Find a decomposition of a $2\times 2$ diagonal matrix into the sum of two diagonal matrix which satisfy some conditions on the coefficients Let $M=\begin{pmatrix} \alpha&0\\
0&\beta
\end{pmatrix} $ be an arbitrary $2\times 2$ real constant diagonal matrix with $\alpha\neq 0$ and $\beta\neq 0$. I want to find a decomposition of $M$ into the sum of two diagonal matrix $P$ and $N$, such that:
$$
P=\begin{pmatrix} a&0\\
0&b
\end{pmatrix},\,N=\begin{pmatrix} c&0\\
0&d
\end{pmatrix},a>0,b>0,c<0,d<0\mbox{ and }ad=bc.
$$
I know that, without the condition $ad=bc$, such a decomposition exists (take the example
$\begin{pmatrix} 1&0\\
0&-1
\end{pmatrix}=\begin{pmatrix} 2&0\\
0&1
\end{pmatrix}+\begin{pmatrix} -1&0\\
0&-2
\end{pmatrix}$).
I need that decomposition in ordre to guarantee that the two operators $P (\nabla \nabla\cdot)N$ and $N (\nabla \nabla\cdot)P$ commute on the space of infinitely differentiable functions of the form $u:\mathbb R^2\to \mathbb R^2;(x,y)\mapsto (u_1(x,y),u_2(x,y))$.

If  $\alpha$ and $\beta$ are both nonnegative then for any $t > 1$,
$$
\begin{pmatrix}
\alpha & 0 \\
0 & \beta 
\end{pmatrix}
=
\begin{pmatrix}
t\alpha & 0 \\
0 & t\beta 
\end{pmatrix}
+
\begin{pmatrix}
(1-t)\alpha & 0 \\
0 & (1-t)\beta 
\end{pmatrix}.
$$
A similar argument works if both are nonpositive.
It's not hard to show there is no solution for
$$
\begin{pmatrix}
1 & 0 \\
0 &  -1 
\end{pmatrix}.
$$
