Nilpotent Matrix And Sequence Properties

Question

The original problem from algebra book asks to prove that if $A$ is a $2\times2$ nilpotent matix then $A^2=0$ . Can this be related with some properties of exact sequences ? I.m. that such matrix is a linear operator with dimension of kernel$=1$ and image$=1$ , then if we look at the sequence $R^2 \rightarrow R^2 \rightarrow R^2 ...$ where all arrows are linear operators , which correspondce to $A$ ,the original statement is about that exactness in last term imply exactness in first term

Answer 1

Remark that if $A$ is nilpotent matrix in $M_{n}(\mathbb{K})$ (where $\mathbb{K}$ is a field) with index of nilpotency $p$,then there is a vector $x$ such that $$x,Ax,A^{2}x\cdots A^{p-1}x$$ are linearly independent, in partcular $p\leq n$ (Since any subset of linearly independent vectors can be completed to a Basis).To prove the first claim, pick a vector $x$ not in $Ker(A^{p-1})$ (which is possible since $p$ is the index of nilpotency, which is defined as the least positive integer such that $A^{k}=0$),and a linear combination $$a_{0}x+a_{1}Ax+a_{2}A^{2}x\cdots +a_{p-1}A^{p-1}x=0$$ For some scalars in $\mathbb{K}$, Apply $A^{p-1}$ to both sides, gives you $a_{0}A^{p-1}(x)=0$,So $a_{0}=0$,then applying $A^{p-2}$ gives you $a_{1}A^{p-1}(x)=0$,hence $a_{0}=0$ , Continuing in this way, u get that all the coefecients are zero, So the vectors $x,Ax,A^{2}x\cdots A^{p-1}x$ are linearly independent.

Answer 2

If $A$ is nilpotent of order $k$ for some $k \ge 1$, then the minimal polynomial for $A$ must divide $x^k$, and its order must be no greater than $2$. Therefore, either $A^2=0$ or $A=0$. In either case $A^2=0$ holds for sure.