \sum_{i=1}^{n} a_{i} x_{i} & 1+\sum_{i=1}^{n} x_{i} \end{array}\right| \\ & =\left(1+\sum_{i=1}^{n} a_{i}\right)\left(1+\sum_{i=1}^{n} x_{i}\right)-n \sum_{i=1}^{n} a_{i} x_{i} \\ & =\sum_{i=1}^{n} a_{i} \cdot \sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} x_{i}-n \sum_{i=1}^{n} a_{i} x_{i}+1 . \end{aligned} $$