The relationship between power and rank drop of a phalanx

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We know that there is $\textrm{rank} (AB) \leq \textrm{min}\{\textrm{rank} (A), \ \textrm{rank} (B) \}$$,$ and if $a$ or $b$ reversible $,$ The unequal number will become equal to $. $
so for any phalanx $a$ and $k$ there is $\textrm{rank} (a^{k+1}) \leq\textrm{rank} (A^K). $ that is $\textrm{rank} (a^k) $ is a monotonically reduced function of $k$ .

but is there any more profound nature of this function? $?$

$\textbf{lemma:}$$\textrm{rank} (ABC) +\textrm{rank} (B) \geq \textrm{rank} (AB) +\textrm{rank} (BC). $
$\textbf{Proof:}$ do Elementary Transformation $$\left ({\begin{array}{*{20}{c}}{abc}&0\\ 0&b\end{array}} \right) \to \left ({\begin {array} {*{20}{c}} {abc}&{ab}\\ 0&b\end{array}} \right) \to \left ({\begin{array}{*{20}{c}}0&{ab}\\ {-BC}&B\end{array}} \right) \to \left ({\begin{array}{*{20}{c}}{ab}&0\\ B&{bc}\end{array}} \right). \ proof. $$

$\textbf{Inference:}$$\textrm{rank} (A^k)-\textrm{rank} (A^{k+1}) \leq \textrm{rank} (A^{k-1})-\textrm{rank} (A^k). $
$\textbf{Proof:}$ The lemma in the $b$ for $a^{k-1}$$,$ $C $ for $a$ can be certified $.$

The above inference shows that $\textrm{rank} (a^k) $ decreases with the increase of the $k$ $,$ and the descending value is also getting smaller $.$

$\textbf{theorem 1:}$ If present $k$ make $\textrm{rank} (a^k) =\textrm{rank} (A^{k+1}) $$,$ $\textrm{rank} (a^k) =\textrm{rank} (A^ {k+1}) =\textrm{rank} (A^{k+2}) =\textrm{rank} (A^{k+3}) =\cdots\cdots$
$\textbf{Proof:}$ use Inference and $\textrm{rank} (a^k) $ is $k$ 's monotone subtraction function $,$ learned $0\leq \textrm{rank} (A^{k+1})-\textrm{rank} (A^{k+2}) \leq \textrm{rank} (a^k)-\textrm{rank} (a^{k+1}) =0$$,$ proof $.$

$\textbf{theorem 2:}$ If $a$ is $n$-step phalanx $,$ then $\textrm{rank} (A^n) =\textrm{rank} (A^{n+1}) =\textrm{rank} (a^{n+2}) =\textrm{ Rank} (A^{n+3}) =\cdots\cdots$
$\textbf{Proof:}$ disprove Law $,$ if $\textrm{rank} (a^n) >\textrm{rank} (A^{n+1}) $$,$ indicates that the rank drop does not stop at $n$ $,$ description $i\leq n$ when $\ Textrm{rank} (A^i) \geq \textrm{rank} (A^{i+1}) +1$$,$ launched $\textrm{rank} (A) \geq \textrm{rank} (a^{n+1}) +n$$.$ because $\TEXTRM {Rank} (a) >\textrm{rank} (a^2) $$,$ description $a$ irreversible $,$ so $\textrm{rank} (a) \leq n-1$$.$ thus derived $\textrm{rank} (a^{n+1}) \leq-1$$,$ There's no way to $.$, so $.$.

The above theorem illustrates that $\textrm{rank} (A^K) $ is a $k$ monotone subtraction function $,$ and the descent value is also getting smaller $.$

A point that will remain constant after a stop in a certain place $,$ and stops the rank drop will certainly reach $.$ at some $s\leq n$

So the relationship between $,$ $\textrm{rank} (a^k) $ and $k$ will be as shown in $:$

The relationship between power and rank drop of a phalanx