LA 小、杂、乱题合辑

${\Large 1.}$(来自丘维声『高等代数』(上)$P_{189,194}$)

$(1).$ 设$A,B$分别是数域${\mathbb F}$上$n\times n,m\times n$矩阵.

证明: 如果$I_n-AB$可逆, 那么$I_m-BA$也可逆; 并求出$(I_m-BA)^{-1}$.

$(2).$ 设$A,B,D$都是数域${\mathbb F}$上$n$级矩阵, 其中$A,D$可逆, 且$B^TA^{-1}B+D^{-1}$也可逆. 证明:

$$(A+BDB^T)^{-1}=A^{-1}-A^{-1}B(B^TA^{-1}B+D^{-1})^{-1}B^TA^{-1}$$

${\bf 解:}$ $(1).$设法找到$m$级矩阵$X,\ s.t.\ (I_m-BA)(I_m+X)=I_m\ \Rightarrow\ -BA+X-BAX=0\ \Rightarrow X-BAX=BA$.

令$X=BYA$, 其中$Y$是待定的$n$级矩阵. 代入上式, 得

$$BYA-BABYA=BA\ \ 即\ \ B(Y-ABY)A=BA$$

如果能找到$Y,\ \ s.t.\ Y-ABY=I_n$, 那么上式成立. 由于$Y-ABY=I_n\ \Leftrightarrow \ (I_n-AB)Y=I_n$,而已知条件$I_n-AB$可逆, 故

$Y=(I_n-AB)^{-1}$. 由此受到启发, 有

\begin{align*}&(I_m-BA)[I_m+B(I_n-AB)^{-1}A]\\ =&I_m+B(I_n-AB)^{-1}A-BA-BAB(I_n-AB)^{-1}A\\
=&I_m-BA+B[(I_n-AB)^{-1}-AB(I_n-AB)^{-1}]A\\ =&I_m-BA+B[(I_n-AB)AB(I_n-AB)^{-1}]A\\
=&I_m-BA+BI_nA\\ =&I_m\end{align*}因此$I_m-BA$可逆, 并且

$$(I_m-BA)^{-1}=I_m+B(I_n-AB)^{-1}A.$$

$(2).$ 事实上,

\[{(A
+ BD{B^T})^{ - 1}} = {[A(I + {A^{ - 1}}BD{B^T})]^{ - 1}} = {(I + {A^{ -
1}}BD{B^T})^{ - 1}}{A^{ - 1}} = {[I - ({A^{ - 1}}BD)( - {B^T})]^{ -
1}}{A^{ - 1}}\]

套用$(1).$的结论可得

\begin{align*}{(A
+ BD{B^T})^{ - 1}}=&{[I - ({A^{ - 1}}BD)( - {B^T})]^{ - 1}}{A^{ -
1}}\\ ^{(1).}= &{[I + ({A^{ - 1}}BD){(I - ( - {B^T})({A^{ -
1}}BD))^{ - 1}}( - {B^T})]^{ - 1}}{A^{ - 1}}\\ =& {[I - ({A^{ -
1}}B){({D^{ - 1}})^{ - 1}}{(I + {B^T}{A^{ - 1}}BD)^{ - 1}}{B^T}]^{ -
1}}{A^{ - 1}}\\ =& {[I - {A^{ - 1}}B{[(I + {B^T}{A^{ - 1}}BD){D^{ -
1}}]^{ - 1}}{B^T}]^{ - 1}}{A^{ - 1}}\\
=&A^{-1}-A^{-1}B(B^TA^{-1}B+D^{-1})^{-1}B^TA^{-1}\end{align*}

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${\Large 2.}$ 两个经典的行列式, 前者取自曾熊的博客; 后者来自张贤科『高等代数学』(第二版) $P_{61-62}$.

\[求\ \ \ \ D_1=\det \left( {\begin{array}{*{20}{c}}1&{\cos {\theta _1}}&{\cos 2{\theta _1}}& \cdots &{\cos \left( {n - 1} \right){\theta _1}}\\1&{\cos {\theta _2}}&{\cos 2{\theta _2}}& \cdots &{\cos \left( {n - 1} \right){\theta _2}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{\cos {\theta _n}}&{\cos 2{\theta _n}}& \cdots &{\cos \left( {n - 1} \right){\theta _n}}\end{array}} \right),\ \ \ D_2=\det \left( {\begin{array}{*{20}{c}}{\sin {\theta _1}}&{\sin 2{\theta _1}}& \cdots &{\sin {n}{\theta _1}}\\{\sin {\theta _2}}&{\sin 2{\theta _2}}& \cdots &{\sin {n}{\theta _2}}\\\vdots & \vdots & \ddots & \vdots \\{\sin {\theta _n}}&{\sin 2{\theta _n}}& \cdots &{\sin {n}{\theta _n}}\end{array}} \right).\]

${\bf 解:}$ 记${\varepsilon _k} = \cos {\theta _k} + i\sin {\theta _k}$, 则$\cos l{\theta _k} = \frac{{\varepsilon _k^l + \bar \varepsilon _k^l}}{2},\ \sin l{\theta _k} = \frac{{\varepsilon _k^l - \bar \varepsilon _k^l}}{2i},\ {\varepsilon _k}{{\bar \varepsilon }_k} = 1,$

\begin{align*}D_1 &= \left| {\begin{array}{*{20}{c}}1&{\cos {\theta _1}}&{\cos 2{\theta _1}}& \cdots &{\cos \left( {n - 1} \right){\theta _1}}\\1&{\cos {\theta _2}}&{\cos 2{\theta _2}}& \cdots &{\cos \left( {n - 1} \right){\theta _2}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{\cos {\theta _n}}&{\cos 2{\theta _n}}& \cdots &{\cos \left( {n - 1} \right){\theta _n}}\end{array}} \right| = \frac{1}{{{2^{n - 1}}}}\left| {\begin{array}{*{20}{c}}1&{{\varepsilon _1} + {{\bar \varepsilon }_1}}&{\varepsilon _1^2 + \bar \varepsilon _1^2}& \cdots &{\varepsilon _1^{n - 1} + \bar \varepsilon _1^{n - 1}}\\1&{{\varepsilon _2} + {{\bar \varepsilon }_2}}&{\varepsilon _2^2 + \bar \varepsilon _2^2}& \cdots &{\varepsilon _2^{n - 1} + \bar \varepsilon _2^{n - 1}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{{\varepsilon _n} + {{\bar \varepsilon }_n}}&{\varepsilon _n^2 + \bar \varepsilon _n^2}& \cdots &{\varepsilon _n^{n - 1} + \bar \varepsilon _n^{n - 1}}\end{array}} \right|\\&= \frac{1}{{{2^{n - 1}}}}\left| {\begin{array}{*{20}{c}}1&{{\varepsilon _1} + {{\bar \varepsilon }_1}}&{{{\left( {{\varepsilon _1} + {{\bar \varepsilon }_1}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _1} + {{\bar \varepsilon }_1}} \right)}^{n - 1}}}\\1&{{\varepsilon _2} + {{\bar \varepsilon }_2}}&{{{\left( {{\varepsilon _2} + {{\bar \varepsilon }_2}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _2} + {{\bar \varepsilon }_2}} \right)}^{n - 1}}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{{\varepsilon _n} + {{\bar \varepsilon }_n}}&{{{\left( {{\varepsilon _n} + {{\bar \varepsilon }_n}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _n} + {{\bar \varepsilon }_n}} \right)}^{n - 1}}}\end{array}} \right| = \frac{1}{{{2^{n - 1}}}}\prod\limits_{1 \le j < i \le n} {\left( {{\varepsilon _i} + {{\bar \varepsilon }_i} - {\varepsilon _j} - {{\bar \varepsilon }_j}} \right)} \\&= \frac{1}{{{2^{n - 1}}}} \times {2^{\frac{{n\left( {n - 1} \right)}}{2}}}\prod\limits_{1 \le j < i \le n} {\left( {\cos {\theta _i} - \cos {\theta _j}} \right)}  = {2^{\frac{{\left( {n - 1} \right)\left( {n - 2} \right)}}{2}}}\prod\limits_{1 \le j < i \le n} {\left( {\cos {\theta _i} - \cos {\theta _j}} \right)} ;\\
注 意\ \  \ \ \ \ &\phantom{=}\varepsilon _k^{n - 1} + \varepsilon _k^{n - 2}{\bar \varepsilon  _1} + \varepsilon _k^{n - 3}{\bar \varepsilon  _k}^2 \cdots  + \varepsilon _k^2\bar \varepsilon  _k^{n - 3} + \varepsilon _k^{}\bar \varepsilon  _k^{n - 1} + \bar \varepsilon  _k^{n - 1} \\
&= \varepsilon _k^{n - 1} + \varepsilon _k^{n - 3} + \varepsilon _k^{n - 5} \cdots  + \bar \varepsilon  _k^{n - 5} + \bar \varepsilon  _k^{n - 3} + \bar \varepsilon  _k^{n - 1} \\
&= (\varepsilon _k^{n - 1} + \bar \varepsilon  _k^{n - 1}) + (\varepsilon _k^{n - 3} + \bar \varepsilon  _k^{n - 3}) + (\varepsilon _k^{n - 5} + \bar \varepsilon  _k^{n - 5}) +  \cdots , \ \ \ 故\\
D_2 &=\left| {\begin{array}{*{20}{c}}
{\sin {\theta _1}}&{\sin 2{\theta _1}}& \cdots &{\sin {n}{\theta _1}}\\
{\sin {\theta _2}}&{\sin 2{\theta _2}}& \cdots &{\sin {n}{\theta _2}}\\
\vdots & \vdots & \ddots & \vdots \\
{\sin {\theta _n}}&{\sin 2{\theta _n}}& \cdots &{\sin {n}{\theta _n}}
\end{array}} \right| =
\frac{1}{{{(2i)^{n}}}}\left| {\begin{array}{*{20}{c}}
{{\varepsilon _1} - {{\bar \varepsilon }_1}}&{\varepsilon _1^2 - \bar \varepsilon _1^2}& \cdots &{\varepsilon _1^{n} - \bar \varepsilon _1^{n}}\\
{{\varepsilon _2} - {{\bar \varepsilon }_2}}&{\varepsilon _2^2 - \bar \varepsilon _2^2}& \cdots &{\varepsilon _2^{n} - \bar \varepsilon _2^{n}}\\
\vdots & \vdots &\ddots & \vdots \\
{{\varepsilon _n} - {{\bar \varepsilon }_n}}&{\varepsilon _n^2 - \bar \varepsilon _n^2}& \cdots &{\varepsilon _n^{n} - \bar \varepsilon _n^{n}}
\end{array}} \right|\\
&= \frac{({\varepsilon _1} - {\bar \varepsilon  _1})({\varepsilon _2} - {\bar \varepsilon  _2}) \cdots ({\varepsilon _n} - {\bar \varepsilon  _n})}{{{(2i)^{n}}}}\left| {\begin{array}{*{20}{c}}
1&{{\varepsilon _1} + {{\bar \varepsilon }_1}}& \cdots &\varepsilon _1^{n - 1} + \varepsilon _1^{n - 2}{\bar \varepsilon  _1} +  \cdots  + \varepsilon _1^{}\bar \varepsilon  _1^{n - 1} + \bar \varepsilon  _1^{n - 1} \\
1&{{\varepsilon _2} + {{\bar \varepsilon }_2}}& \cdots &\varepsilon _2^{n - 1} + \varepsilon _2^{n - 2}{\bar \varepsilon  _2} +  \cdots  + \varepsilon _2^{}\bar \varepsilon  _2^{n - 1} + \bar \varepsilon  _2^{n - 1}\\
\vdots & \vdots &\ddots& \vdots \\
1&{{\varepsilon _n} + {{\bar \varepsilon }_n}}& \cdots &\varepsilon _n^{n - 1} + \varepsilon _n^{n - 2}{\bar \varepsilon  _1} +  \cdots  + \varepsilon _n^{}\bar \varepsilon  _n^{n - 1} + \bar \varepsilon  _n^{n - 1}
\end{array}} \right| \\
&= \frac{({\varepsilon _1} - {\bar \varepsilon  _1})({\varepsilon _2} - {\bar \varepsilon  _2}) \cdots ({\varepsilon _n} - {\bar \varepsilon  _n})}{{{(2i)^{n}}}}
\left|{\begin{array}{*{20}{c}}
1&{{\varepsilon _1} + {{\bar \varepsilon }_1}}&{{{\left( {{\varepsilon _1} + {{\bar \varepsilon }_1}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _1} + {{\bar \varepsilon }_1}} \right)}^{n - 1}}}\\
1&{{\varepsilon _2} + {{\bar \varepsilon }_2}}&{{{\left( {{\varepsilon _2} + {{\bar \varepsilon }_2}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _2} + {{\bar \varepsilon }_2}} \right)}^{n - 1}}}\\
1&{{\varepsilon _3} + {{\bar \varepsilon }_3}}&{{{\left( {{\varepsilon _3} + {{\bar \varepsilon }_3}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _3} + {{\bar \varepsilon }_3}} \right)}^{n - 1}}}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1&{{\varepsilon _n} + {{\bar \varepsilon }_n}}&{{{\left( {{\varepsilon _n} + {{\bar \varepsilon }_n}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _n} + {{\bar \varepsilon }_n}} \right)}^{n - 1}}}
\end{array}} \right| \\
&= \frac{({\varepsilon _1} - {\bar \varepsilon  _1})({\varepsilon _2} - {\bar \varepsilon  _2}) \cdots ({\varepsilon _n} - {\bar \varepsilon  _n})}{{{(2i)^{n}}}}\prod\limits_{1 \le j < i \le n} {\left( {{\varepsilon _i} + {{\bar \varepsilon }_i} - {\varepsilon _j} - {{\bar \varepsilon }_j}} \right)} \\&=\sin{\theta _1} \ldots \sin{\theta _n}\prod\limits_{1 \le j < i \le n}{2(\cos{\theta_i}-\cos{\theta_j})}\\&=2^{\frac{n(n-1)}{2}}\sin{\theta _1} \ldots \sin{\theta _n}\prod\limits_{1 \le j < i \le n}{(\cos{\theta_i}-\cos{\theta_j})} .\end{align*}