Let $n$ be an integer with $n \geq 2$ and let $A \in \mathcal{M}_n(\mathbb{C})$ such that $\operatorname{rank} A \neq \operatorname{rank} A^2.$ Prove that there exists a nonzero matrix $B \in \mathcal{M}_n(\mathbb{C})$ such that $$AB=BA=B^2=0$$ Cornel Delasava
Problem
Source: Romania NMO - 2018
Tags: linear algebra, Matrices, rank, matrix
01.05.2018 13:22
Since $rank(A)>rank(A^2)$, there is $k\in[[2, n]]$ and an invertible $P$ s.t. $P^{-1}AP=diag(J_k,U_{n-k})$ where $J_k$ is the nilpotent Jordan block of dimension $k$. Thus we choose $B=Pdiag({J_k}^{k-1},0_{n-k})P^{-1}$.
26.06.2018 21:05
The result remain the same if we change the hypothesis with $\operatorname{rank} A^k \neq \operatorname{rank} A^p.$ , for some different natural k,p.
28.01.2019 04:23
i cant understand why from $rank(A)>rank(A^2)$ we get there is $k\in[[2, n]]$ and an invertible $P$ s.t. $P^{-1}AP=diag(J_k,U_{n-k})$
29.01.2019 21:45
Since $A$ is not invertible, in the Jordan form of $A$, there are nilpotent jordan blocks. If all these blocks were $[0]$-shaped, then $rank(A)=rank(A^2)$; then at least one block has dimension $\geq 2$.