Unitary matrix problem

[Kolahal Bhattacharya]

Homework Statement

I am to prove that adjoint of(AB)= adjoint of B times adjoint of A

Homework Equations

The Attempt at a Solution

I satrted this as U=adjoint of AB
u_ik=sum(j)[(a_ij)*(b_jk)]
I know then,I may take tarnspose of both sides so that we have:[(u_ki)~]=sum(j){[(b_kj)~][(a_ji)~]}
then [U~]=[B~][A~]
Then,we are done.But,this proves for real elements.I am not sure that this proves also for imaginary elements...
Please help.

 

[nrqed]

Homework Statement

I am to prove that adjoint of(AB)= adjoint of B times adjoint of A

Homework Equations

The Attempt at a Solution

I satrted this as U=adjoint of AB
u_ik=sum(j)[(a_ij)*(b_jk)]
I know then,I may take tarnspose of both sides so that we have:[(u_ki)~]=sum(j){[(b_kj)~][(a_ji)~]}
then [U~]=[B~][A~]
Then,we are done.But,this proves for real elements.I am not sure that this proves also for imaginary elements...
Please help.

The proof works exactlythe same way, you just have to include a complex conjugate of the elements when you take the adjoint. Taking the complex conjugate does not change anything to the indices, so the proof still works...you just have complex conjugates everywhere.

 

[Kolahal Bhattacharya]

OK,I thought of this possibility as I am not using any extra property of complex matrices.
Do I need to write (a_ij)* in those cases?

 

[Dick]

I think you guys are making this too complicated. By definition:

<x,Ay>=<adjoint(A)x,y> for all x,y.

So:

<x,ABy>=<x,A(By)>=<adjoint(A)x,By>=<adjoint(B)*adjoint(A)x,y> for all x,y.