Answer to Question #214229 in Complex Analysis for SID

Question #214229

Let θ be a real number. Then check whether the matrices 


cos θ − sin θ

sin θ cos θ


and 

e

iθ −0

0 e

−iθ 


are similar over the field of complex numbers.


1
Expert's answer
2021-07-07T10:28:39-0400

A=[cosθsinθsinθcosθ]A=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

The characteristic polynomial is


det(AλI)=0\det(A-\lambda I)=0


cosθλsinθsinθcosθλ=0\begin{vmatrix} \cos \theta-\lambda & -\sin \theta \\ \sin \theta & \cos \theta-\lambda \end{vmatrix}=0

cos2θ2λcosθ+λ2+sin2θ=0\cos^2\theta-2\lambda \cos\theta+\lambda^2+\sin^2\theta=0

λ22λcosθ+1=0\lambda^2-2\lambda \cos\theta+1=0

λ=cosθ±cos2θ1\lambda=\cos\theta\pm\sqrt{\cos^2\theta-1}

λ=cosθ±isinθ=e±iθ\lambda=\cos\theta\pm i \sin\theta=e^{\pm i\theta}



λ1=eiθ=cosθ+isinθ\lambda_1=e^{i \theta}=\cos\theta+ i \sin\theta



[isinθsinθsinθisinθ][v1v2]=0\begin{bmatrix} -i\sin\theta & -\sin \theta \\ \sin \theta & -i\sin \theta \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0

isinθ[1ii1][v1v2]=0-i\sin\theta\begin{bmatrix} 1 & -i \\ i & 1 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0

sinθ[1i00][v1v2]=0\sin\theta\begin{bmatrix} 1 & -i \\ 0 & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0

[v1v2]=[i1],sinθ0\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=\begin{bmatrix} i \\ 1 \end{bmatrix}, \sin \theta\not=0


λ2=eiθ=cosθisinθ\lambda_2=e^{-i \theta}=\cos\theta- i \sin\theta



[isinθsinθsinθisinθ][v1v2]=0\begin{bmatrix} i\sin\theta & -\sin \theta \\ \sin \theta & i\sin \theta \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0

isinθ[1ii1][v1v2]=0i\sin\theta\begin{bmatrix} 1 & i \\ - i & 1 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0

sinθ[1i00][v1v2]=0\sin\theta\begin{bmatrix} 1 & i \\ 0 & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0

[v1v2]=[i1],sinθ0\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=\begin{bmatrix} - i \\ 1 \end{bmatrix}, \sin \theta\not=0




Tα1=[cosθsinθsinθcosθ][i1]=[icosθsinθisinθ+cosθ]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}

=(cosθ+isinθ)[i1]=eiθα1=( \cos \theta+i\sin \theta)\begin{bmatrix} i\\ 1 \end{bmatrix}=e^{i\theta}\alpha_1


Tα2=[cosθsinθsinθcosθ][i1]=[icosθsinθisinθ+cosθ]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}

=(cosθisinθ)[i1]=eiθα2=( \cos \theta-i\sin \theta)\begin{bmatrix} -i\\ 1 \end{bmatrix}=e^{-i\theta}\alpha_2


Therefore the following two matrices are similar over the field of complex numbers:


[cosθsinθsinθcosθ]and[eiθ00eiθ]\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}


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