# Thread: Rank of a matrix

1. ## Rank of a matrix

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.

2. ## Re: Rank of a matrix

$0\subseteq \text{Ker}(A)\subseteq \text{Ker}\left(A^2\right)\subseteq \ldots \subseteq \text{Ker}\left(A^n\right)\subseteq \text{Ker}\left(A^{n+1}\right)$

$\dim \text{Ker}\left(A^j\right)$ can't all be distinct since these are integers between 0 and n

so for some $j$ , $\text{Ker}\left(A^j\right)=\text{Ker}\left(A^{j+1} \right)$

3. ## Re: Rank of a matrix

Yes,but maybe from now on it start to increase and ker(a^n) is strictly contained in ker(A^(n+1))?

4. ## Re: Rank of a matrix

It is not true that rank$(A^n)$=rank$(A^{n+1})$ for every non-negative integer n. Idea has proved for you that there is some j with rank$(A^j)$=rank$(A^{j+1})$

5. ## Re: Rank of a matrix

We can prove that

$\ker \left(A^j\right)=\ker \left(A^{j+1}\right)\Longrightarrow \ker \left(A^{j+1}\right)=\ker \left(A^{j+2}\right)$

Let $v\in \ker \left(A^{j+2}\right)$

$A^{j+2}(v)=0\Longrightarrow A(v)\in \ker \left(A^{j+1}\right)=\ker \left(A^j\right)$

$A^j(A(v))=A^{j+1}(v)=0\Longrightarrow v\in \ker \left(A^{j+1}\right)$

Use this and induction to show that $\ker \left(A^n\right)=\ker \left(A^{n+1}\right)$