Does there exists a vector v such that $Av\neq 0$ but $A^{2}v=0$ Let $A$ be a $4\times4$ matrix over $\mathbb C$ such that $\operatorname{rank}(A)=2$ and $A^{3}=A^{2}\neq0$.
Suppose that $A$ is not diagonalizable.
My question is , "Does there exists a vector $v$ such that $Av\neq0$ but $A^{2}v=0$?"
I think not. Since $\operatorname{rank}A=2$ so $A$ is not invertible. So for any vector $v$ if we find $A^{2}v=0$ then we can not multiply both sides by $A^{-1}$.
But I am not sure about it.
Are there any general rule or any theorem to solve this problem or it can be done only setting example?
 A: The question asks you to show that $A$ is nilpotent on some subspace.
Now, from the assumptions you know that the minimal polynomial divides
$$
t^3-t^2 = t^2(t-1).
$$
Hence $0$ and $1$ are the only possible eigenvalues. 
Moreover, the associated Jordan blocks have maximal dimension $2$ (eigenvalue $0$) and $1$ (eigenvalue $1$). It remains to determine size and count of these blocks.
Since $A$ is not diagonalizable, it must have eigenvalue $0$ with multiplicity $\ge 2$. 
Since $rank(A)=2$, the null space of $A$ is two-dimensional. Since $A$ is not diagonalizable, this implies that multiplicity of $0$ is $\ge 3$. Hence the Jordan form of $A$ contains a Jordan block of size $2$, hence there is a vector $v$ such that $Av\ne0$ but $A^2v=0$.
A: Let $A=\begin{bmatrix} 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \end{bmatrix}$.
Then $A^n \neq 0$ for all $n$, $A e_2 =e_1$, but $A^2 e_2 = 0$. $A$ is not diagonalizable (it is in Jordan form).
Modulo permutations, this is the only Jordan form the matrix can have.
$A^3 =A^2$ means any Jordan block of size $>1$ must have eigenvalue zero and size at most 2.
Since $A$ is not diagonalisable, it must have at least one Jordan block of at size at least 2. If there were two Jordan blocks of size 2, then we would have $A^2 = 0$, hence there is exactly one Jordan block of size 2 and hence the remaining blocks have size 1. One of these remaining blocks must be zero, otherwise the rank would be greater than 2, and the remaining block must be non-zero. Since $A^3 = A^2$, the remaining block has eigenvalue 1.
It follows that there always exists some $v$ such that $Av \neq 0$, but $A^2 v= 0$.
A: Hint: You are looking for an element $v \in ker(A^2) - ker(A)$ or you could also look for an $w \in im(A) \cap ker(A) - 0$. Think about why such an element would yield a solution to your problem.
And $A^{-1}$ is only defined for matrices with full rank.
