$A^n\neq0\implies A^k\neq0 \;\forall\;k\in\mathbb{N}$

Question

Let $A$ be an $n\times n$ complex matrix, and suppose that $A^n\neq 0$. Prove that $A^k\neq0\;\forall\;k\in\mathbb{N}$.

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My arguments $\textbf{1}$:

Let $O_{n\times n}$ is the zero matrix.

Since $A^n\neq O_{n\times n}\implies \det(A^n)\neq0\implies (\det(A))^n\neq0 \overset{\det(A)\in\mathbb{C}}{\implies} \det(A)\neq0$. $A\neq O_{n\times n}$ as $\det(O_{n\times n})=0.$ Suppose $A^k=O_{n\times n}$ for some $k\in\mathbb{N}$. $k\neq n$ else this contradicts the statement.

$k<n$ then, $$A^n=A^k\cdot A^{n-k}=O_{n\times n}\;\text{ as}\; A^k=O_{n\times n} \Rightarrow\Leftarrow A^n\neq O_{n\times n}$$

$k>n$ then, $$\begin{align}A^k&=A^n\cdot A^{k-n}\\ &= A^n\cdot A^n\cdot A^{k-2n} \tag{if $k-n>n$}\\ & \quad\qquad\vdots\\ &=A^n\cdot A^n\cdots A^n\cdot A^{k-jn}\tag1\end{align}$$ such that $k-jn\leq n$.

If $k-jn=n$ then $(1)\implies$ $A^k=(A^n)^m$ for some $m\in\mathbb{N}$. $$0=\det(A^k)=\det((A^n)^m)=(\det(A^n))^m\implies \det(A^n)=0\Rightarrow\Leftarrow \det(A^n)\neq0$$

If $k-jn<n$ then, $(1)\implies$ $A^k=(A^n)^m\cdot A^{k-jn}$ for some $m\in\mathbb{N}$.

$$\displaystyle 0=\det(A^k)=\det((A^n)^m\cdot A^{k-jn})=(\det(A^n))^m\cdot (\det(A))^{k-jn}$$ $$\overset{\det(A^n)\neq0}{\implies} (\det(A))^{k-jn}=0\implies \det(A)=0 \Rightarrow\Leftarrow \det(A)\neq0$$

Hence the statement. $\rule{18cm}{0.5pt}$ My arguments $\textbf{2}$:

Since $A^n\neq O_{n\times n}\implies \det(A^n)\neq0\implies (\det(A))^n\neq0 \overset{\det(A)\in\mathbb{C}}{\implies} \det(A)\neq0$. $A\neq O_{n\times n}$ as $\det(O_{n\times n})=0.$ Suppose $A^k=O_{n\times n}$ for some $k\in\mathbb{N}$. $k\neq n$ else this contradicts the statement.

$$\det(A^k)=0\overset{\det(A)\in\mathbb{C}}{\implies} \det(A)=0\Rightarrow\Leftarrow \det(A)\neq0$$

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I feel that my arguments in $\textbf{1}$ are circular while in $\textbf{2}$ are insufficient. Could anyone tell me which parts can be ommitted in $\textbf{1}$ and comment about $\textbf{2}$?

Answer 1

In $\mathbb C$ all matrices are triangularizable. Therefore, there is a basis s.t. $A=D+N$ where $D$ is diagonal and $N$ nilpotent. If $A^n\neq 0$, then $D\neq 0$, and thus $A^k\neq 0$ for all $k$.

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$A\neq 0$ do not implies that $\det(A)\neq 0$. For example, $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and many other matrices...

Answer 2

By contradiction assume that for some $k$ we have $A^k=0$ so the polynomial $x^k$ annihilates $A$ and so the minimal polynomial is $x^p$ for some $p$ and the characteristic polynomial is $x^n$. Using Cayley-Hamilton theorem we get $A^n=0$ which is a contradiction.

Answer 3

Both arguments are wrong at the very start: $A^n\ne0$ does not imply $\det(A^n)\ne0$. (Counterexample: $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$.)

The correct arguments I see all use somewhat more advanced concepts.

Hint: Suppose that $A^k=0$. Show that the minimal polynomial for $A$ must be $m(t)=t^j$ for some $j$. Show that $j\le n$; hence $A^j=0$ implies $A^n=0$.