Notes for AKT-140305/0:41:57

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

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

We show that $\ker \iota ^{*}=\mathrm {im} f^{*}$.

${\begin{aligned}\ker \iota ^{*}&=\{\phi \in V^{*}\;\;|\;\;\iota ^{*}(\phi )=0\}\\&=\{\phi \in V^{*}\;\;|\;\;\phi \circ \iota =0\}\\&=\{\phi \in V^{*}\;\;|\;\;\phi |_{\ker f}=0,\;\phi =f^{*}(\alpha ),\;\alpha \in W^{*}\}\\&=\{\phi \in V^{*}\;\;|\;\;\phi |_{\ker f}=0,\;\phi =\alpha \circ f,\;\alpha \in W^{*}\}\\&=\{f^{*}(\alpha )\in V^{*}\;\;|\;\;(\alpha \circ f)|_{\ker f}=0,\;\alpha \in W^{*}\}\\&=\mathrm {im} f^{*}.\end{aligned}}$