Olympiad problems

2023 Romania National Olympiad, 3

Let $n$ be a natural number $n \geq 2$ and matrices $A,B \in M_{n}(\mathbb{C}),$ with property $A^2 B = A.$ a) Prove that $(AB - BA)^2 = O_{n}.$ b) Show that for all natural number $k$, $k \leq \frac{n}{2}$ there exist matrices $A,B \in M_{n}(\mathbb{C})$ with property stated in the problem such that $rank(AB - BA) = k.$

1995 IMC, 5

Tags: nilpotent , linear algebra

Let $A$ and $B$ be real $n\times n $ matrices. Assume there exist $n+1$ different real numbers $t_{1},t_{2},\dots,t_{n+1}$ such that the matrices $$C_{i}=A+t_{i}B, \,\, i=1,2,\dots,n+1$$ are nilpotent. Show that both $A$ and $B$ are nilpotent.

2024 VJIMC, 2

Tags: nilpotent , linear algebra , matrix

Let $n$ be a positive integer and let $A$, $B$ be two complex nonsingular $n \times n$ matrices such that \[A^2B-2ABA+BA^2=0.\] Prove that the matrix $AB^{-1}A^{-1}B-I_n$ is nilpotent.

2016 Korea USCM, 8

Tags: linear algebra , nilpotent , commutator , matrix

For a $n\times n$ complex valued matrix $A$, show that the following two conditions are equivalent. (i) There exists a $n\times n$ complex valued matrix $B$ such that $AB-BA=A$. (ii) There exists a positive integer $k$ such that $A^k = O$. ($O$ is the zero matrix.)

ICMC 4, 2

Tags: linear algebra , matrices , nilpotent , matrix

Let $A$ be a square matrix with entries in the field $\mathbb Z / p \mathbb Z$ such that $A^n - I$ is invertible for every positive integer $n$. Prove that there exists a positive integer $m$ such that $A^m = 0$. [i](A matrix having entries in the field $\mathbb Z / p \mathbb Z$ means that two matrices are considered the same if each pair of corresponding entries differ by a multiple of $p$.)[/i] [i]Proposed by Tony Wang[/i]

2009 Romania National Olympiad, 2

Tags: abstract algebra , algebra , ring theory , nilpotent , commutative algebra

[b]a)[/b] Show that the set of nilpotents of a finite, commutative ring, is closed under each of the operations of the ring. [b]b)[/b] Prove that the number of nilpotents of a finite, commutative ring, divides the number of divisors of zero of the ring.