$A$ is a complex $n \times n$ matrix with $A^n \neq 0$ then $A^k \neq 0$ $\forall k \in \Bbb{N}$

Question

I was trying this question, $A$ be a complex $n \times n$ matrix and suppose that $A^n \neq 0$ then I have to prove that $A^k \neq 0$ for all $k \in \Bbb{N}$.

I was thinking of taking determinants so that I get $[det(A)]^n \neq 0$ meaning that $det(A) \neq 0$ or that $A$ is non-singular but would that mean $A^k \neq 0 \forall k \in \Bbb{N}$?.

How to proceed with this?

Answer 1

If $A^{k}=0$ for some $k$ then $0$ is the only eigen value since $Ax=\lambda x$ implies $A^{k}x=(\lambda )^{k} x$. Every square matrix satisfies its characteristic equation and the characteristic polynomial is $ (c\lambda )^{n}$ for some constant $c$ because 0 is the only root of this equation. Hence $A^{n}=0$.

Answer 2

If $A^k=0$ for some $k$, then the minimum polynomial of $A$ must be $x^r$ for some $r$. But the minimum polynomial has degree $\le n$, so $r\le n$. Then $A^r=0$ etc.

Answer 3

Let $A^{k} = 0$ for any positive integer k. Then $f(x) = x^{k}$ is an anhilating polynomial of $A$ i.e. $A^{k} = 0$ so minimal polynomial $m(A)$ of $A$ devides $x^{k}$, (because minimal polynomial devides each anihilating polynomial). So characteristic polynomial $\Delta(A)$ of $A$ must be $x^{n}$. (Since $m(A)$ and $\Delta(A)$ have same roots (eigenvalues of $A$) i.e. $A^{n} = 0$, which is a contradiction. Hope this help. :)

Answer 4

As we see from the other answers, this is trivial if we know that the minimal polynomial has degree no larger than $n$, which follows for example from the Cayley-Hamilton theorem, or from the Jordan canonical form (after passing too the algebraic closure of the underlying field).

But C-H is not at all trivial, and the JCF is if anything less trivial, since it implies C-H. As it happens I was led to exactly this question the other day, thinking about how one might prove that the minimal polynomial has degree less than or equal to $n$. Here's a proof that uses nothing at all.

Wrting $Z(A)$ for the nullspace of $A$ to make things easy to type:

Lemma. If $Z(A^{k+1})=Z(A^k)$ then $Z(A^{k+j})=Z(A^k)$ for $j=1,2\dots$.

Proof: By induction on $j$. Suppose that $Z(A^{k+j})=Z(A^k)$ and assume $x\in Z(A^{k+j+1})$. Then $Ax\in Z(A^{k+j})=Z(A^k)$, so $x\in Z(A^{k+1})=Z(A^k)$. QED.

Theorem. If $A^k=0$ then $A^n=0$.

Proof: Let $d_j=\dim(Z(A^j))$. Then $(d_j)$ is a non-decreasing sequence, and the lemma shows that if two consecutive $d_j$ are equal then the sequence is constant from that point on. That is, there exists $N$ such that$$0=d_0<d_1<\dots<d_N=d_{N+1}=\dots.$$Since $A^k=0$, $d_k=n$; hence we must have $d_N=n$. So $N\le n$ (because $d_j\ge j$ for all $j\le N$.)

So $A^N=0$ implies $A^n=0$.