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Let $V$ be a $n$ dimensional vector space spanned by $\{e_{i}\}_{i=1}^{n}$.

Let $T:V\to V$ be a linear operator with matrix transformation $A$. Is there any relationship between the dual operator $T^{*}:V^{*}\to V^{*}$, and the complex conjugate $A^{*}$ of $A$?

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  • $\begingroup$ See math.stackexchange.com/questions/58146/… $\endgroup$ – wj32 Nov 9 '12 at 23:33
  • $\begingroup$ So in part (2) where they explain it, is it just the regular transpose that corresponds to $T^{*}$, not the complex conjugate? $\endgroup$ – roo Nov 9 '12 at 23:35
  • $\begingroup$ I've heard of "dual space" and "dual basis" but not "dual operator". Is there any link to it or short explanation? $\endgroup$ – DonAntonio Nov 9 '12 at 23:36
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    $\begingroup$ @DonAntonio: Usually we define it as $(T^*f)(v)=fTv$. $\endgroup$ – wj32 Nov 9 '12 at 23:38
  • $\begingroup$ @lovinglifein2012: Yes. The difference originates from the fact that the inner product has conjugate symmetry and Riesz representation theorem gives us an anti-isomorphism. $\endgroup$ – wj32 Nov 9 '12 at 23:40
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See Stone and Goldbart, Mathematics for Physics, page 753 for a brief explanation. The dual operator is linear, so it does not have a simple relationship to the complex conjugate of an operator.

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