I'm stuck in this question and pretty lost here.
Question : if $A$ is $4\times 4$ nilpotent matrix such that $\ker A^2 \ne \ker A^3$. Prove or disprove the following :
a) Jordan form of $A$ has only one block
b) Jordan form of $A$ has a block with dimension $\ge 3\times3$
I'm not looking for answers, I want a hint/explanation which information $\ker A^2 \ne \ker A^3$ gives us.
by definition it means that exist $ u \in \mathbb{R^4}$ such that $u \in \ker A^3 \wedge u \notin \ker A^2$ (Because $\ker A^2 ⊆ \ker A^3$)
But I still don't know how to proceed... Thanks in advance
Not sure if I understand it correctly but it can be both.
a) $J$ can have just one block: $$J = \pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0} $$ then if $u=(1,1,1,0)$ we have $A^3u =0$ but $A^2u\ne 0$.
b) $J$ can have block $3\times 3$: $$J = \pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0} $$ then if $u= (0,0,1,0)$ we have $A^3u=0$ and $A^2u \ne 0$.