For any $n \times n$ matrix $A$, is it true that $\ker(A^{n+1}) = \ker(A^{n+2}) = \ker(A^{n+3}) = \dots$ ? If yes, what is the proof and is there a name to this theorem? If not, for what matrices will it be true? How can I find a counterexample in the latter case?

I know that powers of nilpotent matrices increase their kernel's dimension up to $n$ (for the zero matrix) in the first $n$ steps.

But is it necessary that for all singular matrices, all the rank reduction (if it occurs) must be in the initial exponents itself? In other words, is it possible for some matrices to have $\ker(A^{k}) = \ker(A^{k+1}) < \ker(A^{k+1+m})$ for some $m,k > 0$?

  • $\begingroup$ The title isn't supposed to replace the first line of your question. As for the question, the answer depends on how you quantify over $n$. $\endgroup$ – Git Gud Sep 10 '14 at 18:59
  • $\begingroup$ Added the first line. Could you please explain what you mean by 'quantify over n'? $\endgroup$ – allrtaken Sep 10 '14 at 19:07
  • $\begingroup$ Let $P(n)$ be expression in the title before. If you mean $\exists n\in \mathbb NP(n)$, then the statement is true. If you mean $\forall n\in \mathbb NP(n)$, then the statement is false. $\endgroup$ – Git Gud Sep 10 '14 at 19:12
  • $\begingroup$ I meant n to be the dimension of the matrix. $\endgroup$ – allrtaken Sep 10 '14 at 19:21

This is true. To my knowledge, there is no name for this theorem.

You can think of this as a consequence of Jordan canonical form. In particular, we can always write $$ A = S[N \oplus P]S^{-1} $$ Where $N$ is nilpotent and $P$ has full rank. It suffices to show that $N$ has order of nilpotence at most equal to $n$, and that $P$ never reduces in rank.

  • $\begingroup$ I want to check if I understood this correctly. Does the matrix N comprise of Jordan blocks with eigenvalue 0, and P is the matrix comprising of Jordan blocks corresponding to the other eigenvalues? $\endgroup$ – allrtaken Sep 10 '14 at 19:21
  • $\begingroup$ @allrtaken that's exactly right. It is useful to note that a matrix is nilpotent if and only if all of its (complex) eigenvalues are equal to $0$. $\endgroup$ – Omnomnomnom Sep 10 '14 at 20:25
  • $\begingroup$ I am not sure whether this is the right place for asking this question, but can you suggest a good textbook that covers the theory around this? $\endgroup$ – allrtaken Sep 11 '14 at 18:26
  • $\begingroup$ Most linear algebra texts geared towards advanced undergraduates or graduates cover Jordan Canonical form at some point. I, in particular, used Horn and Johnson. Axler's "Linear Algebra Done Right" might be another good bet. If you want some more ideas, you could always post another question on this site. $\endgroup$ – Omnomnomnom Sep 11 '14 at 19:01

An important observation to be made here is that the if for some $k$, we have $\ker(A^k) = \ker(A^{k+1})$, then $\forall j\geq 0, \ker(A^{k+j}) = \ker(A^k)$. To show this, it would be sufficient to show that $\ker(A^{k+2}) = \ker(A^{k+1})$, and the rest would follow from a simple inductive argument.

Note that, we have $\ker(A^{k+1}) \subseteq \ker(A^{k+2})$, and thus it is enough to show that $\ker(A^{k+1}) \supseteq \ker(A^{k+2})$.

For this, consider a vector $v$ such that $v \in \ker(A^{k+2})$, i.e., $A^{k+2}v = 0$. Then, $Av \in \ker(A^{k+1})$ because $A^{k+1}(Av) = 0$. Since $\ker(A^{k+1}) = \ker(A^k)$, we have $Av \in \ker(A^k)$. Thus, $A^{k}(Av) = 0$, and hence $A^{k+1}v = 0$, which implies that $v \in \ker(A^{k+1})$.

Clearly, $\ker(A^{k+2}) \subseteq \ker(A^{k+1})$, and thus $\ker(A^{k+2}) = \ker(A^{k+1})$


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