Matrices: a normal M, a projection Q, Hermitian transpose of product

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nomadreid
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Homework Statement
Given: orthogonal basis, normal matrix M, Projection P onto k-eigenspace, Q orthogonal complement of P, N*= Hermitian transpose of N. Prove: (QMQ)*=Q(M*)Q
Relevant Equations
M*M=MM*, PP=P, Q=(Id-P), Pv = w implies Mw=kw
Establish QMQ=QM and QM*Q = QM*, reducing the problem to
(QM)*=QM*
((Id-P)M)*=(Id-P)M*
(M-PM)*=M*-PM*
Applying to random vector v (ie. |v>),
(M-PM)*v = M*v-PM*v
Not sure where to go from here, although it is probably something that is supposed to be obvious.
Any help would be appreciated.
 
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This thread can be closed, as I understand how to do it now. Simply split the Hermitian transpose into its two parts as transposing then taking complex conjugate; use (AB)T= BTAT on ((QM)Q) twice, then use (this time using * as simply complex conjugate) (AB)*= A*B* twice. Sorry for the inconvenience; I should have seen this the first time.