How to prove $(AB)^{\ast}=B^{\ast}A^{\ast}$?

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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Did you try before asking? – AD. Nov 4 '10 at 14:26
This is much easier if you move beyond matrices and think in terms of linear transformations. Then the definition of * is that <v, Aw> = <A*v, w> where <v, w> is a Hermitian inner product and then the proof is obvious. – Qiaochu Yuan Jan 16 '11 at 14:18

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.
Well I have a hint: There is a direct proof, but it is easier to use the uniqueness result in lemma 1.103, which is following: Let A be an mxn matrix over C. Then $\langle Az, w \rangle = \langle z, A^*w \rangle$ for every $z \in C^n$ and $w \in C^m$. Furthermore $A^*$ is the only matrix with this property. – alvoutila Nov 4 '10 at 13:23
That is a good hint. Why don't you apply it to $(AB)^{\ast}$ and $B^{\ast}A^{\ast}$? – Ross Millikan Nov 4 '10 at 13:27
Uniqueness result is following; to show uniqueness suppose $B=[b_{ij}]$ satisfies $\langle Az, w \rangle = langle z, Bw \rangle$ for all $z \in C^n$ and $w \in C^m$. By formula (1.52)($a_{ij} = \langle Ae_j,e_i \rangle$) – alvoutila Nov 4 '10 at 13:27
$a_{ij} = \langle Ae_j, e_i \rangle = \langle e_j, Be_i \rangle = \upperline{\langle Be_i, e_j \rangle} = \upperline{b_{ij}}$ – alvoutila Nov 4 '10 at 13:29
Yes apply but how? By substituting in the equation $(BA)^{\ast}$? – alvoutila Nov 4 '10 at 13:41