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Thread: Rank of a matrix

  1. #1
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    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.
    Thank's in advance.
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  2. #2
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    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)
    Last edited by Idea; Feb 27th 2017 at 11:54 AM.
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  3. #3
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    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))?
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  4. #4
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    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$(A^{j+1})$
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  5. #5
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    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)
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