I am given a matrix inner product on square matrices defined as $\langle A,B\rangle=tr(AB^t)$, where $M^t$ denotes the transpose. I am asked to prove that this is indeed an inner product. We go by 3 definitions for inner product:
$\langle A+B,C\rangle=\langle A,C\rangle+\langle B,C\rangle$
$\langle A,B\rangle=\overline{\langle B,A\rangle}$
$\langle A,A\rangle\geq0$, in particular, $\langle A,A\rangle=0\iff A=0$
I have proven that the defined inner product fits the first and the last definition, but I am having trouble going through the conjugate symmetry proof. This is what I have so far:
$\overline{\langle B,A\rangle}=\overline{tr(BA^t)}=tr(\overline{BA^t})=tr((\overline{A})\overline{(B^t)})^t$
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1$\begingroup$ For the conjugate symmetry proof , the inner product should be defined as $\langle A,B\rangle=tr(AB^*)$, where $B^*$ is the transconjugate of $B$. $\endgroup$– BernardApr 7, 2015 at 17:48
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Hint: for any compatible matrices $A,B$, we have $$ \operatorname{trace}(AB) = \operatorname{trace}(BA) $$
answered Apr 7, 2015 at 17:44
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$\begingroup$ This is because the $ii$-th entry of $AB$, denoted by $(AB)_{ii}$ is given by $(AB)_{ii}=A_i\cdot B^i$. $\endgroup$– EllyaApr 7, 2015 at 17:58
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$\begingroup$ @ellya you should probably clarify the notation you're using (esp $A_i$ for rows/columns). $\endgroup$ Apr 7, 2015 at 18:00
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1$\begingroup$ $A_i$ is the $i$-th row of $A$, and $B^i$ is the $i$-th column of $B$, and $\cdot$ is the standard inner product. $\endgroup$– EllyaApr 7, 2015 at 18:01