Prove that if two linear transformations of rank 1 $f,g$ have equal kernels and images, i.e. $\mbox{Ker}f=\mbox{Ker}g$, $\mbox{Im}f=\mbox{Im}g$ then $fg=gf$. Any help would be appreciated, I don't know where to start.
2026-04-06 17:45:50.1775497550
Composition of linear transformations
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If $\mbox{Im}f\subset \mbox{Ker}f$ then $fg=gf=0$. Let $\mbox{Im}f=\langle a\rangle\not\subset \mbox{Ker}f$, then $V=\mbox{Im}f\oplus \mbox{Ker}f$ and $f(a)=\alpha a, g(a)=\beta a$ for some $\alpha, \beta$ from the field.
Every $x\in V$ has a form $x=\gamma a+n$ for some $n\in\mbox{Ker}f$. Hence $fg(x)=f(\beta\gamma a)=\alpha\beta\gamma a=gf(x)$.