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

DanielFaraday
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Homework Statement



Prove the following:

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

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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?
 
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Ok, start small. Can you prove <br /> (\alpha A)^*=\bar{\alpha }A^*<br />

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

This feels like trying to prove 1+1=2. It just is! I'm still working on it...
 
Do you think this is a sufficient proof?

<br /> (\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<br />
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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