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総合研究大学院大学 複合科学研究科 統計科学専攻 2020年1月実施 数理 第3問

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Author​

祭音Myyura (Based on miyake's answer refined with GPT 6 Astra)

Description​

区分行列

A=(A11A12A21A22),B=(B11OB21B22)A=\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix},\qquad B=\begin{pmatrix}B_{11}&O\\B_{21}&B_{22}\end{pmatrix}

を考える。A11,A22,B11,B22A_{11},A_{22},B_{11},B_{22} は n×nn\times n 正方行列とする。以下に現れる全ての逆行列は存在すると仮定する。OO は n×nn\times n 零行列、II は n×nn\times n 単位行列とする。

問1​

次を満たす n×nn\times n 行列 Q1Q_1 を A11,A12,A21,A22A_{11},A_{12},A_{21},A_{22} とその逆行列から必要なものを用いて表せ。

(A11A12A21A22)=(A11OA21Q1)(IQ2OI)\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix} =\begin{pmatrix}A_{11}&O\\A_{21}&Q_1\end{pmatrix} \begin{pmatrix}I&Q_2\\O&I\end{pmatrix}

問2​

次を満たす n×nn\times n 行列 Q3Q_3 を B11,B21,B22B_{11},B_{21},B_{22} とその逆行列から必要なものを用いて表せ。

(B11OB21B22)−1=(B11−1OQ3B22−1)\begin{pmatrix}B_{11}&O\\B_{21}&B_{22}\end{pmatrix}^{-1} =\begin{pmatrix}B_{11}^{-1}&O\\Q_3&B_{22}^{-1}\end{pmatrix}

問3​

A−1=(A11A12A21A22)A^{-1}=\begin{pmatrix}A^{11}&A^{12}\\A^{21}&A^{22}\end{pmatrix}

とする。A11,A22A^{11},A^{22} は n×nn\times n 正方行列とする。A11,A12,A21,A22A^{11},A^{12},A^{21},A^{22} を A11,A12,A21,A22A_{11},A_{12},A_{21},A_{22} とその逆行列から必要なものを用いて表せ。

問4​

次の等式を示せ。

(A22−A21A11−1A12)−1=A22−1+A22−1A21(A11−A12A22−1A21)−1A12A22−1.\begin{aligned} &(A_{22}-A_{21}A_{11}^{-1}A_{12})^{-1}\\ &\quad=A_{22}^{-1}+A_{22}^{-1}A_{21} (A_{11}-A_{12}A_{22}^{-1}A_{21})^{-1}A_{12}A_{22}^{-1}. \end{aligned}

题目描述​

考虑分块矩阵

A=(A11A12A21A22),B=(B11OB21B22).A=\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix},\qquad B=\begin{pmatrix}B_{11}&O\\B_{21}&B_{22}\end{pmatrix}.

A11,A22,B11,B22A_{11},A_{22},B_{11},B_{22} 均为 n×nn\times n 方阵。假设以下出现的所有逆矩阵均存在。OO 为 n×nn\times n 零矩阵，II 为 n×nn\times n 单位矩阵。

问1 用 A11,A12,A21,A22A_{11},A_{12},A_{21},A_{22} 及其逆矩阵中所需的矩阵，表示满足下式的 n×nn\times n 矩阵 Q1Q_1。

(A11A12A21A22)=(A11OA21Q1)(IQ2OI)\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix} =\begin{pmatrix}A_{11}&O\\A_{21}&Q_1\end{pmatrix} \begin{pmatrix}I&Q_2\\O&I\end{pmatrix}

问2 用 B11,B21,B22B_{11},B_{21},B_{22} 及其逆矩阵中所需的矩阵，表示满足下式的 n×nn\times n 矩阵 Q3Q_3。

(B11OB21B22)−1=(B11−1OQ3B22−1)\begin{pmatrix}B_{11}&O\\B_{21}&B_{22}\end{pmatrix}^{-1} =\begin{pmatrix}B_{11}^{-1}&O\\Q_3&B_{22}^{-1}\end{pmatrix}

问3 设

A−1=(A11A12A21A22),A^{-1}=\begin{pmatrix}A^{11}&A^{12}\\A^{21}&A^{22}\end{pmatrix},

其中 A11,A22A^{11},A^{22} 为 n×nn\times n 方阵。用 A11,A12,A21,A22A_{11},A_{12},A_{21},A_{22} 及其逆矩阵中所需的矩阵，表示 A11,A12,A21,A22A^{11},A^{12},A^{21},A^{22}。

问4 证明下列等式。

(A22−A21A11−1A12)−1=A22−1+A22−1A21(A11−A12A22−1A21)−1A12A22−1.\begin{aligned} &(A_{22}-A_{21}A_{11}^{-1}A_{12})^{-1}\\ &\quad=A_{22}^{-1}+A_{22}^{-1}A_{21} (A_{11}-A_{12}A_{22}^{-1}A_{21})^{-1}A_{12}A_{22}^{-1}. \end{aligned}

Kai​

問1​

比較のため右辺を乗算すると、

(A11A11Q2A21A21Q2+Q1)=(A11A12A21A22).\begin{pmatrix}A_{11}&A_{11}Q_2\\A_{21}&A_{21}Q_2+Q_1\end{pmatrix} =\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix}.

よって Q2=A11−1A12Q_2=A_{11}^{-1}A_{12} であり、

Q1=A22−A21A11−1A12.\boxed{Q_1=A_{22}-A_{21}A_{11}^{-1}A_{12}}.

以下ではこれを SS と書く。

問2​

BB−1=diag⁡(I,I)BB^{-1}=\operatorname{diag}(I,I) の左下ブロックを比較すると、

B21B11−1+B22Q3=O.B_{21}B_{11}^{-1}+B_{22}Q_3=O.

したがって

Q3=−B22−1B21B11−1.\boxed{Q_3=-B_{22}^{-1}B_{21}B_{11}^{-1}}.

問3​

問1の分解の逆をとり、問2を用いると、

A−1=(I−A11−1A12OI)(A11−1O−S−1A21A11−1S−1)=(A11−1+A11−1A12S−1A21A11−1−A11−1A12S−1−S−1A21A11−1S−1).\begin{aligned} A^{-1} &=\begin{pmatrix}I&-A_{11}^{-1}A_{12}\\O&I\end{pmatrix} \begin{pmatrix}A_{11}^{-1}&O\\-S^{-1}A_{21}A_{11}^{-1}&S^{-1}\end{pmatrix}\\ &=\begin{pmatrix} A_{11}^{-1}+A_{11}^{-1}A_{12}S^{-1}A_{21}A_{11}^{-1}&-A_{11}^{-1}A_{12}S^{-1}\\ -S^{-1}A_{21}A_{11}^{-1}&S^{-1} \end{pmatrix}. \end{aligned}

すなわち、S=A22−A21A11−1A12S=A_{22}-A_{21}A_{11}^{-1}A_{12} として

A11=A11−1+A11−1A12S−1A21A11−1,A12=−A11−1A12S−1,A21=−S−1A21A11−1,A22=S−1.\boxed{\begin{aligned} A^{11}&=A_{11}^{-1}+A_{11}^{-1}A_{12}S^{-1}A_{21}A_{11}^{-1},\\ A^{12}&=-A_{11}^{-1}A_{12}S^{-1},\\ A^{21}&=-S^{-1}A_{21}A_{11}^{-1},\\ A^{22}&=S^{-1}. \end{aligned}}

問4​

T=A11−A12A22−1A21T=A_{11}-A_{12}A_{22}^{-1}A_{21} とおく。A12A22−1A21=A11−TA_{12}A_{22}^{-1}A_{21}=A_{11}-T より

A11−1A12A22−1A21T−1=T−1−A11−1.A_{11}^{-1}A_{12}A_{22}^{-1}A_{21}T^{-1} =T^{-1}-A_{11}^{-1}.

示すべき右辺を R=A22−1+A22−1A21T−1A12A22−1R=A_{22}^{-1}+A_{22}^{-1}A_{21}T^{-1}A_{12}A_{22}^{-1} とすると、

SR=I+A21T−1A12A22−1−A21A11−1A12A22−1−A21A11−1A12A22−1A21T−1A12A22−1=I+A21T−1A12A22−1−A21A11−1A12A22−1−A21(T−1−A11−1)A12A22−1=I.\begin{aligned} SR &=I+A_{21}T^{-1}A_{12}A_{22}^{-1} -A_{21}A_{11}^{-1}A_{12}A_{22}^{-1}\\ &\quad-A_{21}A_{11}^{-1}A_{12}A_{22}^{-1}A_{21}T^{-1}A_{12}A_{22}^{-1}\\ &=I+A_{21}T^{-1}A_{12}A_{22}^{-1} -A_{21}A_{11}^{-1}A_{12}A_{22}^{-1}\\ &\quad-A_{21}(T^{-1}-A_{11}^{-1})A_{12}A_{22}^{-1}=I. \end{aligned}

SS は可逆なので R=S−1R=S^{-1}、すなわち

(A22−A21A11−1A12)−1=A22−1+A22−1A21(A11−A12A22−1A21)−1A12A22−1\boxed{\begin{aligned} &(A_{22}-A_{21}A_{11}^{-1}A_{12})^{-1}\\ &\quad=A_{22}^{-1}\\ &\qquad+A_{22}^{-1}A_{21}(A_{11}-A_{12}A_{22}^{-1}A_{21})^{-1}A_{12}A_{22}^{-1} \end{aligned}}

が得られる。

Reference​

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