$A^{2014}=0$ for a matrix A Let A be a 3*3 matrix and $A^{2014}=0$. Must $A^3$ be the zero matrix? I can work out that I-A is invertible, but I don't know how to proceed further.
 A: This is very much overkill for this problem (Wojowu's suggestion in the comments is a very nice and clean solution).  This answer classifies the possible $A$'s.
Since $A^{2014}=0$, all of the eigenvalues of $A$ must be $0$ (for if $\lambda\not=0$ were an eigenvalue with eigenvector $v$, then $A^{2014}v=\lambda^{2014}v\not=0$).  Considering the Jordan form of $A$, there are three possibilities, there are three independent blocks, two blocks, or one block:
$$ 
\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.
$$
Since $A=PJP^{-1}$, we know that all matrices of the desired form are similar to one of these three matrices.  We see that $A^3=PJ^3P^{-1}$, but $J^3=0$ for all of these matrices, so $A^3=0$.
A: Hint: Let $R(A)$ denote the range (AKA image or column space) of $A$. If $A^k=0$ for some $k$ (and if $A^2 \neq 0$), we must have
$$
R(A^3) \subsetneq R(A^2) \subsetneq R(A) \subsetneq \Bbb R^3
$$
A: Hint: Note that $\operatorname{rank}(A^{n+1})\le\operatorname{rank}(A^n)$. Then show that if $\operatorname{rank}(A^{n+1})=\operatorname{rank}(A^n)$, then all powers $A^k$ with $k\ge n$ have the same rank.
