Finding the normalizer of $\left\{ \left(\begin{matrix} x &0 \\0 & y \end{matrix}\right) : x,y\in \mathbb R\setminus\{0\} \right\}$ I'm having some trouble with the following question:

Let $G=\text{GL}_2(\mathbb R)$. What are the elements of the set: $$N_G \left( \underbrace{\left\{
\left(\begin{matrix} x &0 \\0 & y \end{matrix}\right) : x,y\in \mathbb R\setminus\{0\} \right\}}_S\right)$$

I tried solving this using the fact that if $A = \left(\begin{matrix} a &c \\b & d \end{matrix}\right) \in N_G(S)$ then: $$ASA^{-1} = S$$
And I arrived at the following conditions:

*

*$xbd=ycd$

*$yac=xba$

*$axd \neq yc^2$

*$yad \neq xb^2$
Where $x,y \in \mathbb R \setminus \{0\}$ are fixed constants.
I'm not being able to find all the values of $(a,b,c,d)$ that follow these conditions and this docent's seem like the best approach to me. Is there a better way to solve this?
 A: If $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]\in GL_2(\Bbb R)$, and if $x,y\in\Bbb R\setminus\{0\}$, then$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x&0\\0&y\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\frac1{ad-bc}\begin{bmatrix}a d x-b c y & a b(y-x) \\ c d(x-y) & a d y-b c x\end{bmatrix},\tag1$$which belongs to $S$ if and only if$$adx-bcy,ady-bcx\ne0\tag2$$and$$ab(y-x),cd(x-y)=0.\tag3$$Actually, we only have to make sure that $(3)$ holds, because if it does, then, since the RHS of $(1)$ is invertible, $(2)$ wil have to hold too.
The only matrices such that $(2)$ holds for every $x$ and every $y$ in $\Bbb R\setminus\{0\}$ are those such that $ab=cd=0$. And, since $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]\in GL_2(\Bbb R)$, this means that $b=c=0$ and $a,d\ne0$ or that $a=d=0$ and $b,c\ne0$. So,$$N_{GL_2(\Bbb R)}(S)=\left\{\begin{bmatrix}a&0\\0&d\end{bmatrix}\,\middle|\,a,d\ne0\right\}\cup\left\{\begin{bmatrix}0&b\\c&0\end{bmatrix}\,\middle|\,b,c\ne0\right\}.$$
A: Let $D = \left\{\begin{pmatrix}x & 0\\0 & y\end{pmatrix} \mid x,y\in\mathbb{R}\setminus\{0\}\right\}$ be the set of your matrices and $N$ its normalizer in the GL. For $A = \begin{pmatrix}x & 0\\0 & y\end{pmatrix}\in D$ and $S\in N$, the matrix $SAS^{-1}$ is similar to $A$ and thus has the same eigenvalues. Moreover, the matrix $SAS^{-1}\in D$ is diagonal, leaving only the possibilities $SAS^{-1} = \begin{pmatrix}x & 0\\0 & y\end{pmatrix}$ or $SAS^{-1} = \begin{pmatrix}y & 0\\0 & x\end{pmatrix}$.
From this observation, the rest can be solved in a straightforward way:

*

*Plug in suitable matrices for $A$ to derive that in the first case, $S = \begin{pmatrix}a & 0\\0 & d\end{pmatrix}$ and in the second case, $S = \begin{pmatrix}0 & b\\c & 0\end{pmatrix}$ with $a,b,c,d\in\mathbb{R}\setminus\{0\}$.

*Verify that all these matrices are indeed contained in $N$.

Therefore,
$$N = \{\begin{pmatrix}a & 0 \\ 0 & d\end{pmatrix} \mid a,d\in\mathbb{R}\setminus\{0\}\} \cup \{\begin{pmatrix}0 & b \\ c & 0\end{pmatrix} \mid b,c\in\mathbb{R}\setminus\{0\}\}.$$
