问题及解答

求行列式

\[
D=\begin{vmatrix}
\lambda-1 & -1 & -1 &\cdots & -1\\
-1 & \lambda-1 & -1 &\cdots & -1\\
-1 & -1 &\lambda-1 &\cdots & -1\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
-1 & -1 & -1 & \cdots &\lambda-1
\end{vmatrix}_{n}
\]

设 $c_i\neq 0, i=1,2,\ldots,n$. 求行列式

\[
D=\begin{vmatrix}
\lambda-1 & -\frac{c_1}{c_2} & -\frac{c_1}{c_3} &\cdots & -\frac{c_1}{c_n}\\
-\frac{c_2}{c_1} & \lambda-1 & -\frac{c_2}{c_3} &\cdots & -\frac{c_2}{c_n}\\
-\frac{c_3}{c_1} & -\frac{c_3}{c_2} &\lambda-1 &\cdots & -\frac{c_3}{c_n}\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
-\frac{c_n}{c_1} & -\frac{c_n}{c_2} & -\frac{c_n}{c_3} & \cdots &\lambda-1
\end{vmatrix}_{n}
\]

这两个行列式都等于 $\lambda^{n-1}(\lambda-n)$.

\[
\begin{split}
D&=\begin{vmatrix}
\lambda-1 & -1 & -1 &\cdots & -1\\
-1 & \lambda-1 & -1 &\cdots & -1\\
-1 & -1 &\lambda-1 &\cdots & -1\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
-1 & -1 & -1 & \cdots &\lambda-1
\end{vmatrix}_n\xlongequal[i=2,\ldots,n]{c_1+c_i}
\begin{vmatrix}
\lambda-n & -1 & -1 &\cdots & -1\\
\lambda-n & \lambda-1 & -1 &\cdots & -1\\
\lambda-n & -1 &\lambda-1 &\cdots & -1\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
\lambda-n & -1 & -1 & \cdots &\lambda-1
\end{vmatrix}_n\\
&=(\lambda-n)\begin{vmatrix}
1 & -1 & -1 &\cdots & -1\\
1 & \lambda-1 & -1 &\cdots & -1\\
1 & -1 &\lambda-1 &\cdots & -1\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
1 & -1 & -1 & \cdots &\lambda-1
\end{vmatrix}_n\xlongequal[i=2,3,\ldots,n]{c_i+c_1}
(\lambda-n)\begin{vmatrix}
1 & 0 & 0 &\cdots & 0\\
1 & \lambda & 0 &\cdots & 0\\
1 & 0 &\lambda &\cdots & 0\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
1 & 0 & 0 & \cdots &\lambda
\end{vmatrix}_n\\
=(\lambda-n)\lambda^{n-1}.
\end{split}
\]

这表明矩阵

\[
A=\begin{pmatrix}
1 & 1 & \cdots & 1\\
1 & 1 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 1\\
\end{pmatrix}_n
\]

的特征值是 $0$ ($n-1$ 重) 和 $n$.

\[
\begin{split}
D&=\begin{vmatrix}
\lambda-1 & -\frac{c_1}{c_2} & -\frac{c_1}{c_3} &\cdots & -\frac{c_1}{c_n}\\
-\frac{c_2}{c_1} & \lambda-1 & -\frac{c_2}{c_3} &\cdots & -\frac{c_2}{c_n}\\
-\frac{c_3}{c_1} & -\frac{c_3}{c_2} &\lambda-1 &\cdots & -\frac{c_3}{c_n}\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
-\frac{c_n}{c_1} & -\frac{c_n}{c_2} & -\frac{c_n}{c_3} & \cdots &\lambda-1
\end{vmatrix}_{n}\\
&\xlongequal{\text{提取每列的公因子}}\frac{1}{c_1}\frac{1}{c_2}\cdots\frac{1}{c_n}
\begin{vmatrix}
c_1(\lambda-1) & -c_1 & -c_1 & \cdots & -c_1\\
-c_2 & c_2(\lambda-1) & -c_2 &\cdots & -c_2\\
-c_3 & -c_3 & c_3(\lambda-1) &\cdots & -c_3\\
\vdots & \vdots &\vdots &\ddots &\vdots\\
-c_n & -c_n & -c_n &\cdots & c_n(\lambda-1)\\
\end{vmatrix}_n\\
&\xlongequal{\text{注意每行有公因子}c_i}\begin{vmatrix}
\lambda-1 & -1 & -1 &\cdots & -1\\
-1 & \lambda-1 & -1 &\cdots & -1\\
-1 & -1 &\lambda-1 &\cdots & -1\\
\vdots &\vdots &\vdots  &\ddots & \vdots\\
-1 & -1 & -1 & \cdots &\lambda-1
\end{vmatrix}_{n}
\end{split}
\]