行列式

行列式 (英：determinant) とは、正方行列に対して定義される量であり、$n$-次正方行列 $A^{n\times n}$ の行列式は次式で定義される。ここで $S_n$ は対称群、$\sigma$ は置換、$\mathrm{sgn}(\sigma)$ は置換符号である。

\[ \det(A) = \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\prod_{i=1}^n A_{i\sigma(i)} \]

二次の行列式：

\[ \begin{aligned} A &= \begin{bmatrix} a_{11} & a_{12} \cr a_{21} & a_{22} \cr \end{bmatrix} \cr \cr \det(A) &= \sum_{\sigma\in S_2}\mathrm{sgn}(\sigma)\prod_{i=1}^2 A_{i\sigma(i)} \cr &= \mathrm{sgn}(\sigma_1)\prod_{i=1}^2 A_{i\sigma_1(i)} + \mathrm{sgn}(\sigma_2)\prod_{i=1}^2 A_{i\sigma_2(i)} \cr \cr \sigma_1 &= \begin{pmatrix} 1 & 2 \cr 1 & 2 \cr \end{pmatrix} \cr \sigma_2 &= \begin{pmatrix} 1 & 2 \cr 2 & 1 \cr \end{pmatrix} \cr \cr \det(A) &= \prod_{i=1}^2 A_{i\sigma_1(i)} - \prod_{i=1}^2 A_{i\sigma_2(i)} \cr &= a_{11}a_{22} - a_{12}a_{21} \end{aligned} \]

三次の行列式：

\[ \begin{aligned} A &= \begin{bmatrix} a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23} \cr a_{31} & a_{32} & a_{33} \cr \end{bmatrix} \cr \cr \det(A) &= \sum_{\sigma\in S_3}\mathrm{sgn}(\sigma)\prod_{i=1}^3 A_{i\sigma(i)} \cr &= \mathrm{sgn}(\sigma_1)\prod_{i=1}^3 A_{i\sigma_1(i)} + \mathrm{sgn}(\sigma_2)\prod_{i=1}^3 A_{i\sigma_2(i)} + \mathrm{sgn}(\sigma_3)\prod_{i=1}^3 A_{i\sigma_3(i)} \cr &\quad + \mathrm{sgn}(\sigma_4)\prod_{i=1}^3 A_{i\sigma_4(i)} + \mathrm{sgn}(\sigma_5)\prod_{i=1}^3 A_{i\sigma_5(i)} + \mathrm{sgn}(\sigma_6)\prod_{i=1}^3 A_{i\sigma_6(i)} \cr \cr \sigma_1 &= \begin{pmatrix} 1 & 2 & 3 \cr 1 & 2 & 3 \cr \end{pmatrix} \cr \sigma_2 &= \begin{pmatrix} 1 & 2 & 3 \cr 1 & 3 & 2 \cr \end{pmatrix} \cr \sigma_3 &= \begin{pmatrix} 1 & 2 & 3 \cr 2 & 1 & 3 \cr \end{pmatrix} \cr \sigma_4 &= \begin{pmatrix} 1 & 2 & 3 \cr 2 & 3 & 1 \cr \end{pmatrix} \cr \sigma_5 &= \begin{pmatrix} 1 & 2 & 3 \cr 3 & 1 & 2 \cr \end{pmatrix} \cr \sigma_6 &= \begin{pmatrix} 1 & 2 & 3 \cr 3 & 2 & 1 \cr \end{pmatrix} \cr \cr \det(A) &= \sum_{\sigma\in S_3}\mathrm{sgn}(\sigma)\prod_{i=1}^3 A_{i\sigma(i)} \cr &= \prod_{i=1}^3 A_{i\sigma_1(i)} - \prod_{i=1}^3 A_{i\sigma_2(i)} - \prod_{i=1}^3 A_{i\sigma_3(i)} \cr &\quad + \prod_{i=1}^3 A_{i\sigma_4(i)} + \prod_{i=1}^3 A_{i\sigma_5(i)} - \prod_{i=1}^3 A_{i\sigma_6(i)} \cr &= a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} \cr &\quad + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} \end{aligned} \]

小行列式

$n$-次正方行列 $M$ から $i$ 行 $j$ 列を取り除いた小行列の行列式を小行列式 (英：minor determinant) という。このとき小行列を $M_{ij}$ で表す。

\[ \begin{aligned} M_{ij} &= \det\overbrace{\begin{bmatrix} m_{11} & \cdots & \cancel{m_{1j}} & \cdots & m_{1n} \cr \vdots & \ddots & \vdots & & \vdots \cr \cancel{m_{i1}} & \cdots & \cancel{m_{ij}} & \cdots & \cancel{m_{in}}\cr \vdots & & \vdots & \ddots & \vdots \cr m_{n1} & \cdots & \cancel{m_{nj}} & \cdots & m_{nn} \end{bmatrix}}^{n\times n} \cr &= \det\overbrace{\begin{bmatrix} m_{11} & \cdots & m_{1(j-1)} & m_{1(j+1)} & \cdots & m_{1n} \cr \vdots & \ddots & \vdots & \vdots & & \vdots \cr m_{(i-1)1} & \cdots & m_{(i-1)(j-1)} & m_{(i-1)(j+1)} & \cdots & m_{(i-1)n} \cr m_{(i+1)1} & \cdots & m_{(i+1)(j-1)} & m_{(i+1)(j+1)} & \cdots & m_{(i+1)n} \cr \vdots & & \vdots & \vdots & \ddots & \vdots \cr m_{n1} & \cdots & m_{n(j-1)} & m_{n(j+1)} & \cdots & m_{nn} \cr \end{bmatrix}}^{(n-1)\times(n-1)} \end{aligned} \]

行列式の諸定理

\[ \def\b{\boldsymbol} \begin{aligned} \text{(D1)} &: \det(A^{(i\updownarrow j)}) = -\det(A) \cr \text{(D2)} &: \b{a}_i=\b{a}_j\rArr\det(A) = 0 \quad (A=\lbrack \b{a}_1 ~\cdots ~\b{a}_n\rbrack^\top) \cr \text{(D3)} &: \det(\lbrack\b{a}_1 ~\cdots ~\b{b}_i ~\cdots ~\b{a}_n\rbrack^\top) + \det(\lbrack\b{a}_1 ~\cdots ~\b{c}_i ~\cdots ~\b{a}_n\rbrack^\top) = \det(\lbrack\b{a}_1 ~\cdots ~\b{b}_i+\b{c}_i ~\cdots ~\b{a}_n\rbrack^\top) \cr \text{(D4)} &: \det(\lbrack\b{a}_1 ~\cdots ~c\b{a}_i ~\cdots ~\b{a}_n\rbrack^\top) = c\det(A) \cr \end{aligned} \]

$\bold{(D1)}$ の証明：

\[ \begin{aligned} \det(A^{(i\updownarrow j)}) &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\big(A^{(i\updownarrow j)}_{1\sigma(1)}\cdots A^{(i\updownarrow j)}_{i\sigma(i)}\cdots A^{(i\updownarrow j)}_{j\sigma(j)}\cdots A^{(i\updownarrow j)}_{n\sigma(n)}\big) \cr &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\big(A_{1\sigma(1)}\cdots A_{j\sigma(i)}\cdots A_{i\sigma(j)}\cdots A_{n\sigma(n)}\big) \cr \end{aligned} \]

ここで $i,j$ のみを入れ替える互換 $\xi$ を定義すると、

\[ \begin{gathered} \xi(k) = \begin{cases} i & \text{if }k=j \cr j & \text{if }k=i \cr k & \text{otherwise} \end{cases} \end{gathered} \tag{1} \]

すると $(1)$ より、

\[ \begin{aligned} \sigma(i) &= \sigma(\xi(j)) \cr \sigma(j) &= \sigma(\xi(i)) \cr \sigma(k) &= \sigma(\xi(k)) \quad (k\ne i,k\ne j) \end{aligned} \]

\[ \begin{aligned} \det(A^{(i\updownarrow j)}) &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\big(A_{1\sigma(\xi(1))}\cdots A_{j\sigma(\xi(j))}\cdots A_{i\sigma(\xi(i))}\cdots A_{n\sigma(\xi(n))}\big) \cr &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\prod_{k=1}^n A_{k\sigma(\xi(k))} \cr \end{aligned} \]

置換の符号も同様に $(1)$ から次式の関係が得られる。また互換は全単射であることから $\sigma\circ\xi$ の集合全体は $S_n$ と一致する。よって、

\[ \mathrm{sgn}(\sigma) = -\mathrm{sgn}(\sigma\circ\xi) \]

\[ \begin{gathered} \begin{aligned} \det(A^{(i\updownarrow j)}) &= -\sum_{(\sigma\circ\xi)\in S_n}\mathrm{sgn}(\sigma\circ\xi)\prod_{k=1}^n A_{k\sigma(\xi(k))} \cr &= -\det(A) \cr \end{aligned} \cr \cr \therefore \det(A^{(i\updownarrow j)}) = -\det(A) \end{gathered} \]

$\bold{(D2)}$ の証明：

行列 $A$ の二つの行 $\boldsymbol a_i, \boldsymbol a_j$ が共に等しいことから、

\[ \begin{aligned} A &= A^{(i\updownarrow j)} \cr \det(A) &= \det(A^{(i\updownarrow j)}) \cr \end{aligned} \]

定理 $\text{(D3)}$ により、

\[ \begin{gathered} \det(A) = -\det(A) \cr \cr \therefore\det(A) = 0 \end{gathered} \]

$\bold{(D3)}$ の証明：

\[ \def\b{\boldsymbol} \begin{gathered} A = \begin{bmatrix}\b{a}_1\cr\vdots\cr\b{a}_i\cr\vdots\cr\b{a}_n\end{bmatrix} = \begin{bmatrix}\b{a}_1\cr\vdots\cr\b{b}_i+\b{c}_i\cr\vdots\cr\b{a}_n\end{bmatrix} \cr \cr \begin{aligned} \det(A) &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\prod_{i=1}^n A_{i\sigma(i)} \cr &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)( a_{1\sigma(1)}\cdots a_{i\sigma(i)}\cdots a_{n\sigma(n)}) \cr &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)( a_{1\sigma(1)}\cdots (b_{i\sigma(i)}+c_{i\sigma(i)})\cdots a_{n\sigma(n)}) \cr &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)( a_{1\sigma(1)}\cdots b_{i\sigma(i)}\cdots a_{n\sigma(n)}) \cr &\quad + \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)( a_{1\sigma(1)}\cdots c_{i\sigma(i)}\cdots a_{n\sigma(n)}) \cr &= \det\begin{bmatrix}\b{a}_1\cr\vdots\cr\b{b}_i\cr\vdots\cr\b{a}_n\end{bmatrix} + \det\begin{bmatrix}\b{a}_1\cr\vdots\cr\b{c}_i\cr\vdots\cr\b{a}_n\end{bmatrix} \end{aligned} \cr \cr \therefore \det\begin{bmatrix}\b{a}_1\cr\vdots\cr\b{b}_i\cr\vdots\cr\b{a}_n\end{bmatrix} + \det\begin{bmatrix}\b{a}_1\cr\vdots\cr\b{c}_i\cr\vdots\cr\b{a}_n\end{bmatrix} = \det\begin{bmatrix}\b{a}_1\cr\vdots\cr\b{b}_i+\b{c}_i\cr\vdots\cr\b{a}_n\end{bmatrix} \end{gathered} \]

$\bold{(D4)}$ の証明：

\[ \def\b{\boldsymbol} \begin{gathered} \begin{aligned} \det(\lbrack\b{a}_1 ~\cdots ~c\b{a}_i ~\cdots ~\b{a}_n\rbrack^\top) &= \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)( a_{1\sigma(1)}\cdots ca_{i\sigma(i)}\cdots a_{n\sigma(n)}) \cr &= c\sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)( a_{1\sigma(1)}\cdots a_{i\sigma(i)}\cdots a_{n\sigma(n)}) \cr &= \det(A) \end{aligned} \cr \cr \therefore \det(\lbrack\b{a}_1 ~\cdots ~c\b{a}_i ~\cdots ~\b{a}_n\rbrack^\top) = c\det(A) \end{gathered} \]

積の行列式

\[ \det(AB) = \det(A)\det(B) \]

積の行列式の証明：

\( \begin{aligned} \det(AB) &= \det\begin{bmatrix} \sum_{i} a_{1i}b_{i1} & \cdots &\sum_{i} a_{1i}b_{in} \\ \vdots & \ddots & \vdots \\ \sum_{i} a_{ni}b_{i1} & \cdots &\sum_{i} a_{ni}b_{in} \\ \end{bmatrix} \\ &= \det\begin{bmatrix} (a_{11}b_{11}+a_{12}b_{21}+\cdots + a_{1n}b_{n1}) & \cdots & \sum_i a_{1i}b_{in} \\ \vdots & \ddots & \vdots \\ (a_{n1}b_{11}+a_{n2}b_{21}+\cdots + a_{nn}b_{n1}) & \cdots & \sum_i a_{ni}b_{in} \end{bmatrix} \\ &= \sum_{j_1}\det\begin{bmatrix} a_{1j_1}b_{j_11} & \sum_i a_{1i}b_{i2} & \cdots & \sum_i a_{1i}b_{in} \\ \vdots & \vdots & \ddots & \vdots \\ a_{nj_1}b_{j_11} & \sum_i a_{ni}b_{i2} & \cdots & \sum_i a_{ni}b_{in} \end{bmatrix} \quad \because\text{(D4)} \\ &= \sum_{j_1}\cdots\sum_{j_n}\det\begin{bmatrix} a_{1j_1}b_{j_11} & \cdots & a_{1j_n}b_{j_nn} \\ \vdots & \ddots & \vdots \\ a_{nj_1}b_{j_11} & \cdots & a_{nj_n}b_{j_nn} \end{bmatrix} \quad \because\text{(D4)} \\ &= \sum_{j_1}\cdots\sum_{j_n}\left\lbrace b_{j_11}\cdots b_{j_nn}\det\begin{bmatrix} a_{1j_1} & \cdots & a_{1j_n} \\ \vdots & \ddots & \vdots \\ a_{nj_1} & \cdots & a_{nj_n} \end{bmatrix}\right\rbrace \quad \because\text{(D5)} \\ \end{aligned} \) 

ここで、$j_k = j_l$ が等しい場合を考える。このとき定理 $\text{(D3)}$ より、

\[ \det\begin{bmatrix} a_{1j_1} & \cdots & a_{1j_k} & \cdots & a_{1j_l} & \cdots & a_{1j_n} \\ \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{nj_1} & \cdots & a_{nj_k} & \cdots & a_{nj_l} & \cdots & a_{nj_n} \end{bmatrix} = 0 \quad (j_k=j_l) \]

となることから、$j_1,j_2,\ldots,j_n$ の間で重複がない組合せの集まりは、置換全ての集まりと同値である。 \( \begin{pmatrix} 1 & 2 & \cdots & n \cr j_1 & j_2 & \cdots & j_n \end{pmatrix} = \sigma \in S_n \) とすると、 \( \begin{aligned} \det(AB) &= \sum_{\sigma\in S_n}\left\lbrace\prod_i b_{\sigma(i)i}\cdot\det\begin{bmatrix} a_{1\sigma(1)} & \cdots & a_{1\sigma(1)} \\ \vdots & \ddots & \vdots \\ a_{n\sigma(1)} & \cdots & a_{n\sigma(n)} \end{bmatrix}\right\rbrace \\ &= \sum_{\tau\in S_n}\left\lbrace \prod_i b_{i\tau(i)}\cdot\mathrm{sgn}(\tau)\det\begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix}\right\rbrace \quad \because\text{(D2)} \\ &= \det(A)\sum_{\tau\in S_n}\mathrm{sgn}(\tau)\prod_i b_{i\tau(i)} \\ &= \det(A)\det(B) \\ \\ \therefore \det(AB) &= \det(A)\det(B) \end{aligned} \)

逆行列の行列式

\[ \det(A^{-1}) = \frac{1}{\det(A)} \]

逆行列の行列式の証明：

\[ \begin{aligned} AA^{-1} = A^{-1}A &= I \\ \det(AA^{-1}) = \det(A^{-1}A) &= I \\ \det(A)\det(A^{-1}) &= \det(I) \quad \because\text{積の行列式より} \\ \det(A)\det(A^{-1}) &= 1 \\ \\ \therefore \det(A^{-1}) &= \frac{1}{\det(A)} \end{aligned} \]

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• 対称群
• 単位行列