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So I'm trying to prove whether

$\operatorname*{adjoint}(AB) = \operatorname*{adjoint}(B)\operatorname*{adjoint}(A)$.

Here, for any matrix $C$, the matrix $\operatorname*{adjoint}(C)$ is defined as the complex conjugate of the transposed matrix of $C$.

My tactic is to look at the ijth entry of both matrices and compare, but the fact that AB is being transposed is confusing me.

How should I do this proof? Is there a better way?

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  • $\begingroup$ What's your definition of adjoint? $\endgroup$ – wj32 Apr 30 '13 at 4:31
  • $\begingroup$ er, conjugate and transposed? $\endgroup$ – praks5432 Apr 30 '13 at 4:44
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Let's look at the $(i,j)$th entry of $(AB)^\dagger$ (I use "dagger" for adjoint):

$$ (AB)^\dagger_{i,j}=\overline{(AB)^T_{i,j}}=\overline{\sum_{k} A_{j,k}B_{k,i}}=\overline{\sum_{k}B_{k,i}A_{j,k}}=\sum_{k}\overline{B_{k,i}}\overline{A_{j,k}}=((B)^\dagger(A)^\dagger)_{i,j} $$ To explain, the $(i,j)$th component of $(AB)^T$ will be the $(j,i)$th component of $AB$, which is the ("real") inner product of the $j$th row of $A$ with the $i$th column of $B$. Putting the conjugate over top the sum, and simply re-ordering, we see that this is the same as taking the ("real") inner product of the $i$th column of $B$ with the $j$th row of $A$, in other words, taking the ("real") inner product of the $i$th row of $B^T$ with the $j$th column of $A^T$. The conjugate also drops into the sum.

It's a lot of words but it comes down to this: rows become columns and the $(i,j)$th entry of a product is a (real) inner product of a row with a column. I keep emphasizing "real" inner product to mean:

$$ \sum_{i=1}^na_ib_i $$ instead of the complex inner product which is: $$ \sum_{i=1}^na_i\overline{b_i} $$

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  • $\begingroup$ kind of - I'm having trouble visualizing this though. $\endgroup$ – praks5432 Apr 30 '13 at 5:03

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