# Dual operator relationship with complex conjugate.

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 $A$?

• – wj32 Nov 9 '12 at 23:33
• So in part (2) where they explain it, is it just the regular transpose that corresponds to $T^{*}$, not the complex conjugate? – roo Nov 9 '12 at 23:35
• I've heard of "dual space" and "dual basis" but not "dual operator". Is there any link to it or short explanation? – DonAntonio Nov 9 '12 at 23:36
• @DonAntonio: Usually we define it as $(T^*f)(v)=fTv$. – wj32 Nov 9 '12 at 23:38
• @lovinglifein2012: Yes. The difference originates from the fact that the inner product has conjugate symmetry and Riesz representation theorem gives us an anti-isomorphism. – wj32 Nov 9 '12 at 23:40