名古屋大学 情報学研究科 複雑系科学専攻 2022年8月実施 数1 [1] Author 思齐塾 , 祭音Myyura
Description 出典：名古屋大学公式問題
相似変換の定義（題意の要約）：Q T Q = I Q^TQ=I Q T Q = I 2 × 2 2\times2 2 × 2 Q Q Q λ > 0 \lambda>0 λ > 0 h ∈ R 2 h\in\mathbb R^2 h ∈ R 2 r ′ = λ Q r + h r'=\lambda Qr+h r ′ = λ Q r + h λ = 1 \lambda=1 λ = 1
任意の2次直交行列 A A A − 1 ≤ a ≤ 1 -1 \leq a \leq 1 − 1 ≤ a ≤ 1 a a a
(ア) ( a ∓ 1 − a 2 ± 1 − a 2 a ) \begin{pmatrix} a & \mp \sqrt{1-a^2} \\ \pm \sqrt{1-a^2} & a \end{pmatrix} ( a ± 1 − a 2 ∓ 1 − a 2 a ) ( a ± 1 − a 2 ± 1 − a 2 − a ) \begin{pmatrix} a & \pm \sqrt{1-a^2} \\ \pm \sqrt{1-a^2} & -a \end{pmatrix} ( a ± 1 − a 2 ± 1 − a 2 − a )
(一つの行列の中の複号は同順) に限られることを証明せよ。つまり、 A T A = I A^T A = I A T A = I A A A
题目描述
证明任意 2 2 2 A A A − 1 ≤ a ≤ 1 -1\le a\le1 − 1 ≤ a ≤ 1 a a a
b = 1 − a 2 , ε ∈ { 1 , − 1 } , b=\sqrt{1-a^2},\qquad \varepsilon\in\{1,-1\}, b = 1 − a 2 , ε ∈ { 1 , − 1 } , 则
A = ( a − ε b ε b a ) 或 A = ( a ε b ε b − a ) . (ア) A=
\begin{pmatrix}
a&-\varepsilon b\\
\varepsilon b&a
\end{pmatrix}
\quad\text{或}\quad
A=
\begin{pmatrix}
a&\varepsilon b\\
\varepsilon b&-a
\end{pmatrix}. \tag{ア} A = ( a ε b − ε b a ) 或 A = ( a ε b ε b − a ) . ( ア ) 也就是说，证明每个满足 A T A = I A^{\mathsf T}A=I A T A = I 2 × 2 2\times2 2 × 2
Kai Let the first column of A A A u = ( a , b ) T u=(a,b)^T u = ( a , b ) T a 2 + b 2 = 1 a^2+b^2=1 a 2 + b 2 = 1 − 1 ≤ a ≤ 1 -1\le a\le1 − 1 ≤ a ≤ 1 b = ± 1 − a 2 b=\pm\sqrt{1-a^2} b = ± 1 − a 2 u u u R 2 \mathbb R^2 R 2 ( − b , a ) T (-b,a)^T ( − b , a ) T ( b , − a ) T (b,-a)^T ( b , − a ) T
A = ( a − b b a ) or A = ( a b b − a ) , b = ± 1 − a 2 . \boxed{
A=\begin{pmatrix}a&-b\\b&a\end{pmatrix}
\quad\text{or}\quad
A=\begin{pmatrix}a&b\\b&-a\end{pmatrix},
\qquad b=\pm\sqrt{1-a^2}.
} A = ( a b − b a ) or A = ( a b b − a ) , b = ± 1 − a 2 . This proves the two stated families, including a = ± 1 a=\pm1 a = ± 1 b = 0 b=0 b = 0 A T A = I A^TA=I A T A = I 1 1 1 − 1 -1 − 1