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.
How to prove $(AB)^{\ast}=B^{\ast}A^{\ast}$?
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linear-algebra
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1Did you try before asking? – 2010-11-04
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0This is much easier if you move beyond matrices and think in terms of linear transformations. Then the definition of * is that
= where – 2011-01-16is a Hermitian inner product and then the proof is obvious.
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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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0Well 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. – 2010-11-04
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0That is a good hint. Why don't you apply it to $(AB)^{\ast}$ and $B^{\ast}A^{\ast}$? – 2010-11-04
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0Uniqueness 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)($a_{ij} = \langle Ae_j,e_i \rangle$) – 2010-11-04
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0$a_{ij} = \langle Ae_j, e_i \rangle = \langle e_j, Be_i \rangle = \upperline{\langle Be_i, e_j \rangle} = \upperline{b_{ij}}$ – 2010-11-04
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0Yes apply but how? By substituting in the equation $(BA)^{\ast}$? – 2010-11-04
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0You work on $\langle B^{\ast}A^{\ast}z,w\rangle$, transferring the matrices to the other side one at a time. Use associativity to say $A*\astz$ is some vector $y$. – 2010-11-04
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0now I have put $\langle B*A*z,w \rangle = \langle A*z,Bw \rangle = \langle z,ABw \rangle. question: how to show that (AB)^{\ast}=B^{\ast} A^{\ast} from what I have achieved so far? – 2010-11-04
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0$\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.