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\begin{equation}L^m=0 \implies KerL^m=V \implies L^m(u)=0 \tag{1}\end{equation}

$$L^{m-1}(u)=v\neq 0 \tag{2}$$ Multiplying both sides of (2) from the left with $L$ we get $L^m(u)=L(v)$, and replacing $L^m(u)=0$ from (1), we get $L(v)=0$ which proves that $v=L^{m-1}(u)\in KerL$

How do I prove $v=L^{m-1}(u)\in ImL$

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2 Answers 2

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Since $L \neq 0$ we have that $m \geq 2$. So we can note $L^{m-1}u = L(L^{m-2}u) \in \mbox{Im}L$

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Since $L\neq 0,$ we have $m\geq 2.$ Now let $z= L^{m-2}(u).$ Then we have $v=L(z).$

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