Is (\alpha AB)^* Equal to \bar{\alpha }B^*A^* for All n × n Complex Matrices?

[DanielFaraday]

Homework Statement

Prove the following:

For every n × n complex matrices A and B, $$(\alpha AB)^*=\bar{\alpha }B^*A^*$$.

Homework Equations

None

The Attempt at a Solution

Okay, I'm just getting started on this problem. All the ideas I have come up with so far involve using two "test" matrices. The problem with this is that it doesn't prove it for any n × n matrix. Does this matter?

 

[Office_Shredder]

Ok, start small. Can you prove $$(\alpha A)^*=\bar{\alpha }A^*$$

If you can, then you can just focus on proving that A and B swap like that

 

[DanielFaraday]

Hmm...

This feels like trying to prove 1+1=2. It just is! I'm still working on it...

 

[DanielFaraday]

Do you think this is a sufficient proof?

$$(\alpha AB)^*=\bar{\alpha }\overline{AB}=\bar{\alpha }\left(\bar{A}\right)\left(\bar{B}\right)=\bar{\alpha }\left(\left(\bar{A}\right)^T\right)^T\left(\left(\bar{B}\right)^T\right)^T=\bar{\alpha }\left(A^*\right)^T\left(B^*\right)^T=\bar{\alpha }\left(B^*A^*\right)^T$$