Kronecker積のディターミナント
$A=[a_{ij}] \in \mathrm{M}_{n}(\mathbb{K}),\,B=[b_{k\ell}] \in \mathrm{M}_{m}(\mathbb{K})$とし，$A,B$から定まる線型変換をそれぞれ$t_{A},t_{B}$とする：
$$
t_{A} \colon \mathbb{K}^{n} \to \mathbb{K}^{n},\ t_{B} \colon \mathbb{K}^{m} \to \mathbb{K}^{m}.$$
$$ t_{A}\otimes t_{B} = (t_{A} \otimes \id_{\mathbb{K}^{m}})\circ(\id_{\mathbb{K}^{n}}\otimes t_{B}).$$
右辺を$t$とおくと，基底$(e_{i} \otimes f_{k})_{(i,k)}$の像について
$$
t(e_{i}\otimes f_{k}) = (t_{A}\otimes\id_{\mathbb{K}^{m}})(e_{i}\otimes(t_{B}f_{k})) = (t_{A}e_{i})\otimes(t_{B}f_{k}) = (t_{A}\otimes t_{B})(e_{i}\otimes f_{k})$$
が成り立つので，結論を得る．
基底ベクトル$e_{i} \otimes f_{k}$を逆辞書式順序で並べたとき，$t_{A} \otimes \id_{\mathbb{K}^{m}}$の表現行列について
$$
\left[t_{A}\otimes \id_{\mathbb{K}^{m}}: \frac{(e_{1}\otimes f_{1},\ldots,e_{n}\otimes f_{1};\ldots;e_{1} \otimes f_{m},\ldots,e_{n} \otimes f_{m})}{(e_{1}\otimes f_{1},\ldots,e_{n}\otimes f_{1};\ldots;e_{1} \otimes f_{m},\ldots,e_{n} \otimes f_{m})}\right] = \underbrace{A \oplus\cdots\oplus A}_{m}$$
が成り立つ．
各$\ell \in \{1,\ldots,m\}$に対して，
$$
(t_{A}\otimes\id_{\mathbb{K}^{m}})(e_{j}\otimes f_{\ell}) = (t_{A}e_{j})\otimes(\id_{\mathbb{K}^{m}}f_{\ell}) = \left(\sum_{i=1}^{n}e_{i}a_{ij}\right)\otimes f_{\ell} = \sum_{i=1}^{n} (e_{i}\otimes f_{\ell})a_{ij}$$
より
$$
\begin{bmatrix}
(t_{A}\otimes\id_{\mathbb{K}^{m}})(e_{1}\otimes f_{\ell}) & \cdots & (t_{A}\otimes\id_{\mathbb{K}^{m}})(e_{n} \otimes f_{\ell})
\end{bmatrix} = \begin{bmatrix}
e_{1}\otimes f_{\ell} & \cdots & e_{n} \otimes f_{\ell}
\end{bmatrix}A $$
が成り立つので，結論を得る．
基底ベクトル$e_{i} \otimes f_{k}$を辞書式順序で並べたとき，$\id_{\mathbb{K}^{n}}\otimes t_{B}$の表現行列について
$$
\left[\id_{\mathbb{K}^{n}}\otimes t_{B}: \frac{(e_{1} \otimes f_{1},\ldots,e_{1} \otimes f_{m};\ldots;e_{n} \otimes f_{1},\ldots,e_{n} \otimes f_{m})}{(e_{1} \otimes f_{1},\ldots,e_{1} \otimes f_{m};\ldots;e_{n} \otimes f_{1},\ldots,e_{n} \otimes f_{m})}\right] = \underbrace{B \oplus\cdots\oplus B}_{n}$$
が成り立つ．
各$j\in\{1,\ldots,n\}$に対して，
$$
(\id_{\mathbb{K}^{n}}\otimes t_{B})(e_{j} \otimes f_{\ell}) = (\id_{\mathbb{K}^{n}}e_{j}) \otimes (t_{B}f_{\ell}) = e_{j} \otimes \left(\sum_{k=1}^{m} f_{k} b_{k\ell}\right) = \sum_{k=1}^{m} (e_{j} \otimes f_{k})b_{k\ell}$$
より
$$
\begin{bmatrix}
(\id_{\mathbb{K}^{n}}\otimes t_{B})(e_{j} \otimes f_{1}) & \cdots & (\id_{\mathbb{K}^{n}}\otimes t_{B})(e_{j}\otimes f_{m})
\end{bmatrix} = \begin{bmatrix}
e_{j}\otimes f_{1} & \cdots & e_{j} \otimes f_{m}
\end{bmatrix}B$$
が成り立つので，結論を得る．
$$ \det(A \otimes B) = (\det{A})^{m}(\det{B})^{n}.$$
$t_{A} \otimes t_{B}$の表現行列として$A \otimes B$を定義していたので，上の補題群より
$$
\det(A \otimes B) = \det(t_{A} \otimes t_{B}) = \det(t_{A}\otimes\id_{\mathbb{K}^{m}}) \det(\id_{\mathbb{K}^{n}}\otimes t_{B}) = (\det{A})^{m}(\det{B})^{n}$$
を得る．
