I came across a difficult question and was unable to prove it.
Let $Y$ be a 6 dimensional vector space over some field $\mathbb F$ and $T\in L(Y, Y)$.
Suppose there are two vectors $\mathbf{u}$, $\mathbf{v}$ such that $$T^3(\mathbf{u}) = T^3(\mathbf{v}) = \mathbf{0} \quad \quad \textrm{ where } T^2(\mathbf{u}) \textrm{ and } T^2(\mathbf{v}) \textrm{ are linearly independent. } $$
Let $$N_{1} = \text{span}\{\mathbf{u},\mathbf{v}\} \quad \textrm{ and } \quad N_{2} = \text{span}\{T(\mathbf{u}), T(\mathbf{v})\} \quad \textrm{ and } \quad N_{3} = \text{span}\{T^2(\mathbf{u}),T^2(\mathbf{v})\}$$
Prove that $Y = N_{1} ⊕ N_{2} ⊕ N_{3}$
As $Y$ is a $6$ dimensional linear space over $\mathbb F$, it is sufficient to prove that $\{u,v,T(u),T(v),T^2(u),T^2(v)\}$ is linearly independent. Suppose that $a_u,a_v,b_u,b_v,c_u,c_v \in \mathbb F$ are such that
$$a_u u +a_v v + b_u T(u)+ b_v T(v) +c_u T^2(u) + c_v T^2(v)=0.$$
Applying $T^2$ on both sides of this equality, you get $a_u=a_v=0$ according to the hypothesis. Then applying $T$ to the remaining terms, $b_u=b_v=0$. And applying $T$ again, $c_u=c_v=0$.
This allows to conclude to the desired conclusion.