Let $A$ be a nilpotent matrix and suppose $A^6 = 0$ but $A^5 \neq 0$. Further, suppose the vector spaces $\mathcal V_1, \ldots, \mathcal V_6$ satisfy $$\ker(A^{i}) = \mathcal V_1 \oplus \mathcal V_2 \oplus \cdots \oplus \mathcal V_i.$$ Is it possible that $\dim(\mathcal V_3) = 0$?
I think that maybe we have to show $\mathcal V_3$ contains only zero vector or $\ker(A^2) = \ker(A^3)$ or something else, but don't know how to prove it. Can anyone give some hints?
Let $V$ be a vector space, not necessarily finite dimensional, and $T : V \to V$ a linear operator. Then $\ker(T^{k}) \subseteq \ker(T^{k+1})$ for all $k \ge 0$. Suppose that for some $k$, one has equality, $\ker(T^{k}) = \ker(T^{k+1})$. I claim that $\ker(T^l) = \ker(T^k) $ for all $l \ge k+1$. One proves this by induction as follows. The base case $l = k+1$ holds by hypothesis. Suppose for a particular $l \ge k+1$ one has $\ker(T^l) = \ker(T^k) $. Let $v \in \ker(T^{l + 1})$. Then $T v \in \ker(T^l) = \ker(T^k)$, and thus $0 = T^k(T v) = T^{k+1} v$, so $v \in \ker(T^{k+1}) = \ker(T^k)$. This shows $\ker(T^{l+1}) \subseteq \ker(T^k)$. The opposite containment is obvious, so $\ker(T^{l+1}) = \ker(T^k)$. This completes the induction.
In the specific problem, if $V_3 = 0$, then $\ker(A^3) = \ker(A^2)$. Hence $\ker(A^l) = \ker(A^2)$ for all $l \ge 3$. In particular, $\ker(A^6) = \ker(A^5) = \ker(A^2)$, contradicting the hypothesis.