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Let $A$ be a real matrix dimension $n \times n$ with rank $m What are conditions for this matrix that for any natural $k$: $\text{rank}(A)=\text{rank}(A^k)$?

Is it sufficient, for example, that $\text{rank}(A)=\text{rank}(A^2)$ ?

2 Answers 2

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Yes, it is sufficient. Considering $A$ as a linear transformation, the ranges $\text{Ran}(A^2) \subseteq \text{Ran}(A)$. If the ranks are equal, the subspaces are equal. Thus $A$ maps $\text{Ran}(A)$ onto itself, and then $A^k$ must do the same for any positive integer $k$.

Another equivalent condition is that the kernel and range of $A$ intersect only at $0$, i.e. there is no vector $v$ such that $Av \ne 0$ but $A^2 v = 0$.

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    Hmm, why $A^k $ must do the same? I'm not sure of this ...Maybe they are also other equivalent conditions ? – 2017-02-02
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    Every member $y$ of $\text{Ran}(A)$ can be written as $A^2 x = A(Ax)$ for some $x$. But then $Ax$ can also be written as $A^2 w$, which says $y = A(Ax) = A^3 w$. And so on... – 2017-02-02
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    Robert you have forestalled my question about example of $A^3$.. I'm convinced by you :) – 2017-02-02
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    I wonder now also whether if $rank(A^2)$k $: $rank(A^{k})< rank(A^{k-1}) $.. – 2017-02-02
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Every Matrix has the "chance" of decreasing the rank. The rank-deficiency means, that there are some vectors, that will be mapped to zero. Everything, that is zero, will stay zero for ever. It does not matter, how many Matrices you can make up, the vector will stay at zero.

So, whenever a multiplication with a matrix decreases the rank, you loose that forever. So for $A^k$ to have the same rank as $A$ you must never loose any.

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    When you say 'never' you mean for example that the condition could be as good as also for example $rank(A^7)=rank(A^8)$ or even more generally $rank(A^{p_1})=rank(A^{p_2})$ ? – 2017-02-02
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    Exactly. You can also do that, by using the rank of a multiplicaton. $rank(A\cdot B)\leq \text{min}(rank(A), rank(B))$. Expand your $A^{p_1}$ and use this and the fact that on $\text{min}$ is enough. – 2017-02-02
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    I don't see exactly the way of proving it in this more general case. Could you expand it a little in your answer ? – 2017-02-02
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    It's basically using the fact, that if $rank(A^2)=rank(A)$, than it's the same for $rank(A^{p_1})=rank(A) = rank(A^{p_2})$. You are solving the general case by using the basic case. – 2017-02-02
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    but if I would know only for example $rank(A^{11})=rank(A^{27})$ ( I assume that additionally it should be not equal $0$) not knowing whether $rank(A)=rank(A^2)$ ? – 2017-02-02
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    AARGH, I'm so sorry... I misunderstood your last question! Since it is (for example) possible for $rank(A)=1$ and $rank(A^2)=0$, you can not deduce rank equality by arbitrary powers... However: If you know the rank deficiency of your matrix ($n-m$) that is maximum of rank-loss to occur in one multiplcation. That can lead somewhere. – 2017-02-02
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    Not exactly , I'm only want to know whether we can infer from $rank(A^{11})=rank(A^{27})$ that $rank(A)=rank(A^2)$ .. – 2017-02-02
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    Not in general! The other way round it's true! – 2017-02-02