If $\operatorname{rank}(A) = \operatorname{rank}(A^2)$, thn $\operatorname{rank}(A) = \operatorname{rank}(A^n)$, if $A$ has $2$ rows and $2$ columns? How to prove that if $\operatorname{rank}(A) = \operatorname{rank} (A^2)$, then $\operatorname{rank}(A) = \operatorname{rank}(A^n)$, if $A$ has $2$ rows and $2$ columns?
 A: There are only three cases, $rank(A)=0,1,2$. Note that for $2\times 2$ matrices $A$ the following holds true;

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*$rank(A)=0$ if and only if $A=0$

*$rank(A)=1$ if and only if $A=fg^\top$ for nonzero column vectors $f,g$

*$rank(A)=2$ if and only if $\det A\not=0$.

Thus if $rank(A)=rank(A^2)=0$ then $A=0$, and clearly so is $A^n$, hence $rank(A^n)=0$.
If $rank(A)=rank(A^2)=2$ then $\det A\not = 0$, hence $\det(A^n)=(\det A)^n\not =0$ and also $rank(A^n)=2$.
Finally, if $rank(A)=rank(A^2)=1$, then $A=fg^\top$, as mentioned above. We note that $g^\top f\not=0$; otherwise, $A^2=0$, which contradicts the hypothesis. Thus $A^n=(fg^\top)^n=(g^\top f)^{n-1}f g^\top$ which is also rank $1$ for all $n\geq 1$. Hence $rank(A^n)=1=rank(A)$ for all $n\geq 1$.
A: Since $\operatorname{im}(AB)\subset \operatorname{im}(A)$ for any $A, B$, and since a proper subspace ha a smaller dimension, the condition on ranks can be recast in terms of images: if $\operatorname{im}(A)=\operatorname{im}(A^2)$, then $\operatorname{im}(A)=\operatorname{im}(A^n)$ for all $n$.  This is true for any square matrix (not just $2\times 2$).
Let us proceed by induction.  For the sake of notation, let $A:V\to V$, and we write $A^n(V)$ for the image of $A$. By assumption $A^2(V)=A(V)$. Applying $A^{n-1}$ to both sides, we get $A^{n+1}(V)=A^n(V)$.  If we know that $A^n(V)=A(V)$, then $A^{n+1}(V)=A^n(V)=A(V)$.  Thus, but induction, the proposition will be true for all $n$.
It is perhaps worth noting that we can strengthen this result quite easily.  If $A^N(V)=A^{N+1}(V)$, for some $N$, then $A^n(V)=A^{N}(V)$ for all $n>N$. The proof is identical.
