ビネ・コーシーとラグランジュの恒等式

ビネ・コーシーとラグランジュの恒等式

(1)ビネ・コーシーの恒等式

\[ \left(\sum_{i=1}^{n}a_{i}c_{i}\right)\left(\sum_{j=1}^{n}b_{j}d_{j}\right)-\left(\sum_{i=1}^{n}a_{i}d_{i}\right)\left(\sum_{j=1}^{n}b_{j}c_{j}\right)=\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)\left(c_{i}d_{j}-c_{j}d_{i}\right) \]

\(n=3\)のとき、

\begin{align*} \left|\begin{array}{cc} \boldsymbol{a}\cdot\boldsymbol{c} & \boldsymbol{a}\cdot\boldsymbol{d}\\ \boldsymbol{b}\cdot\boldsymbol{c} & \boldsymbol{b}\cdot\boldsymbol{d} \end{array}\right| & =\left(\boldsymbol{a}\cdot\boldsymbol{c}\right)\left(\boldsymbol{b}\cdot\boldsymbol{d}\right)-\left(\boldsymbol{a}\cdot\boldsymbol{d}\right)\left(\boldsymbol{b}\cdot\boldsymbol{c}\right)\\ & =\left(\boldsymbol{a}\times\boldsymbol{b}\right)\cdot\left(\boldsymbol{c}\times\boldsymbol{d}\right) \end{align*}

(2)ラグランジュの恒等式

\[ \left(\sum_{i=1}^{n}a_{i}^{\;2}\right)\left(\sum_{i=1}^{n}b_{i}^{\;2}\right)-\left(\sum_{i=1}^{n}a_{i}b_{i}\right)^{2}=\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2} \]

複素数とすると、

\[ \left(\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right)\left(\sum_{j=1}^{n}\left|b_{j}\right|^{2}\right)-\left|\sum_{i=1}^{n}a_{i}\overline{b_{i}}\right|^{2}=\sum_{1\leq i<j\leq n}\left|a_{i}b_{j}-a_{j}b_{i}\right|^{2} \]

\(n=3\)のとき、複素ベクトルとすると、

\[ \left|\boldsymbol{a}\right|^{2}\left|\boldsymbol{b}\right|^{2}-\left|\boldsymbol{a}\cdot\boldsymbol{b}\right|^{2}=\left|\boldsymbol{a}\times\boldsymbol{b}\right|^{2} \]

(1)

\begin{align*} \left(\sum_{i=1}^{n}a_{i}c_{i}\right)\left(\sum_{j=1}^{n}b_{j}d_{j}\right)-\left(\sum_{i=1}^{n}a_{i}d_{i}\right)\left(\sum_{j=1}^{n}b_{j}c_{j}\right) & =\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}c_{i}b_{j}d_{j}-\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}d_{i}b_{j}c_{j}\\ & =\left(\sum_{1\leq i<j\leq n}+\sum_{1\leq j<i\leq n}+\sum_{1\leq i=j\leq n}\right)a_{i}c_{i}b_{j}d_{j}-\left(\sum_{1\leq i<j\leq n}+\sum_{1\leq j<i\leq n}+\sum_{1\leq i=j\leq n}\right)a_{i}d_{i}b_{j}c_{j}\\ & =\sum_{1\leq i<j\leq n}\left(a_{i}c_{i}b_{j}d_{j}+a_{j}c_{j}b_{i}d_{i}\right)+\sum_{1\leq i=j\leq n}a_{i}c_{i}b_{j}d_{j}-\sum_{1\leq i<j\leq n}\left(a_{i}d_{i}b_{j}c_{j}+a_{j}d_{j}b_{i}c_{i}\right)-\sum_{1\leq i=j\leq n}a_{i}d_{i}b_{j}c_{j}\\ & =\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}c_{i}d_{j}+a_{j}b_{i}c_{j}d_{i}\right)-\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}c_{j}d_{i}+a_{j}b_{i}c_{i}d_{j}\right)\\ & =\sum_{1\leq i<j\leq n}\left\{ a_{i}b_{j}\left(c_{i}d_{j}-c_{j}d_{i}\right)-a_{j}b_{i}\left(c_{i}d_{j}-c_{j}d_{i}\right)\right\} \\ & =\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)\left(c_{i}d_{j}-c_{j}d_{i}\right) \end{align*}

(1)-2

\(n=3\)のとき、

\begin{align*} \left(\sum_{i=1}^{3}a_{i}c_{i}\right)\left(\sum_{j=1}^{3}b_{j}d_{j}\right)-\left(\sum_{i=1}^{3}a_{i}d_{i}\right)\left(\sum_{j=1}^{3}b_{j}c_{j}\right) & =\left(\boldsymbol{a}\cdot\boldsymbol{c}\right)\left(\boldsymbol{b}\cdot\boldsymbol{d}\right)-\left(\boldsymbol{a}\cdot\boldsymbol{d}\right)\left(\boldsymbol{b}\cdot\boldsymbol{c}\right)\\ & =\left\{ \left(\boldsymbol{a}\cdot\boldsymbol{c}\right)\boldsymbol{b}-\left(\boldsymbol{b}\cdot\boldsymbol{c}\right)\boldsymbol{a}\right\} \cdot\boldsymbol{d}\\ & =\left\{ \left(\boldsymbol{a}\times\boldsymbol{b}\right)\times\boldsymbol{c}\right\} \cdot\boldsymbol{d}\\ & =\left(\boldsymbol{a}\times\boldsymbol{b}\right)\cdot\left(\boldsymbol{c}\times\boldsymbol{d}\right)\\ & =\sum_{1\leq i<j\leq3}\left(a_{i}b_{j}-a_{j}b_{i}\right)\left(c_{i}d_{j}-c_{j}d_{i}\right) \end{align*}

(2)

(1)で\(c_{i}=a_{i}\;,\;d_{i}=b_{i}\)とすると、

\begin{align*} \LHS & =\left(\sum_{i=1}^{n}a_{i}a_{i}\right)\left(\sum_{j=1}^{n}b_{j}b_{j}\right)-\left(\sum_{i=1}^{n}a_{i}b_{i}\right)\left(\sum_{j=1}^{n}b_{j}a_{j}\right)\\ & =\left(\sum_{i=1}^{n}a_{i}^{\;2}\right)\left(\sum_{j=1}^{n}b_{j}^{\;2}\right)-\left(\sum_{i=1}^{n}a_{i}b_{i}\right)^{2} \end{align*}

\begin{align*} \RHS & =\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)\left(a_{i}b_{j}-a_{j}b_{i}\right)\\ & =\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2} \end{align*}

これより、

\[ \left(\sum_{i=1}^{n}a_{i}^{\;2}\right)\left(\sum_{j=1}^{n}b_{j}^{\;2}\right)-\left(\sum_{i=1}^{n}a_{i}b_{i}\right)^{2}=\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2} \]

(2)-2

(1)で\(c_{i}=\overline{a_{i}}\;,\;d_{i}=\overline{b_{i}}\)とすると、

\begin{align*} \LHS & =\left(\sum_{i=1}^{n}a_{i}\overline{a_{i}}\right)\left(\sum_{j=1}^{n}b_{j}\overline{b_{j}}\right)-\left(\sum_{i=1}^{n}a_{i}\overline{b_{i}}\right)\left(\sum_{j=1}^{n}b_{j}\overline{a_{j}}\right)\\ & =\left(\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right)\left(\sum_{j=1}^{n}\left|b_{j}\right|^{2}\right)-\left(\sum_{i=1}^{n}a_{i}\overline{b_{i}}\right)\overline{\left(\sum_{j=1}^{n}a_{i}\overline{b_{i}}\right)}\\ & =\left(\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right)\left(\sum_{j=1}^{n}\left|b_{j}\right|^{2}\right)-\left|\sum_{i=1}^{n}a_{i}\overline{b_{i}}\right|^{2} \end{align*}

\begin{align*} \RHS & =\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)\left(\overline{a_{i}b_{j}}-\overline{a_{j}b_{i}}\right)\\ & =\sum_{1\leq i<j\leq n}\left|a_{i}b_{j}-a_{j}b_{i}\right|^{2} \end{align*}

これより、

\[ \left(\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right)\left(\sum_{j=1}^{n}\left|b_{j}\right|^{2}\right)-\left|\sum_{i=1}^{n}a_{i}\overline{b_{i}}\right|^{2}=\sum_{1\leq i<j\leq n}\left|a_{i}b_{j}-a_{j}b_{i}\right|^{2} \]

(2)-3

\(n=3\)のとき、

\begin{align*} \LHS & =\left(\sum_{i=1}^{3}\left|a_{i}\right|^{2}\right)\left(\sum_{j=1}^{3}\left|b_{j}\right|^{2}\right)-\left|\sum_{i=1}^{3}a_{i}\overline{b_{i}}\right|^{2}\\ & =\left|\boldsymbol{a}\right|^{2}\left|\boldsymbol{b}\right|^{2}-\left|\boldsymbol{a}\cdot\boldsymbol{b}\right|^{2} \end{align*}

\begin{align*} \RHS & =\sum_{1\leq i<j\leq3}\left|a_{i}b_{j}-a_{j}b_{i}\right|^{2}\\ & =\left|\boldsymbol{a}\times\boldsymbol{b}\right|^{2} \end{align*}

これより、

\[ \left|\boldsymbol{a}\right|^{2}\left|\boldsymbol{b}\right|^{2}-\left|\boldsymbol{a}\cdot\boldsymbol{b}\right|^{2}=\left|\boldsymbol{a}\times\boldsymbol{b}\right|^{2} \]

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ビネ・コーシーとラグランジュの恒等式

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