広島大学 先進理工系科学研究科 情報科学プログラム 2021年8月実施 専門科目I 問題1 Author samparker, 祭音Myyura
Description (1) A = [ 0 − α α 0 ] A = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} A = [ 0 α − α 0 ] α ≠ 0 \alpha \neq 0 α = 0
(2) A k A^k A k k k k
(3) 実正方行列 X X X
exp ( X ) = ∑ k = 0 ∞ 1 k ! X k = E + X + 1 2 ! X 2 + ⋯ \exp(X) = \sum_{k=0}^{\infty} \frac{1}{k!} X^k = E + X + \frac{1}{2!} X^2 + \cdots exp ( X ) = k = 0 ∑ ∞ k ! 1 X k = E + X + 2 ! 1 X 2 + ⋯ で定義する。これは、すべての行列 X X X E E E
exp ( A ) = [ cos α − sin α sin α cos α ] \exp(A) = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} exp ( A ) = [ cos α sin α − sin α cos α ] となることを示せ。
(1) Find all the eigenvalues of a matrix A = [ 0 − α α 0 ] A = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} A = [ 0 α − α 0 ] α ≠ 0 \alpha \neq 0 α = 0
(2) Find A k A^k A k k k k
(3) For real square matrix X X X
exp ( X ) = ∑ k = 0 ∞ 1 k ! X k = E + X + 1 2 ! X 2 + ⋯ \exp(X) = \sum_{k=0}^{\infty} \frac{1}{k!} X^k = E + X + \frac{1}{2!} X^2 + \cdots exp ( X ) = k = 0 ∑ ∞ k ! 1 X k = E + X + 2 ! 1 X 2 + ⋯ It is known that this function converges for all matrices X X X E E E
exp ( A ) = [ cos α − sin α sin α cos α ] \exp(A) = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} exp ( A ) = [ cos α sin α − sin α cos α ] Kai (1)
Eigenvalues
[ − λ − α α λ ] = 0 ⇒ λ 2 + α 2 = 0 ⇒ { λ 1 = i α λ 2 = − i α \begin{bmatrix}
-\lambda & -\alpha \\
\alpha & \lambda
\end{bmatrix}
= 0
\Rightarrow
\lambda^2 + \alpha^2 = 0
\Rightarrow
\begin{cases}
\lambda_1 = i\alpha \\
\lambda_2 = -i \alpha
\end{cases} [ − λ α − α λ ] = 0 ⇒ λ 2 + α 2 = 0 ⇒ { λ 1 = i α λ 2 = − i α the corresponding eigenvectors
{ v 1 = ( i , 1 ) v 2 = ( − i , 1 ) \begin{cases}
v_1 = (i, 1) \\
v_2 = (-i, 1)
\end{cases} { v 1 = ( i , 1 ) v 2 = ( − i , 1 ) (2)
u 1 = 1 2 ( i 1 ) , u 2 = 1 2 ( − i 1 ) , u_1 = \frac{1}{\sqrt{2}} \begin{pmatrix}
i \\ 1
\end{pmatrix},
u_2 = \frac{1}{\sqrt{2}} \begin{pmatrix}
-i \\ 1
\end{pmatrix}, u 1 = 2 1 ( i 1 ) , u 2 = 2 1 ( − i 1 ) , U = 1 2 ( i − i 1 1 ) , U − 1 = 1 2 ( − i 1 i 1 ) , U = \frac{1}{\sqrt{2}} \begin{pmatrix}
i & -i \\ 1 & 1
\end{pmatrix},
U^{-1} = \frac{1}{\sqrt{2}} \begin{pmatrix}
-i & 1 \\ i & 1
\end{pmatrix}, U = 2 1 ( i 1 − i 1 ) , U − 1 = 2 1 ( − i i 1 1 ) , Then,
A k = 1 2 ( i − i 1 1 ) ( ( i α ) k 0 0 ( − i α ) k ) ( − i 1 i 1 ) = 1 2 ( ( − i α ) k + ( i α ) k i ( i α ) k − i ( − i α ) k i ( − i α ) k − i ( i α ) k ( − i α ) k + ( i α ) k ) \begin{aligned}
A^k &= \frac{1}{2} \begin{pmatrix}
i & -i \\ 1 & 1
\end{pmatrix}
\begin{pmatrix}
(i\alpha)^k & 0 \\ 0 & (-i\alpha)^k
\end{pmatrix}
\begin{pmatrix}
-i & 1 \\ i & 1
\end{pmatrix} \\
&= \frac{1}{2}\begin{pmatrix}
(-i\alpha)^k + (i\alpha)^k & i(i\alpha)^k - i(-i\alpha)^k \\
i(-i\alpha)^k -i(i\alpha)^k & (-i\alpha)^k + (i\alpha)^k
\end{pmatrix}
\end{aligned} A k = 2 1 ( i 1 − i 1 ) ( ( i α ) k 0 0 ( − i α ) k ) ( − i i 1 1 ) = 2 1 ( ( − i α ) k + ( i α ) k i ( − i α ) k − i ( i α ) k i ( i α ) k − i ( − i α ) k ( − i α ) k + ( i α ) k ) (3)
The Cayley-Hamilton Theorem guarantees that
A 2 + α 2 E = 0 A^2 + \alpha^2 E = 0 A 2 + α 2 E = 0 so that A 2 = − α 2 E A^2 = -\alpha^2 E A 2 = − α 2 E A A A A 3 = − α 2 A A^3 = -\alpha^2 A A 3 = − α 2 A A 4 = − α 2 A 2 = α 4 E A^4 = -\alpha^2 A^2 = \alpha^4 E A 4 = − α 2 A 2 = α 4 E exp ( A ) \exp(A) exp ( A ) E E E A A A
exp ( A ) = E + A + 1 2 ! A 2 + 1 3 ! A 3 + ⋯ = E + A − α 2 2 ! E − α 2 3 ! A + α 4 4 ! E + α 4 5 ! A + ⋯ = ( 1 − α 2 2 ! + α 4 4 ! − ⋯ ) E + ( 1 − α 2 3 ! + α 4 5 ! − ⋯ ) A = ( cos α ) E + sin α α A = [ cos α − sin α sin α cos α ] \begin{aligned}
\exp(A) &= E + A + \frac{1}{2!} A^2 + \frac{1}{3!}A^3 + \cdots \\
&= E + A - \frac{\alpha^2}{2!}E - \frac{\alpha^2}{3!}A + \frac{\alpha^4}{4!}E + \frac{\alpha^4}{5!}A + \cdots \\
&= \left(1 - \frac{\alpha^2}{2!} + \frac{\alpha^4}{4!} - \cdots \right)E + \left(1 - \frac{\alpha^2}{3!} + \frac{\alpha^4}{5!} - \cdots \right)A \\
&= (\cos \alpha) E + \frac{\sin \alpha}{\alpha} A \\
&= \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}
\end{aligned} exp ( A ) = E + A + 2 ! 1 A 2 + 3 ! 1 A 3 + ⋯ = E + A − 2 ! α 2 E − 3 ! α 2 A + 4 ! α 4 E + 5 ! α 4 A + ⋯ = ( 1 − 2 ! α 2 + 4 ! α 4 − ⋯ ) E + ( 1 − 3 ! α 2 + 5 ! α 4 − ⋯ ) A = ( cos α ) E + α sin α A = [ cos α sin α − sin α cos α ]