1
$\begingroup$

I want to show that any $X\in SU(2)$ is conjugate to a matrix of the \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix} for $\theta\in \mathbb{R}$.

So I guess I want to find $K\in SU(2)$ st. $X=K\begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}K^{-1}$. I've tried to diagonalize $X$ and taking the $\exp$ function, but it doesn't seem to get me anywhere. Any hint is appriciated.

$\endgroup$
3
$\begingroup$

If $X\in SU(2)$, then its characteristic polynomial is a quadratic polynomial. Therefore, it has some root $\lambda\in\mathbb C$. Let $v$ be a unit vector that is an eigenvector of $X$ with eigenvalue $\lambda$. Let $w$ be an unit vector orthogonal to $v$. Since $X$ is unitary, $w$ is also an eigenvector of $X$; let $\mu$ be its eigenvalue. Then$$1=\det X=\lambda\mu.$$So, $\lambda\neq0$ and $\mu=\frac1\lambda$. Furthermore, since $X$ is unitary, $v$ and $Xv(=\lambda v)$ have the same norm (which is $1$). So, $\lvert\lambda\rvert=1$, which means that $\lambda=e^{i\theta}$ for some $\theta\in\mathbb R$. So, $\mu=\frac1\lambda=\frac1{e^{i\theta}}=e^{-i\theta}$. Can you take it from here?

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.