Can there exist a set of 4 square complex matrices $A,B,C,D$ where at least 2 matrices of the four are non-zero, such that for some NxN the following 6 equations are obeyed:
$$A^*D+C^TC^\dagger = 0 \tag{1}$$ $$B^*B+A^TD^\dagger = 0 \tag{2}$$ $$A^*B+A^TC^\dagger = 0 \tag{3}$$ $$A^*D+B^*C^\dagger = 0 \tag{4}$$ $$A^TD+C^TB = 0 \tag{5}$$ $$B^*D+C^TD^\dagger = 0 \tag{6}$$
Where the superscripts $T$ gives the transpose of the matrix, $ * $ turns all elements of the matrix into their complex conjugate and $\dagger$ gives the hermitian adjoint of the matrix which is equivalent to action of both $T$ and $*$ in any order.
I've tried very hard to solve them using gamma matrices in 4x4 but nothing works out. I expect it's impossible. I don't know anything about higher dimensional gamma matices, which representations exist and how they behave under $*,T,\dagger$ perhaps you someone can offer me some helpful insight.
Assume that $A,B,C,D$ are real and symmetric, so that $$A=A^*=A^T=A^\dagger$$ and the same for $B,C,D$. Hence, the equations become
$$AD+C^2 = 0 \tag{1}$$ $$B^2+AD = 0 \tag{2}$$ $$A(B+C) = 0 \tag{3}$$ $$AD+BC = 0 \tag{4}$$ $$AD+CB = 0 \tag{5}$$ $$(B+C)D = 0 \tag{6}$$
Now, take $C=-B$: Equation $3$ and $6$ are satisfied, Equation $1$ and $2$ become the same, and also Equation $4$ and $5$, for any choice of $A,B,D$: $$AD+B^2 = 0 \tag{1}$$ $$B^2+AD = 0 \tag{2}$$ $$0 = 0 \tag{3}$$ $$AD-B^2 = 0 \tag{4}$$ $$AD-B^2 = 0 \tag{5}$$ $$0 = 0 \tag{6}$$
The problem is now easy to solve: it suffices to take $A,D$ such that $AD=0$, for example the diagonals matrices $$A=\begin{pmatrix}1 & \\ & 0_{N-1} \end{pmatrix}\qquad D=\begin{pmatrix} 0 & \\ & 1_{N-1} \end{pmatrix}$$ and $B=0$.