Notes for AKT-140305/0:41:57

Showing that [math]\displaystyle{ (\ker f)^* = V^*/\mathrm{im}f^* }[/math] for a linear map [math]\displaystyle{ f : V \rightarrow W, V\; \mathrm{and}\; W }[/math] are vector spaces. (Leo algknt and Jesse had a discussion.)

Let [math]\displaystyle{ \iota : \ker f \rightarrow V }[/math] be the inclusion of [math]\displaystyle{ \ker f }[/math] into [math]\displaystyle{ V }[/math], then [math]\displaystyle{ V^*/\ker \iota^* \cong (\ker f)^* }[/math].

We show that [math]\displaystyle{ \ker \iota^* = \mathrm{im} f^* }[/math].

[math]\displaystyle{ \begin{align} \ker \iota^* &= \{\phi \in V^* \;\;|\;\; \iota^*(\phi) = 0\}\\ &= \{\phi \in V^* \;\;|\;\; \phi\circ\iota = 0\}\\ &= \{\phi \in V^* \;\;|\;\; \phi|_{\ker f} = 0, \; \phi = f^*(\alpha), \; \alpha \in W^* \}\\ &= \{\phi \in V^* \;\;|\;\; \phi|_{\ker f} = 0, \; \phi = \alpha\circ f, \; \alpha \in W^* \}\\ &= \{f^*(\alpha) \in V^* \;\;|\;\; (\alpha\circ f)|_{\ker f} = 0, \; \alpha \in W^* \}\\ &= \mathrm{im} f^*. \end{align} }[/math]