I didn't check all your work, but my initial reaction: all four matrices have to be complex matrices, that's the initial condition. If they're not complex, the conjugate transpose is the same as the transpose, and they're just simple symmetric matrices.
Am I wrong?
I just realized something: If AB=(AB)*=B*A*=BA, then AB commutes. But that is only true if A, B, and AB are all Hermitian.
Alternatively, we can say that if A, B are both Hermitian and AB commutes, then AB=BA=B*A*=(AB)*, so we know that AB is Hermitian. That's why the property I cited from...
Sorry about forgetting the brackets. I fixed that.
Okay, (XY)*=Y*X*. So we can rewrite as AD-(CB)*=I. But how does that help?
My main problem is that we don't know that any of the individual matrices A, B, C, D are Hermitian. If we could prove that, we're done. B*C* is the same as BC...
Homework Statement
Let A, B, C, D be nxn complex matrices such that AB and CD are Hermitian, i.e., (AB)*=AB and (CD)*=CD.
Show that AD-B*C*=I implies that DA-BC=I
The symbol * indicates the conjugate transpose of a matrix, i.e., M* is the conjugate transpose of M.
I refers to the identity...