Nilpotent matrix problem

Question

For an $n\times n$ matrix $A$ we have $A^N=0$ for some $N$. Prove that $A^n=0$.

If $ N \le n$ then the solution is trivial, but what if $N \gt n$ ?

Answer 1

Minimal polynomial of $A$ divides $x^N$. So eigen-values of $A$ are only $0$. So the characteristic polynomial of $A$ is $x^n$. By Cayley-Hamilton Theorem we have, $A^n=0$.

Answer 2

Let $A^n\not=0$, choose $m\in \{n+1,...,N\}$ for which $A^m=0$ but, $A^{m-1}\not=0$. Choose $v\in \Bbb F^n$ with $A^nv\not=0$. Then, $\{Av,....,A^nv\}$ forms a basis of $\Bbb F^n$ : $$\lambda_1Av+...+\lambda_nA^nv=0$$$$\implies A^{m-n+j}(\lambda_1Av+...+\lambda_nA^nv)=0,\forall j=n-1,...,0$$$$\implies\lambda_1=0,...,\lambda_n=0.$$

Now note that, each of the vector $Av,...,A^nv$ are eigenvectors of $A^{m-1}$ with eigenvalues $0$. So eigen space of $A^{m-1}$ corresponding to the eigen value $0$ is $\Bbb F^n$.

That is $A^{m-1}w=0,\forall w\in \Bbb F^n=\text{span}\{ Av,...,A^nv\}$. So, $A^{m-1}=0$, a contradiction to our earlier assumption.

So for all $m\in \{n+1,...,N\}$ for which $A^m=0$ implies $A^{m-1}=0$. Hence inductively we have, $A^n=0$

Answer 3

Since $t^N$ is the annihilating polynomial, so the possible choices of minimal polynomial is any one of the $t,t^2, t^3,...,t^n$. For any choice $t^n=0$ always hold.