練習問題メモ 3

Contents

練習問題メモ 3#

3.1#

$A$ を $k \times l$ 行列、 $B$ を $k \times n$ 行列、 $C$ を $m \times n$ 行列とする。分割された $(k + m) \times (l + n)$ 行列に対する次の計算をせよ。 $(\begin{array}{cc} A & B \\ O_{m, l} & C \end{array}) + (\begin{array}{cc} A & - B \\ O_{m, l} & - 2 C \end{array})$

\begin{array}{r} (\begin{array}{cc} A & B \\ O_{m, l} & C \end{array}) + (\begin{array}{cc} A & - B \\ O_{m, l} & - 2 C \end{array}) = (\begin{array}{cc} A + A & B - B \\ O_{m, l} - O_{m, l} & C - 2 C \end{array}) = (\begin{array}{cc} 2 A & O_{k, n} \\ O_{m, l} & - C \end{array}) \end{array}

3.2#

$A$ を $m \times n$ 行列とする。分割された $(m + n)$ 次の正方行列 $(\begin{array}{cc} E_{m} & A \\ O_{n, m} & E_{n} \end{array})$ の3乗を計算せよ。

\begin{array}{r} {(\begin{array}{cc} E_{m} & A \\ O_{n, m} & E_{n} \end{array})}^{3} = (\begin{array}{cc} E_{m}^{3} & 3 A \\ O_{n, m} & E_{n}^{3} \end{array}) = (\begin{array}{cc} E_{m} & 3 A \\ O_{n, m} & E_{n} \end{array}) \end{array}

\begin{array}{r} \underset{(\begin{array}{c} E_{m} & 2 A \\ O_{n, m} & E_{n} \end{array})}{\underset{⏟}{(\begin{array}{c} E_{m} & A \\ O_{n, m} & E_{n} \end{array}) (\begin{array}{c} E_{m} & A \\ O_{n, m} & E_{n} \end{array})}} (\begin{array}{c} E_{m} & A \\ O_{n, m} & E_{n} \end{array}) = (\begin{array}{c} E_{m} & 3 A \\ O_{n, m} & E_{n} \end{array}) \end{array}

3.3#

4次の正方行列 $I, J, K$ を $ $I = (\begin{array}{cccc} 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 1 \\ 0 & 0 & 1 & 0 \end{array}), J = (\begin{array}{cccc} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{array}), K = (\begin{array}{cccc} 0 & 0 & 0 & - 1 \\ 0 & 0 & - 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array})$ $により定める。$ I, J, K $を 2 次の正方行列を用いて分割することにより、積$ I^2, J^2 $,$ K^2, I J, J I, J K, K J, K I, I K$ を計算せよ。

\begin{array}{r} A = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}), B = (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}), C = (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) \end{array}

とおくと

\begin{array}{r} I = (\begin{array}{c} A & O \\ O & A \end{array}), J = (\begin{array}{c} O & B \\ - B & O \end{array}), K = (\begin{array}{c} O & - C \\ C & O \end{array}) \end{array}

$I^{2}$

\begin{array}{r} I^{2} = (\begin{array}{c} A & O \\ O & A \end{array}) (\begin{array}{c} A & O \\ O & A \end{array}) = (\begin{array}{c} A^{2} & O \\ O & A^{2} \end{array}) \end{array}

\begin{array}{r} A^{2} = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} - 1 & 0 \\ 0 & - 1 \end{array}) = - E_{2} \end{array}

なので

\begin{array}{r} I^{2} = (\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{array}) \end{array}

$J^{2}$

\begin{array}{r} J^{2} = (\begin{array}{c} O & B \\ - B & O \end{array}) (\begin{array}{c} O & B \\ - B & O \end{array}) = (\begin{array}{c} B (- B) & O \\ O & - B B \end{array}) = (\begin{array}{c} - 1 \cdot B^{2} & O \\ O & - 1 \cdot B^{2} \end{array}) \end{array}

\begin{array}{r} B^{2} = (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) = (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) = E_{2} \end{array}

\begin{array}{r} J^{2} = (\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{array}) \end{array}

$K^{2}$

\begin{array}{r} K^{2} = (\begin{array}{c} O & - C \\ C & O \end{array}) (\begin{array}{c} O & - C \\ C & O \end{array}) = (\begin{array}{c} - 1 \cdot C^{2} & O \\ O & - 1 \cdot C^{2} \end{array}) \end{array}

\begin{array}{r} C^{2} = (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) = E_{2} \end{array}

\begin{array}{r} K^{2} = (\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{array}) \end{array}

$I J$

\begin{array}{r} I J = (\begin{array}{c} I_{s} & O \\ O & I_{s} \end{array}) (\begin{array}{c} O & J_{1} \\ J_{2} & O \end{array}) = (\begin{array}{c} O & I_{s} J_{1} \\ I_{s} J_{2} & O \end{array}) \end{array}

\begin{array}{r} I_{s} J_{1} = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) = (\begin{array}{c} 0 & - 1 \\ - 1 & 0 \end{array}) \end{array}

\begin{array}{r} I_{s} J_{2} = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) (\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}) = (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) \end{array}

\begin{array}{r} I J = (\begin{array}{c} 0 & 0 & 0 & - 1 \\ 0 & 0 & - 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}) \end{array}

$J I$

\begin{array}{r} J I = (\begin{array}{c} O & J_{1} \\ J_{2} & O \end{array}) (\begin{array}{c} I_{s} & O \\ O & I_{s} \end{array}) = (\begin{array}{c} O & J_{1} I_{s} \\ J_{2} I_{s} & O \end{array}) \end{array}

\begin{array}{r} J_{1} I_{s} = (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) \end{array}

\begin{array}{r} J_{2} I_{s} = (\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}) (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} 0 & - 1 \\ - 1 & 0 \end{array}) \end{array}

\begin{array}{r} J I = (\begin{array}{c} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 0 \\ - 1 & 0 & 0 & 0 \end{array}) \end{array}

$J K$

\begin{array}{r} J K = (\begin{array}{c} O & J_{1} \\ J_{2} & O \end{array}) (\begin{array}{c} O & K_{1} \\ K_{2} & O \end{array}) = (\begin{array}{c} J_{1} K_{2} & O \\ O & J_{2} K_{1} \end{array}) \end{array}

\begin{array}{r} J_{1} K_{2} = (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) \end{array}

\begin{array}{r} J_{2} K_{1} = (\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}) (\begin{array}{c} 0 & - 1 \\ - 1 & 0 \end{array}) = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) \end{array}

\begin{array}{r} J K = (\begin{array}{c} 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 1 \\ 0 & 0 & 1 & 0 \end{array}) \end{array}

$K J$

\begin{array}{r} K J = (\begin{array}{c} O & K_{1} \\ K_{2} & O \end{array}) (\begin{array}{c} O & J_{1} \\ J_{2} & O \end{array}) = (\begin{array}{c} K_{1} J_{2} & O \\ O & K_{2} J_{1} \end{array}) \end{array}

\begin{array}{r} K_{2} J_{1} = (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) = (\begin{array}{c} 0 & 1 \\ - 1 & 0 \end{array}) \end{array}

\begin{array}{r} K_{1} J_{2} = (\begin{array}{c} 0 & - 1 \\ - 1 & 0 \end{array}) (\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}) = (\begin{array}{c} 0 & 1 \\ - 1 & 0 \end{array}) \end{array}

\begin{array}{r} K J = (\begin{array}{c} 0 & 1 & 0 & 0 \\ - 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & - 1 & 0 \end{array}) \end{array}

$K I$

\begin{array}{r} K I = (\begin{array}{c} O & K_{1} \\ K_{2} & O \end{array}) (\begin{array}{c} I_{s} & O \\ O & I_{s} \end{array}) = (\begin{array}{c} O & K_{1} I_{s} \\ K_{2} I_{s} & O \end{array}) \end{array}

\begin{array}{r} K_{1} I_{s} = (\begin{array}{c} 0 & - 1 \\ - 1 & 0 \end{array}) (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) \end{array}

\begin{array}{r} K_{2} I_{s} = (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}) \end{array}

\begin{array}{r} K I = (\begin{array}{c} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{array}) \end{array}

$I K$

\begin{array}{r} I K = (\begin{array}{c} I_{s} & O \\ O & I_{s} \end{array}) (\begin{array}{c} O & K_{1} \\ K_{2} & O \end{array}) = (\begin{array}{c} O & I_{s} K_{1} \\ I_{s} K_{2} & O \end{array}) \end{array}

\begin{array}{r} I_{s} K_{1} = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) (\begin{array}{c} 0 & - 1 \\ - 1 & 0 \end{array}) = (\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}) \end{array}

\begin{array}{r} I_{s} K_{2} = (\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}) (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) = (\begin{array}{c} - 1 & 0 \\ 0 & 1 \end{array}) \end{array}

\begin{array}{r} I K = (\begin{array}{c} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & - 1 \\ - 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}) \end{array}

3.4#

$A_{12}, A_{13}, A_{22}, A_{23}, A_{33}, B_{11}, B_{12}, B_{13}, B_{23}, B_{33}, C_{11}, C_{12}, C_{13}, C_{22}, C_{23}$ を $n$ 次の正方行列、 $O$ を $n$ 次の零行列とする。次の計算をせよ。 $(\begin{array}{ccc} O & A_{12} & A_{13} \\ O & A_{22} & A_{23} \\ O & O & A_{33} \end{array}) (\begin{array}{ccc} B_{11} & B_{12} & B_{13} \\ O & O & B_{23} \\ O & O & B_{33} \end{array}) (\begin{array}{ccc} C_{11} & C_{12} & C_{13} \\ O & C_{22} & C_{23} \\ O & O & O \end{array})$

\begin{array}{r} (\begin{array}{c} O & A_{12} & A_{13} \\ O & A_{22} & A_{23} \\ O & O & A_{33} \end{array}) (\begin{array}{c} B_{11} & B_{12} & B_{13} \\ O & O & B_{23} \\ O & O & B_{33} \end{array}) = (\begin{array}{c} O & O & A_{12} B_{23} + A_{13} B_{33} \\ O & O & A_{22} B_{23} + A_{23} B_{33} \\ O & O & A_{33} B_{33} \end{array}) \end{array}

\begin{array}{r} (\begin{array}{c} O & O & A_{12} B_{23} + A_{13} B_{33} \\ O & O & A_{22} B_{23} + A_{23} B_{33} \\ O & O & A_{33} B_{33} \end{array}) (\begin{array}{c} C_{11} & C_{12} & C_{13} \\ O & C_{22} & C_{23} \\ O & O & O \end{array}) = (\begin{array}{c} O & O & O \\ O & O & O \\ O & O & O \end{array}) = O \end{array}

3.5#

$a, b$ を異なる数、 $X_{11}, X_{12}, X_{21}, X_{22}$ をそれぞれ $m$ 次の正方行列、 $m \times n$ 行列、 $n \times m$ 行列、 $n$ 次の正方行列とし、 $(m + n)$ 次の正方行列 $A$ および $X$ を

$\begin{array}{r} A = (\begin{array}{cc} a E_{m} & O_{m, n} \\ O_{n, m} & b E_{n} \end{array}), X = (\begin{array}{ll} X_{11} & X_{12} \\ X_{21} & X_{22} \end{array}) \end{array}$

により定める。 $A$ と $X$ が可換となるのは $X_{12}$ および $X_{21}$ が零行列のときであることを示せ。

\begin{array}{r} A X = (\begin{array}{c} a E_{m} & O_{m, n} \\ O_{n, m} & b E_{n} \end{array}) (\begin{array}{c} X_{11} & X_{12} \\ X_{21} & X_{22} \end{array}) = (\begin{array}{c} a E_{m} X_{11} & a E_{m} X_{12} \\ b E_{n} X_{21} & b E_{n} X_{22} \end{array}) = (\begin{array}{c} a X_{11} & a X_{12} \\ b X_{21} & b X_{22} \end{array}) \end{array}

\begin{array}{r} X A = (\begin{array}{c} X_{11} & X_{12} \\ X_{21} & X_{22} \end{array}) (\begin{array}{c} a E_{m} & O_{m, n} \\ O_{n, m} & b E_{n} \end{array}) = (\begin{array}{c} a X_{11} E_{m} & b X_{12} E_{n} \\ a X_{21} E_{m} & b X_{22} E_{n} \end{array}) = (\begin{array}{c} a X_{11} & b X_{12} \\ a X_{21} & b X_{22} \end{array}) \end{array}

となり、 $a, b$ が異なる数であるため、 $A X$ と $X A$ の対角成分は異なる値になっている。

$X_{12}$ および $X_{21}$ が零行列であれば対角成分は零行列になるため、 $A X = X A$ となる