Find a decomposition of a $2\times 2$ diagonal matrix into the sum of two diagonal matrix which satisfy some conditions on the coefficients

Question

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 $u:\mathbb R^2\to \mathbb R^2;(x,y)\mapsto (u_1(x,y),u_2(x,y))$.

Answer 1

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 $$ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$

Answer 2

I'll complement the other answer by showing there is no solution for $(1,-1)$.

Suppose there were, then the matrix formed by $a,b,c,d$ would have det 0, since $a,b,c,d$ are non-zero, the dimension of the row space must be $1$. Thus, $(c,d) = r(a,b)$.

Furthermore, $a+c = (r+1)a = 1$, which implies $a=1/(r+1)$. By similar reasoning, $b = -1/(r+1)$. The signs of $a$ and $b$ are different, so we can't have a solution respecting $a,b > 0$.