If A is a Hermitian matrix, and B is a complex matrix of same order as A, then $\displaystyle BAB^* $ is also a Hermitian matrix.
Anyone know how to prove this other than just writing it all out by hand?
Definition: A is Hermitian iff A* = A where A* = (A bar)^T
Remember the following facts
1. (AB)* = B* A*
2. (A*)* = A
Need to show that (BAB*)* = BAB*
By 1. we see that (BAB*)* = (B*)*(A*)(B*)
By 2. we see that (B*)*(A*)(B*) = B(A*)(B*) since (B*)* = B
Since A is Hermitian, A* = A. Thus, B(A*)(B*) = BAB*.
Therefore,
(BAB*)* = (B*)*(A*)(B*) = B(A*)(B*) = BAB*. Thus, BAB* is Hermitian.