# Show every $X\in SU(2)$ is conjugate to a matrix $\begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}$

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.

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?