Let $A$ be an $n \times n$ matrix for which $A^k = 0$ for some $k > n$. Show that $A^n = 0$.

Question

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Let $A$ be an $n \times n$ matrix for which $A^k=0$ for some $k>n$. Show that $A^n=0$.

Answer 1

Think of $A$ as of a linear function, $A: \mathbb R^n \to \mathbb R^n$. Note a view things:

For all $k: A(A^k(\mathbb R^n)) \subset A^k(\mathbb R^n)$.

If there is $k_0$ St. $A^{k_0+1} (\mathbb R^n) =A^{k_0} (\mathbb R^n) $, then $A^{k_0+l} (\mathbb R^n) =A^{k_0} (\mathbb R^n) $ for all $l \in \mathbb N$.

So you've got a nested sequence $\mathbb R^n = A^0(\mathbb R^n)\supset A^1(\mathbb R^n) \supset A^2(\mathbb R^n) \supset ...$.

$A^k(\mathbb R^n)$ means the image of $\mathbb R^n$ under $A^k$. Each time you apply $A$, the resulting image space can have lower dimension as the space in the previous step, or it can have the same Dimension . If there is a step $k_0$ in which the dimension stays the same, then it never decrease again in all steps that follow.

Answer 2

The only eigen value of $A$ is $0$ and $A$ satisfies its characteristic polynomial. [ This means that if $p$ is the characteristic polynomial then $p(A)=0$. Any polynomial $p$ of degree $n$ with $0$ as the only root is of the type $p(x)=cx^{n}$. Hence $A^{n}=0$].