Show Hermitian Identity: (AB)^+ = A^+ B^+

[evlyn]

Homework Statement

Show that (AB)^+ = A^+ B^+ using index notation

Homework Equations

+ is the Hermitian transpose

The Attempt at a Solution

I know that AB = Ʃa_ik b_kj summed over k

so (AB)^+ = (Ʃa_ik b_kj)^+ = Ʃ (a_ik b_kj)^+ = Ʃ (a_ik)^+(b_kj)^+ = A^+ B^+

I am not really sure if this makes sense, I don't know if it is acceptable to distribute the transpose within the sum.

 

[Dick]

(AB)^+ is equal to (B^+)(A^+), not (A^+)(B^+). That may be a sign something is going wrong. (A^+)_ij=(A_ji)*, where * is complex conjugate. Start from there.

 

[evlyn]

That helped.

So now using the definition (usually a good thing) I have:

(AB)^+_ij = [AB_ji]^* = A_ji ^* B_ji^* = B_ij^+ A_ij^+ = B^+ A^+

I know that the complex conjugate is distributive can I just assume that for the proof?

 

[Dick]

That helped.

So now using the definition (usually a good thing) I have:

(AB)^+_ij = [AB_ji]^* = A_ji ^* B_ji^* = B_ij^+ A_ij^+ = B^+ A^+

I know that the complex conjugate is distributive can I just assume that for the proof?

Yes, you can use (xy)^*=(x^*)(y^*). That's fine. But now you've lost the matrix product part. AB is a product. (AB)_ij isn't equal to A_ij B_ij.