<s>[INST] Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$ Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $rank A=2$ and $A^3=A^2\neq 0$.Suppose that $A$ is not diagonalisable. Then 
Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$
My try:$\dim \operatorname{Im}(A)+\dim \ker (A)=4$ so $\dim(\ker A)=\dim (\operatorname{Im}A)=2$.
Now $A$ satisfies $x^3=x^2\implies x^2(x-1)=0$ thus $0,1$ are the only eigen values of $A$. Since $A$ has rank $2$ so geometric multiplicity of $0$ is $2$ but $A$ is not diagonalizable thus algebraic multiplicity of $0$ is $3$.
So the characteristic polynomial will be $x^4-x^3=0$
Obviously $A$ will have a Jordan block of size $2$ corresponding to 0
$$\   A  \text{ will have Jordan form as } \pmatrix{1&0&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0}  
$$
How should I use these  information to conclude my result??
 [/INST] There are only two possible Jordan forms: 
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1}, \quad \pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&1\\0&0&0&1}$$ only the first first one satisfies $A^2 = A^3.$
$\bf p.s.$
the nonzero vector $v$ you are looking for is in $\ker(A^2)\setminus \ker(A)$ we have $dim(\ker(A^2) = 3,dim(\ker(A) = 2.$
step 1.  find a basis of $3$ vector for $ker(A^2)$ by row reducing $A$ if you must.
step 2. find the vector $v$ in the basis so that $Av \neq 0.$
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