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Could someone tell me what I am missing here?

I want to show that for linear maps $L_1, L_2$, $v\in \ker(L_1)+\ker(L_2) \implies (L_1+L_2)v = 0$.

$v\in \ker(L_1)+\ker(L_2)$

$\implies \exists a,b\in V : L_1(a)=0=L_2(b)\wedge v=a+b$

$\implies (L_1+L_2)v = L_1(a)+L_2(a)+L_1(b)+L_2(b)$

$\implies (L_1+L_2)v = L_2(a)+L_1(b)$...

But then...? Thanks.

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Consider $K^2$ , the projections $\pi_1:K^2\longrightarrow V^2$ $\pi_1(x,y)=(x,0)$ and $\pi_2:K^2\longrightarrow K^2$ $\pi_1(x,y)=(0,y)$. Note that $\pi_1+\pi_2=Id_V$ and (1,1)=(1,0)+(0,1) with $(0,1)\in \ker \pi_1$ and $(1,0)\in\ker \pi_2$, son $(1,1)\in\ker\pi_1+\ker\pi_2$ and $(\pi_1+\pi_2)(1,1)=(1,1)\neq(0,0)$

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