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Let A and B be nxn matrices over C. How to prove $(AB)^{\ast}=B^{\ast}A^{\ast}$? $A^{\ast}$ is complex conjugate transpose( if matrix $A$ is real then $A^{\ast} = A^{t}$. This is homework and is in the area of linear algebra.

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    This is much easier if you move beyond matrices and think in terms of linear transformations. Then the definition of * is that = where is a Hermitian inner product and then the proof is obvious.2011-01-16

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Do you have the theorem that $(AB)^t=B^tA^t$ for real matrices? Can you follow that proof through for the complex case (using conjugates)? Without going back to look, I think it works.

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    $\langle (AB)^\ast z, w\rangle = \langle z, ABw\rangle = \langle A^\ast z,Bw \rangle = \langle B^\ast A^\ast z,w \rangle$2010-11-05
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One simple way to prove is to use outer product form for AB write A as [a1 | a2 | a3 ----| an] and B as [b1 |-------| bn]* and do the operations.