Hi,
If A is an nxn matrix then the ranks of A^n and A^(n+1) are equal.
I need help to prove this assertion.
Thank's in advance.
$\displaystyle 0\subseteq \text{Ker}(A)\subseteq \text{Ker}\left(A^2\right)\subseteq \ldots \subseteq \text{Ker}\left(A^n\right)\subseteq \text{Ker}\left(A^{n+1}\right)$
$\displaystyle \dim \text{Ker}\left(A^j\right)$ can't all be distinct since these are integers between 0 and n
so for some $\displaystyle j$ , $\displaystyle \text{Ker}\left(A^j\right)=\text{Ker}\left(A^{j+1} \right)$
We can prove that
$\displaystyle \ker \left(A^j\right)=\ker \left(A^{j+1}\right)\Longrightarrow \ker \left(A^{j+1}\right)=\ker \left(A^{j+2}\right)$
Let $\displaystyle v\in \ker \left(A^{j+2}\right)$
$\displaystyle A^{j+2}(v)=0\Longrightarrow A(v)\in \ker \left(A^{j+1}\right)=\ker \left(A^j\right)$
$\displaystyle A^j(A(v))=A^{j+1}(v)=0\Longrightarrow v\in \ker \left(A^{j+1}\right)$
Use this and induction to show that $\displaystyle \ker \left(A^n\right)=\ker \left(A^{n+1}\right)$