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  • 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$n imes n,m imes n$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:

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

     

    ${f 解:}$ $(1).$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

    egin{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_mend{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

    egin{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*}


     


     

    ${Large 2.}$ 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( {egin{array}{*{20}{c}}1&{cos { heta _1}}&{cos 2{ heta _1}}& cdots &{cos left( {n - 1} ight){ heta _1}}\1&{cos { heta _2}}&{cos 2{ heta _2}}& cdots &{cos left( {n - 1} ight){ heta _2}}\vdots & vdots & vdots &{}& vdots \1&{cos { heta _n}}&{cos 2{ heta _n}}& cdots &{cos left( {n - 1} ight){ heta _n}}end{array}} ight), D_2=det left( {egin{array}{*{20}{c}}{sin { heta _1}}&{sin 2{ heta _1}}& cdots &{sin {n}{ heta _1}}\{sin { heta _2}}&{sin 2{ heta _2}}& cdots &{sin {n}{ heta _2}}\vdots & vdots & ddots & vdots \{sin { heta _n}}&{sin 2{ heta _n}}& cdots &{sin {n}{ heta _n}}end{array}} ight).]

     ${f 解:}$ 记${varepsilon _k} = cos { heta _k} + isin { heta _k}$, 则$cos l{ heta _k} = frac{{varepsilon _k^l + ar varepsilon _k^l}}{2}, sin l{ heta _k} = frac{{varepsilon _k^l - ar varepsilon _k^l}}{2i}, {varepsilon _k}{{ar varepsilon }_k} = 1,$

    egin{align*}D_1 &= left| {egin{array}{*{20}{c}}1&{cos { heta _1}}&{cos 2{ heta _1}}& cdots &{cos left( {n - 1} ight){ heta _1}}\1&{cos { heta _2}}&{cos 2{ heta _2}}& cdots &{cos left( {n - 1} ight){ heta _2}}\vdots & vdots & vdots &{}& vdots \1&{cos { heta _n}}&{cos 2{ heta _n}}& cdots &{cos left( {n - 1} ight){ heta _n}}end{array}} ight| = frac{1}{{{2^{n - 1}}}}left| {egin{array}{*{20}{c}}1&{{varepsilon _1} + {{ar varepsilon }_1}}&{varepsilon _1^2 + ar varepsilon _1^2}& cdots &{varepsilon _1^{n - 1} + ar varepsilon _1^{n - 1}}\1&{{varepsilon _2} + {{ar varepsilon }_2}}&{varepsilon _2^2 + ar varepsilon _2^2}& cdots &{varepsilon _2^{n - 1} + ar varepsilon _2^{n - 1}}\vdots & vdots & vdots &{}& vdots \1&{{varepsilon _n} + {{ar varepsilon }_n}}&{varepsilon _n^2 + ar varepsilon _n^2}& cdots &{varepsilon _n^{n - 1} + ar varepsilon _n^{n - 1}}end{array}} ight|\&= frac{1}{{{2^{n - 1}}}}left| {egin{array}{*{20}{c}}1&{{varepsilon _1} + {{ar varepsilon }_1}}&{{{left( {{varepsilon _1} + {{ar varepsilon }_1}} ight)}^2}}& cdots &{{{left( {{varepsilon _1} + {{ar varepsilon }_1}} ight)}^{n - 1}}}\1&{{varepsilon _2} + {{ar varepsilon }_2}}&{{{left( {{varepsilon _2} + {{ar varepsilon }_2}} ight)}^2}}& cdots &{{{left( {{varepsilon _2} + {{ar varepsilon }_2}} ight)}^{n - 1}}}\vdots & vdots & vdots &{}& vdots \1&{{varepsilon _n} + {{ar varepsilon }_n}}&{{{left( {{varepsilon _n} + {{ar varepsilon }_n}} ight)}^2}}& cdots &{{{left( {{varepsilon _n} + {{ar varepsilon }_n}} ight)}^{n - 1}}}end{array}} ight| = frac{1}{{{2^{n - 1}}}}prodlimits_{1 le j < i le n} {left( {{varepsilon _i} + {{ar varepsilon }_i} - {varepsilon _j} - {{ar varepsilon }_j}} ight)} \&= frac{1}{{{2^{n - 1}}}} imes {2^{frac{{nleft( {n - 1} ight)}}{2}}}prodlimits_{1 le j < i le n} {left( {cos { heta _i} - cos { heta _j}} ight)}  = {2^{frac{{left( {n - 1} ight)left( {n - 2} ight)}}{2}}}prodlimits_{1 le j < i le n} {left( {cos { heta _i} - cos { heta _j}} ight)} ;\
    注 意   &phantom{=}varepsilon _k^{n - 1} + varepsilon _k^{n - 2}{ar varepsilon  _1} + varepsilon _k^{n - 3}{ar varepsilon  _k}^2 cdots  + varepsilon _k^2ar varepsilon  _k^{n - 3} + varepsilon _k^{}ar varepsilon  _k^{n - 1} + ar varepsilon  _k^{n - 1} \
    &= varepsilon _k^{n - 1} + varepsilon _k^{n - 3} + varepsilon _k^{n - 5} cdots  + ar varepsilon  _k^{n - 5} + ar varepsilon  _k^{n - 3} + ar varepsilon  _k^{n - 1} \
    &= (varepsilon _k^{n - 1} + ar varepsilon  _k^{n - 1}) + (varepsilon _k^{n - 3} + ar varepsilon  _k^{n - 3}) + (varepsilon _k^{n - 5} + ar varepsilon  _k^{n - 5}) +  cdots , 故\
    D_2 &=left| {egin{array}{*{20}{c}}
    {sin { heta _1}}&{sin 2{ heta _1}}& cdots &{sin {n}{ heta _1}}\
    {sin { heta _2}}&{sin 2{ heta _2}}& cdots &{sin {n}{ heta _2}}\
    vdots & vdots & ddots & vdots \
    {sin { heta _n}}&{sin 2{ heta _n}}& cdots &{sin {n}{ heta _n}}
    end{array}} ight| =
    frac{1}{{{(2i)^{n}}}}left| {egin{array}{*{20}{c}}
    {{varepsilon _1} - {{ar varepsilon }_1}}&{varepsilon _1^2 - ar varepsilon _1^2}& cdots &{varepsilon _1^{n} - ar varepsilon _1^{n}}\
    {{varepsilon _2} - {{ar varepsilon }_2}}&{varepsilon _2^2 - ar varepsilon _2^2}& cdots &{varepsilon _2^{n} - ar varepsilon _2^{n}}\
    vdots & vdots &ddots & vdots \
    {{varepsilon _n} - {{ar varepsilon }_n}}&{varepsilon _n^2 - ar varepsilon _n^2}& cdots &{varepsilon _n^{n} - ar varepsilon _n^{n}}
    end{array}} ight|\
    &= frac{({varepsilon _1} - {ar varepsilon  _1})({varepsilon _2} - {ar varepsilon  _2}) cdots ({varepsilon _n} - {ar varepsilon  _n})}{{{(2i)^{n}}}}left| {egin{array}{*{20}{c}}
    1&{{varepsilon _1} + {{ar varepsilon }_1}}& cdots &varepsilon _1^{n - 1} + varepsilon _1^{n - 2}{ar varepsilon  _1} +  cdots  + varepsilon _1^{}ar varepsilon  _1^{n - 1} + ar varepsilon  _1^{n - 1} \
    1&{{varepsilon _2} + {{ar varepsilon }_2}}& cdots &varepsilon _2^{n - 1} + varepsilon _2^{n - 2}{ar varepsilon  _2} +  cdots  + varepsilon _2^{}ar varepsilon  _2^{n - 1} + ar varepsilon  _2^{n - 1}\
    vdots & vdots &ddots& vdots \
    1&{{varepsilon _n} + {{ar varepsilon }_n}}& cdots &varepsilon _n^{n - 1} + varepsilon _n^{n - 2}{ar varepsilon  _1} +  cdots  + varepsilon _n^{}ar varepsilon  _n^{n - 1} + ar varepsilon  _n^{n - 1}
    end{array}} ight| \
    &= frac{({varepsilon _1} - {ar varepsilon  _1})({varepsilon _2} - {ar varepsilon  _2}) cdots ({varepsilon _n} - {ar varepsilon  _n})}{{{(2i)^{n}}}}
    left|{egin{array}{*{20}{c}}
    1&{{varepsilon _1} + {{ar varepsilon }_1}}&{{{left( {{varepsilon _1} + {{ar varepsilon }_1}} ight)}^2}}& cdots &{{{left( {{varepsilon _1} + {{ar varepsilon }_1}} ight)}^{n - 1}}}\
    1&{{varepsilon _2} + {{ar varepsilon }_2}}&{{{left( {{varepsilon _2} + {{ar varepsilon }_2}} ight)}^2}}& cdots &{{{left( {{varepsilon _2} + {{ar varepsilon }_2}} ight)}^{n - 1}}}\
    1&{{varepsilon _3} + {{ar varepsilon }_3}}&{{{left( {{varepsilon _3} + {{ar varepsilon }_3}} ight)}^2}}& cdots &{{{left( {{varepsilon _3} + {{ar varepsilon }_3}} ight)}^{n - 1}}}\
    vdots & vdots & vdots & ddots & vdots \
    1&{{varepsilon _n} + {{ar varepsilon }_n}}&{{{left( {{varepsilon _n} + {{ar varepsilon }_n}} ight)}^2}}& cdots &{{{left( {{varepsilon _n} + {{ar varepsilon }_n}} ight)}^{n - 1}}}
    end{array}} ight| \
    &= frac{({varepsilon _1} - {ar varepsilon  _1})({varepsilon _2} - {ar varepsilon  _2}) cdots ({varepsilon _n} - {ar varepsilon  _n})}{{{(2i)^{n}}}}prodlimits_{1 le j < i le n} {left( {{varepsilon _i} + {{ar varepsilon }_i} - {varepsilon _j} - {{ar varepsilon }_j}} ight)} \&=sin{ heta _1} ldots sin{ heta _n}prodlimits_{1 le j < i le n}{2(cos{ heta_i}-cos{ heta_j})}\&=2^{frac{n(n-1)}{2}}sin{ heta _1} ldots sin{ heta _n}prodlimits_{1 le j < i le n}{(cos{ heta_i}-cos{ heta_j})} .end{align*}

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  • 原文地址:https://www.cnblogs.com/poorich/p/4263556.html
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