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I am having some confusion over how I would attack a proof of the properties of matrix adjoints. Here is an example:

Let $A,B\in M_{n\times n}(\mathbb{F})$ with adjoints $A^*$ and $B^*$. Prove $(A+B)^*=A^*+B^*$.

Now I know what I would do for the similar problem:

Let $S\colon V\to V$ and $T\colon V\to V$ be linear operators, and $V$ has inner product $\langle\cdot,\cdot\rangle$. Prove $(S+T)^*=S^*+T^*$.

This would be done by:

\begin{align} \langle(S+T)^*(x),y\rangle& =\langle x,(S+T)(y)\rangle \\ & =\langle x,S(y)+T(y)\rangle \\ & =\langle x,S(y)\rangle+\langle x,T(y)\rangle \\ & =\langle S^*(x),y\rangle+\langle T^*(x),y\rangle \\ & =\langle S^*(x)+T^*(x),y\rangle \\ & =\langle(S^*+T^*)(x),y\rangle. \end{align}

I am confused though, as to how the first proof (with the matrix adjoint) would be done differently than when it is done with the linear operator adjoint.

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  • $\begingroup$ I'm assuming you use $\Bbb F$ to denote an arbitrary field, in which case, how do we define the "conjugate" for an element $x\in\Bbb F$ ? $\endgroup$ – learner Nov 23 '15 at 23:28
  • $\begingroup$ $\mathbb{F}$ in this case is just $\mathbb{R}$ or $\mathbb{C}$ $\endgroup$ – user3784030 Nov 24 '15 at 0:24
  • $\begingroup$ You should've mentioned that explicitly then. In that case, I'd have to go with uniquesolution's answer. The simplest way to prove this is to just consider the $ij^{\textrm{th}}$ element of both $(A+B)^\ast$ and $A^\ast+B^\ast$. You can easily verify they're the same that way. $\endgroup$ – learner Nov 24 '15 at 8:23
  • $\begingroup$ Just note that if $A=(a_{ij})$ and $B=(b_{ij})$, then $(A+B)^\ast=(\overline{a_{ji}+b_{ji}})$ and $A^\ast+B^\ast=(\overline{a_{ji}}+\overline{b_{ji}})$. All you have to show now is that $\overline{a_{ji}+b_{ji}}=\overline{a_{ji}}+\overline{b_{ji}}~\forall~(i,j)\in X^2$ where $X=\{1,2,\ldots,n\}$. This is trivial since $\overline{x+y}=\overline{x}+\overline{y}~\forall~x,y\in\Bbb C$. $$~$$ To prove the above-mentioned identity, take $x=\alpha+i\beta$ and $y=\gamma+i\delta$ where $\alpha,\beta,\gamma,\delta\in\Bbb R$ and $i$ denotes the imaginary unit in this brief proof. $\endgroup$ – learner Nov 24 '15 at 8:42
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It seems that for matrices the easiest procedure is to write down the $(i,j)$'th entry of $(A+B)^*$ and convince oneself that it is equal to the $(i,j$)'th entry of $A^*+B^*$.

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