Prove that $A^n=0$ if and only if $A^2=0$

Question

Let $A$ be a $2×2$ matrix and $n$ be an integer exceeding $2$. Show that $A^n=0$ if and only if $A^2=0$ where $0$ denotes null matrix of second order.

Proving the 'if part' is easy.

Given, $A^2=0$

Then, pre-multiplying(or post-multiplying) both sides by $A$ for $n-2$ times consecutively, we get

$A^n=0$.

But I cannot prove the 'only if' part, i.e., given $A^n=0$, how can I show that $A^2=0$?

Please anyone help me solve it. Thanks in advance.

Show that the kernel of $A$ has dimension either $1$ or $2$, and that if $\ker{A^2}$ has dimension $1$, then, for all $p \geq 1$, $A^p$ and $A$ have the same kernel. — Aphelli, Nov 08 '19 at 16:49
https://math.stackexchange.com/questions/3014429/all-nilpotent-matrices-2-times-2-satisfies-a2-0 https://math.stackexchange.com/questions/1200829/all-nilpotent-2-times-2-matrices — Jean-Claude Arbaut, Nov 08 '19 at 16:51

score 3 · Answer 1 · answered Nov 08 '19 at 16:50

3

Since $A$ has an eigenvalue of $0$, the characteristic polynomial is of the form $$x(x-a)$$ for some $a$. Thus $$A^2-aA = 0$$ If $a\neq 0$, then $$A^2=aA$$ And therefore $$A^n=a^{n-1}A$$ hence is never $0$. Thus $A^2=0$.

answered Nov 08 '19 at 16:50

Matt Samuel

58,164

$A^n=a^{n-1}A$ is never $0$ assuming $A\ne 0$, but if $A=0$ then $A^2=0$ and we are done, right? – J. W. Tanner Nov 08 '19 at 17:05
1

@J.W.Tanner Right, if $A=0$ there is nothing to prove. – Matt Samuel Nov 08 '19 at 17:06
Why did you delete your answer here? – Ekhin Taylor R. Wilson Nov 08 '19 at 17:16
1

@Ekhin Well it seemed not to answer your question since it was something you already knew. – Matt Samuel Nov 08 '19 at 17:18
@MattSamuel However the 'no subset homeomorphic to $\mathbb R^n$ is an embedded Lie subgroup' part is not something new, as I was about to further comment on. Are you answering 'no' to both questions? – Ekhin Taylor R. Wilson Nov 08 '19 at 17:21
@EkhinTaylorR.Wilson It can be an embedded Lie subgroup, but it is a subgroup at all only if the connected components of $G$ are all homeomorphic to $\mathbb{R}^n$. – Matt Samuel Nov 08 '19 at 17:22

Prove that $A^n=0$ if and only if $A^2=0$

1 Answers1