I have the problem of calculating the normalizer of $\begin{bmatrix} \lambda & 0 \\ 0 & \lambda^{- 1} \end{bmatrix} $ in the group $\begin{bmatrix} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{bmatrix} $
this, to solve the problem of:
For all $ g \in SL_2 (\mathbb {R}) $ there is a decomposition $ g = \begin{bmatrix} \cos (\theta_1) & -\sin (\theta_1) \\ \sin (\theta_1) & \cos (\theta_1) \end{bmatrix} \begin{bmatrix} \lambda & 0 \\ 0 & \lambda^{- 1} \end{bmatrix} \begin{bmatrix} \cos (\theta_2) & -\sin (\theta_2) \\ \sin (\theta_2) & \cos (\theta_2) \end{bmatrix} $ which is unique except conjugation by elements of the normalizer.
I know that the normalizer $ N $ is generated by $ N = \langle \begin {bmatrix} 0 & 1 \\ -1 & 0 \end {bmatrix} \rangle $
I was able to show $ \langle \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \rangle \subseteq N $
Does anyone have an idea of how to show the other contention?
You can do this directly by definition. Write $R(\theta)$ for your rotation matrix. You need $$ R(\theta)\begin{bmatrix} \lambda&0\\0&\lambda^{-1}\end{bmatrix}=\begin{bmatrix}\lambda&0\\0&\lambda^{-1}\end{bmatrix} R(\theta). $$ This is $$ \begin{bmatrix} \lambda \cos (\theta) & -\lambda^{-1}\sin (\theta) \\ \lambda\sin (\theta) & \lambda^{-1}\cos (\theta) \end{bmatrix}=\begin{bmatrix} \lambda\cos (\theta) & -\lambda\sin (\theta) \\ \lambda^{-1}\sin (\theta) & \lambda^{-1}\cos (\theta) \end{bmatrix}. $$ The diagonal entries give you nothing, they are always equal. The non-diagonal entries give you the equality $$\tag1 \lambda\sin\theta=\lambda^{-1}\sin\theta. $$ So, if $\lambda=\pm1$, then any $\theta $ works and the normalizer consists of all matrices in the group. When $\lambda^2\ne1$, the equality in $(1)$ forces $\sin\theta=0$, so $\theta=0$ and the normalizer is $\{I\}$.