Four conditions imply invertibility of a filter - or not? I have an image (which is just three matrices $R,G,B$ in the usual red/green/blue configuration) and four transformations, that is four matrices $A,B,C,D$ which act on the image by means of multiplication. So $A$ for example is both a matrix and a transformation acting on $R$ and returning $AR$.
I've managed to determine the following about the filters:

*

*$AC-BD$ leaves the image unchanged so $AC-BD=I$ the identity matrix.

*$AD+BC$ makes the image flat so $AD+BC=0$ the zero matrix.

*$CA-DB=I$

*$DA+CB=0$
Based on way $AC$ works on the image, I suspect it is almost always invertible with $det(AC)\geq 0$ in every case. Is there a way to prove this using 1) to 4)?

A special case . . .

The claim holds if $A,B$ commute.

Proof:

Assume $A,B,C,D$ are square matrices of the same size with real entries such that the given equations are satisfied, and suppose $A,B$ commute.

We want to show that $\det(AC)\ge 0$.

If $A$ is singular, then $\det(AC)=\det(A)\det(C)=0$, and we're done.

So assume $A$ is nonsingular.

From the equation $AD+BC=0$, we get $D=-A^{-1}BC$.
\begin{align*}
\text{Then}\;\;&
AC-BD=I
\\[4pt]
\implies\;&
AC-B\bigl(-A^{-1}BC\bigr)=I
\\[4pt]
\implies\;&
AC+A^{-1}B^2C=I
\\[4pt]
\implies\;&
A^2C+B^2C=A
\\[4pt]
\implies\;&
(A^2+B^2)C=A
\\[4pt]
\end{align*}
hence $A^2+B^2$ and $C$ are both nonsingular.

Since $C$ is nonsingular, we have $\det(C^2) > 0$.

Since $A,B$ commute, we have $A^2+B^2=(A+iB)(A-iB)$.

Since $A^2+B^2$ is nonsingular, we get that $\det(A+iB),\det(A-iB)$ are nonzero.

Since $A,B$ are real matrices, we get that $\det(A+iB),\det(A-iB)$ are complex conjugates. 

It follows that $\det(A^2+B^2) > 0$.
\begin{align*}
\text{Then}\;\;&
(A^2+B^2)C=A
\\[4pt]
\implies\;&
(A^2+B^2)C^2=AC
\\[4pt]
\implies\;&
\det(AC) > 0
\\[4pt]
\end{align*}
which completes the proof.
