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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?

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    $\begingroup$ The normlizer of a single matrix is the same as its centralizer. Is that what you meant to ask? $\endgroup$ – Derek Holt May 21 at 7:35
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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 $\{I\}$.

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    $\begingroup$ Isn't this the centraliser rather than the normaliser? $\endgroup$ – verret May 21 at 1:54
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    $\begingroup$ Yes and no. The question is not clear, my understanding was that it is about a single matrix. After seeing your question maybe OP meant the set of all such matrices. $\endgroup$ – Martin Argerami May 21 at 2:42

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