行列の演習問題とその解答

$K$を可換体として，$K$に成分をもつ$2$次正方行列の全体を${\mathbb{M}}_2(K)$で表す．$\mathbf{A}=[a_{ij}]\in {\mathbb{M}}_2(K)$を固定して，${\mathrm{C}}_{\mathbf{A}}:=\{\mathbf{B}\in {\mathbb{M}}_2(K)\mid \mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}\}$とする．行列の基本変形を "$\to$" で表す．

$\mathbf{B}=[b_{ij}]\in {\mathbb{M}}_2(K)$について，$\mathbf{B}\in {\mathrm{C}}_{\mathbf{A}}$というのは任意の$1\le i,j\le 2$に対して
$$a_{i1}{b}_{1j}+a_{i2}{b}_{2j}=a_{1j}{b}_{i1}+a_{2j}{b}_{i2}$$
が成り立つことと同値である．$(i,j)=(1,1),(2,2)$のときこれは$a_{21}{b}_{12}-a_{12}{b}_{21}=0$と表され，$(i,j)=(1,2),(2,1)$のときは$a_{12}{b}_{11}+(a_{22}-a_{11})b_{12}-a_{12}{b}_{22}=0$及び$a_{21}{b}_{11}+(a_{22}-a_{11})b_{21}-a_{21}{b}_{22}=0$と表されるので，これは$[\begin{array}{c}{b}_{11}\\ b_{12}\\ b_{21}\\ b_{22}\end{array}]$が$[\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]$の解空間
$$\{\mathbf{x}\in K^4|[\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]\mathbf{x}=[\begin{array}{c}0\\ 0\\ 0\end{array}]\}$$
に属していることと同値である．

$a_{12}=a_{21}=0$かつ$a_{11}=a_{22}$の場合:

$[\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]=[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}]$より${\mathrm{C}}_{\mathbf{A}}={\mathbb{M}}_2(K)$を得る．

$a_{12}=a_{21}=0$かつ$a_{11}\ne a_{22}$の場合:

$$[\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]=[\begin{array}{cccc}0& 0& 0& 0\\ 0& a_{22}-a_{11}& 0& 0\\ 0& 0& a_{22}-a_{11}& 0\end{array}]\to [\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}]$$
より
$$\begin{array}{rl}{\mathrm{C}}_{\mathbf{A}}& =\{[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}]\in {\mathbb{M}}_2(K)|(b_{12}=0)\wedge (b_{21}=0)\}\\ & =\{[\begin{array}{cc}{b}_{11}& 0\\ 0& b_{22}\end{array}]\in {\mathbb{M}}_2(K)|b_{11},b_{22}\in K\}\\ & =K[\begin{array}{cc}1& 0\\ 0& 0\end{array}]\oplus K[\begin{array}{cc}0& 0\\ 0& 1\end{array}]\end{array}$$
を得る．

$a_{12}=0$かつ$a_{21}\ne 0$の場合:

$$\begin{array}{rl}[\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]& =[\begin{array}{cccc}0& a_{21}& 0& 0\\ 0& a_{22}-a_{11}& 0& 0\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]\\ & \to [\begin{array}{cccc}0& 1& 0& 0\\ 0& a_{22}-a_{11}& 0& 0\\ 1& 0& (a_{22}-a_{11})/a_{21}& -1\end{array}]\\ & \to [\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 0& 0\\ 1& 0& (a_{22}-a_{11})/a_{21}& -1\end{array}]\\ & \to [\begin{array}{cccc}1& 0& (a_{22}-a_{11})/a_{21}& -1\\ 0& 1& 0& 0\\ 0& 0& 0& 0\end{array}]\end{array}$$
より
$$\begin{array}{rl}{\mathrm{C}}_{\mathbf{A}}& =\{[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}]\in {\mathbb{M}}_2(K)|(b_{11}+\frac{a_{22}-a_{11}}{a_{21}}{b}_{21}-b_{22}=0)\wedge (b_{12}=0)\}\\ & =\{[\begin{array}{cc}{b}_{11}& 0\\ b_{21}& b_{11}+[(a_{22}-a_{11})/a_{21}]b_{21}\end{array}]\in {\mathbb{M}}_2(K)|b_{11},b_{21}\in K\}\\ & =K[\begin{array}{cc}1& 0\\ 0& 1\end{array}]\oplus K[\begin{array}{cc}0& 1\\ 0& (a_{22}-a_{11})/a_{21}\end{array}]\end{array}$$
を得る．

$a_{12}\ne 0$かつ$a_{21}=0$の場合:

$$\begin{array}{rl}[\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]& =[\begin{array}{cccc}0& 0& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ 0& 0& a_{22}-a_{11}& 0\end{array}]\\ & \to [\begin{array}{cccc}0& 0& 1& 0\\ 1& (a_{22}-a_{11})/a_{12}& 0& -1\\ 0& 0& a_{22}-a_{11}& 0\end{array}]\\ & \to [\begin{array}{cccc}0& 0& 1& 0\\ 1& (a_{22}-a_{11})/a_{12}& 0& -1\\ 0& 0& 0& 0\end{array}]\\ & \to [\begin{array}{cccc}1& (a_{22}-a_{11})/a_{12}& 0& -1\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}]\end{array}$$
より
$$\begin{array}{rl}{\mathrm{C}}_{\mathbf{A}}& =\{[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}]\in {\mathbb{M}}_2(K)|(b_{11}+\frac{a_{22}-a_{11}}{a_{12}}{b}_{12}-b_{22}=0)\wedge (b_{21}=0)\}\\ & =\{[\begin{array}{cc}{b}_{11}& b_{12}\\ 0& b_{11}+[(a_{22}-a_{11})/a_{12}]b_{12}\end{array}]|b_{11},b_{12}\in K\}\\ & =K[\begin{array}{cc}1& 0\\ 0& 1\end{array}]\oplus K[\begin{array}{cc}0& 1\\ 0& (a_{22}-a_{11})/a_{12}\end{array}]\end{array}$$
を得る．

$a_{12}\ne 0$かつ$a_{21}\ne 0$の場合:

$$\begin{array}{rl}[\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ a_{12}& a_{22}-a_{11}& 0& -a_{12}\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]& \to [\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ 1& (a_{22}-a_{11})/a_{12}& 0& -1\\ a_{21}& 0& a_{22}-a_{11}& -a_{21}\end{array}]\\ & \to [\begin{array}{cccc}0& a_{21}& -a_{12}& 0\\ 1& (a_{22}-a_{11})/a_{12}& 0& -1\\ 0& -a_{21}(a_{22}-a_{11})/a_{12}& a_{22}-a_{11}& 0\end{array}]\\ & \to [\begin{array}{cccc}0& 1& -a_{12}/a_{21}& 0\\ 1& (a_{22}-a_{11})/a_{12}& 0& -1\\ 0& -a_{21}(a_{22}-a_{11})/a_{12}& a_{22}-a_{11}& 0\end{array}]\\ & \to [\begin{array}{cccc}0& 1& -a_{12}/a_{21}& 0\\ 1& 0& (a_{22}-a_{11})/a_{21}& -1\\ 0& 0& 0& 0\end{array}]\\ & \to [\begin{array}{cccc}1& 0& (a_{22}-a_{11})/a_{21}& -1\\ 0& 1& -a_{12}/a_{21}& 0\\ 0& 0& 0& 0\end{array}]\end{array}$$
より
$$\begin{array}{rl}{\mathrm{C}}_{\mathbf{A}}& =\{[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}]\in {\mathbb{M}}_2(K)|(b_{11}+\frac{a_{22}-a_{11}}{a_{12}}{b}_{12}-b_{22}=0)\wedge (b_{12}-\frac{a_{12}}{a_{21}}{b}_{21}=0)\}\\ & =\{[\begin{array}{cc}{b}_{11}& b_{12}\\ (a_{21}/a_{12})b_{12}& b_{11}+[(a_{22}-a_{11})/a_{12}]b_{12}\end{array}]|b_{11},b_{12}\in K\}\\ & =K[\begin{array}{cc}1& 0\\ 0& 1\end{array}]\oplus K[\begin{array}{cc}0& 1\\ a_{21}/a_{12}& (a_{22}-a_{11})/a_{12}\end{array}]\end{array}$$
を得る．

${\mathbf{C}}_1,{\mathbf{C}}_2\in {\mathbb{M}}_2(K)$に対して，${\mathbf{C}}_1$と${\mathbf{C}}_2$が可換なら$K{\mathbf{C}}_1\oplus K{\mathbf{C}}_2$の任意の$2$元も可換になることに注意すれば，以下を得る:

まとめ

上の問題は Twitter でフォロワーの方が呟いていたものを簡単にした (一般の次元で考えていたものを$2$次元にした) ものです．ただ計算しただけなので，線形代数を使った見通しが良い解答などがあれば是非教えてください．(追記: $K=\mathbb{C}$の場合について，delta さんが Jordan 標準形を使った証明をコメントで与えてくださいました．ありがとうございます．)

高次元の場合の反例について

解答にある主張のうち「(1)$\Rightarrow$(2)$\iff$(3)」は次元を$2$以上の整数$n$に広げても成り立ちますが，「(1)$\Leftarrow$(3)」については，$K$の標数が$2$でなく$n$が$4$以上の偶数のときに反例があります (こちらの user1551 氏の回答 を参考にしました)．$n=4$の場合には，例えば$\mathbf{A}=[\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}]$が反例になります．実際，$\mathbf{A}$は単位行列の定数倍ではなく，${\mathbf{B}}_1=[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -1\end{array}]$と${\mathbf{B}}_2=[\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\\ 0& 1& 0& 0\end{array}]$は
$$\begin{array}{rl}\mathbf{A}{\mathbf{B}}_1& =[\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& 1\end{array}]={\mathbf{B}}_1\mathbf{A},\\ \mathbf{A}{\mathbf{B}}_2& =[\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& -1\\ 1& 0& 0& 0\\ 0& -1& 0& 0\end{array}]={\mathbf{B}}_2\mathbf{A},\end{array}$$
より共に$\mathbf{A}$と可換ですが，
$${\mathbf{B}}_1{\mathbf{B}}_2=[\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 1\\ 0& -1& 0& 0\end{array}]\ne [\begin{array}{cccc}0& 0& -1& 0\\ 0& 0& 0& -1\\ 1& 0& 0& 0\\ 0& 1& 0& 0\end{array}]={\mathbf{B}}_2{\mathbf{B}}_1,$$
よりこれらは可換ではないので，$\mathbf{A}$は (3) をみたすが (1) をみたさない例になっています．$n=6,8,10,12,14,\dots$の場合にも同様の手法により反例が作れます．