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Let $U,V,W$ be vector spaces over a field $K$ and $\varphi: U\to V,$ $\psi: V\to W$ linear maps. Given that $\text{ker}(\psi)=\text{im}(\varphi)$ show that $\ker(\varphi^{tr})=\operatorname{im}(\psi^{tr})$

I tried proving this by double inclusion. So far I have:

$"\subseteq"$

Let $x \in \text{im}(\psi^{tr}),$ then I know there exists $\alpha \in W^*$ such that $\alpha \circ \psi=x$ and $\varphi^{tr}(\alpha \circ \psi)=\alpha \circ \psi \circ \varphi$ is the zero mapping as $\ker(\psi)=\operatorname{im}(\varphi)$.

Can someone help me with the other inclusion?

Anne Bauval
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Mathbds
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  • Are you familiar with the relationships between kernels and images of a linear map and its dual / transpose? – blargoner Sep 01 '23 at 20:09
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    $\ker(\varphi^{tr})={f\in V^\mid f(\operatorname{im}\varphi)=0},$ and $\operatorname{im}(\psi^{tr})={f\in V^\mid f(\ker\psi)=0}.$ – Anne Bauval Sep 01 '23 at 20:21

1 Answers1

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If $\alpha \in \ker(\varphi^{tr})$, then $\alpha\circ\varphi$ is the zero map. By $\ker(\psi)=Im(\varphi)$, it follows that $\alpha|_{\ker(\psi)}=0$, it means that $\alpha \in (\ker \psi)^0$ (the annihilator of $\psi$) from This Question it follows that $\alpha \in Im(\psi^{tr})$.

Ygor Arthur
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