A = [ cos θ − sin θ sin θ cos θ ] A=\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix} A = [ cos θ sin θ − sin θ cos θ ] The characteristic polynomial is
det ( A − λ I ) = 0 \det(A-\lambda I)=0 det ( A − λ I ) = 0
∣ cos θ − λ − sin θ sin θ cos θ − λ ∣ = 0 \begin{vmatrix}
\cos \theta-\lambda & -\sin \theta \\
\sin \theta & \cos \theta-\lambda
\end{vmatrix}=0 ∣ ∣ cos θ − λ sin θ − sin θ cos θ − λ ∣ ∣ = 0
cos 2 θ − 2 λ cos θ + λ 2 + sin 2 θ = 0 \cos^2\theta-2\lambda \cos\theta+\lambda^2+\sin^2\theta=0 cos 2 θ − 2 λ cos θ + λ 2 + sin 2 θ = 0
λ 2 − 2 λ cos θ + 1 = 0 \lambda^2-2\lambda \cos\theta+1=0 λ 2 − 2 λ cos θ + 1 = 0
λ = cos θ ± cos 2 θ − 1 \lambda=\cos\theta\pm\sqrt{\cos^2\theta-1} λ = cos θ ± cos 2 θ − 1
λ = cos θ ± i sin θ = e ± i θ \lambda=\cos\theta\pm i \sin\theta=e^{\pm i\theta} λ = cos θ ± i sin θ = e ± i θ
λ 1 = e i θ = cos θ + i sin θ \lambda_1=e^{i \theta}=\cos\theta+ i \sin\theta λ 1 = e i θ = cos θ + i sin θ
[ − i sin θ − sin θ sin θ − i sin θ ] [ v 1 v 2 ] = 0 \begin{bmatrix}
-i\sin\theta & -\sin \theta \\
\sin \theta & -i\sin \theta
\end{bmatrix}\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=0 [ − i sin θ sin θ − sin θ − i sin θ ] [ v 1 v 2 ] = 0
− i sin θ [ 1 − i i 1 ] [ v 1 v 2 ] = 0 -i\sin\theta\begin{bmatrix}
1 & -i \\
i & 1
\end{bmatrix}\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=0 − i sin θ [ 1 i − i 1 ] [ v 1 v 2 ] = 0
sin θ [ 1 − i 0 0 ] [ v 1 v 2 ] = 0 \sin\theta\begin{bmatrix}
1 & -i \\
0 & 0
\end{bmatrix}\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=0 sin θ [ 1 0 − i 0 ] [ v 1 v 2 ] = 0
[ v 1 v 2 ] = [ i 1 ] , sin θ ≠ 0 \begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=\begin{bmatrix}
i \\
1
\end{bmatrix}, \sin \theta\not=0 [ v 1 v 2 ] = [ i 1 ] , sin θ = 0
λ 2 = e − i θ = cos θ − i sin θ \lambda_2=e^{-i \theta}=\cos\theta- i \sin\theta λ 2 = e − i θ = cos θ − i sin θ
[ i sin θ − sin θ sin θ i sin θ ] [ v 1 v 2 ] = 0 \begin{bmatrix}
i\sin\theta & -\sin \theta \\
\sin \theta & i\sin \theta
\end{bmatrix}\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=0 [ i sin θ sin θ − sin θ i sin θ ] [ v 1 v 2 ] = 0
i sin θ [ 1 i − i 1 ] [ v 1 v 2 ] = 0 i\sin\theta\begin{bmatrix}
1 & i \\
- i & 1
\end{bmatrix}\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=0 i sin θ [ 1 − i i 1 ] [ v 1 v 2 ] = 0
sin θ [ 1 i 0 0 ] [ v 1 v 2 ] = 0 \sin\theta\begin{bmatrix}
1 & i \\
0 & 0
\end{bmatrix}\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=0 sin θ [ 1 0 i 0 ] [ v 1 v 2 ] = 0
[ v 1 v 2 ] = [ − i 1 ] , sin θ ≠ 0 \begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}=\begin{bmatrix}
- i \\
1
\end{bmatrix}, \sin \theta\not=0 [ v 1 v 2 ] = [ − i 1 ] , sin θ = 0
T α 1 = [ cos θ − sin θ sin θ cos θ ] [ i 1 ] = [ i cos θ − sin θ i sin θ + cos θ ] T\alpha_1=\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}\begin{bmatrix}
i \\
1
\end{bmatrix}=\begin{bmatrix}
i\cos \theta-\sin \theta \\
i\sin \theta+\cos \theta
\end{bmatrix} T α 1 = [ cos θ sin θ − sin θ cos θ ] [ i 1 ] = [ i cos θ − sin θ i sin θ + cos θ ]
= ( cos θ + i sin θ ) [ i 1 ] = e i θ α 1 =( \cos \theta+i\sin \theta)\begin{bmatrix}
i\\
1
\end{bmatrix}=e^{i\theta}\alpha_1 = ( cos θ + i sin θ ) [ i 1 ] = e i θ α 1
T α 2 = [ cos θ − sin θ sin θ cos θ ] [ − i 1 ] = [ − i cos θ − sin θ − i sin θ + cos θ ] T\alpha_2=\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}\begin{bmatrix}
- i \\
1
\end{bmatrix}=\begin{bmatrix}
-i\cos \theta-\sin \theta \\
- i\sin \theta+\cos \theta
\end{bmatrix} T α 2 = [ cos θ sin θ − sin θ cos θ ] [ − i 1 ] = [ − i cos θ − sin θ − i sin θ + cos θ ]
= ( cos θ − i sin θ ) [ − i 1 ] = e − i θ α 2 =( \cos \theta-i\sin \theta)\begin{bmatrix}
-i\\
1
\end{bmatrix}=e^{-i\theta}\alpha_2 = ( cos θ − i sin θ ) [ − i 1 ] = e − i θ α 2
Therefore the following two matrices are similar over the field of complex numbers:
[ cos θ − sin θ sin θ cos θ ] and [ e i θ 0 0 e − i θ ] \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix} \text{and} \begin{bmatrix}
e^{i\theta} & 0 \\
0 & e^{-i\theta}
\end{bmatrix} [ cos θ sin θ − sin θ cos θ ] and [ e i θ 0 0 e − i θ ]
#340153 on Dec 2023
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