By using change of basis formula, prove that for any $\theta \in \mathbb R$, $M$ and $M'$ are similar/conjugate, where $ M\,=\, \left[ {\begin{array}{cc} \text{cos} \theta & -\text{sin} \theta \\ \text{sin} \theta & \text{cos} \theta \\ \end{array} } \right] $ and $M'\,=\, \left[ {\begin{array}{cc} e^{i\theta} & 0 \\ 0 & e^{-i \theta} \\ \end{array} } \right]. $

Here is my attempt:

Let $\theta \in \mathbb{R}$ and $B$ be the standard basis of $\mathbb{C}^2$. Consider the linear transformation given by $$T_\theta: \mathbb{C}^2 \rightarrow \mathbb{C}^2$$ with respect to $B$ and defined by $$T_\theta(e_1)=(cos~\theta, ~sin~ \theta)=cos~\theta.e_1+sin~\theta.e_2$$ and $$T_\theta(e_2)=(-sin~\theta, ~cos~ \theta)=-sin~\theta.e_1+cos~\theta.e_2$$ Then $[T_\theta]_B^B\,=\, \left[ {\begin{array}{cc} \text{cos} \theta & -\text{sin} \theta \\ \text{sin} \theta & \text{cos} \theta \\ \end{array} } \right] $.

Consider $\alpha_1=(1,-i)~ \text{ and } \alpha_2=(-1,-i)$. One can prove that $\{\alpha_1,\alpha_2\}$ forms a basis. Let $$T_\theta(\alpha_1)=T_\theta((1,-i))=T_\theta(e_1-ie_2)= \left(\begin{array}{c} cos~\theta +isin~\theta \\ sin~\theta-icos~\theta \end{array} \right) =(cos~\theta+isin~\theta) \left( \begin{array}{c} 1\\ -i \end{array} \right) $$ Similarly, $$T_\theta(\alpha_2)=(cos~\theta-isin~\theta).\left( \begin{array}{c} -1\\ -i \end{array} \right)$$ Hence $[T_\theta]_{B'}=\, \left[ {\begin{array}{cc} e^{i\theta} & 0 \\ 0 & e^{-i \theta} \\ \end{array} } \right].$

So $M$ and $M'$ are matrix respectively of the same linear transformation with respect to $B$ and $B'$. Hence $M$ and $M'$ are similar.

I would like to know whether the above calculation is correct or not.

  • $\begingroup$ Do you know about eigenvalues and eigenvectors? $\endgroup$
    – amd
    Jun 22 '18 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.