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Given two dual vector spaces $V$ and $V^*$ as well as linear maps $\phi: V \to V$ and $\phi^*: V^* \to V^*$ such that $\phi$ and $\phi^*$ are dual, i.e. there exists a non-degenerate bilinear form such that $$ \langle \phi^*(v^*), v \rangle = \langle v^*, \phi(v) \rangle, \quad \forall v^* \in V^*, \quad \forall v \in V. $$ Also $\phi$ and $\phi^*$ are projection operators, i.e. $\phi(\phi(v)) = \phi(v)$ and $\phi^*(\phi^*(v)) = \phi^*(v)$.

I am trying to work out how to show that $\def\Im{\operatorname{Im}} \Im \phi = (\ker \phi^*)^\perp$.

Since $(\ker \phi^*)^\perp = (\Im \phi)^{\perp\perp}$ this is equivalent to showing that $\Im \phi = (\Im \phi)^{\perp\perp}$.

It follows from the definition of the orthogonal complement that $\Im \phi \subseteq (\Im \phi)^{\perp\perp}$, so now I am trying to show that $\Im \phi \supseteq (\Im \phi)^{\perp\perp}$.

Given $v \in (\Im\phi)^{\perp\perp}$ how can I show that $v \in \Im\phi$?

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  • $\begingroup$ It's not clear what you're asking. Are you trying to prove that $\operatorname{Im} \phi = (\ker \phi^*)^\perp$ or the double perp identity? $\endgroup$ – Sammy Black Oct 26 '15 at 20:42
  • $\begingroup$ They are the same thing for dual maps. $(\ker \phi^*)^\perp = (\text{Im } \phi)^{\perp\perp}$ $\endgroup$ – Anfänger Oct 26 '15 at 21:59

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