I am told $m(x)$ divides $x^2(x-1)$, $rank(A) = n-1$ & $rank(A^2) = n-2$. So I have the base form: $$J_2(0)^a \oplus J_1(0)^b\oplus J_1(1)^c$$So I know the geometric multiplicity of the $0$ eigenvalue is $1$ i.e. $rank(A) = n-1 = n-a-b$.
However a step in the solution for this problem says that since $rank(A^2) = n-2 = n-2a-b$. I am unsure of why this is the case, surely this does not correspond to the geometric multiplicity again?
So what does $rank(A^2)$ actually tell us apart from the rank of the JCF?
In general, if the maximum size of any Jordan block of an $\ n\times n\ $ matrix $\ M\ $ corresponding to an eigenvalue $\ \lambda\ $ is $\ m\ $, and the matrix has $\ k_s\ $ Jordan blocks of size $\ s\ $ corresponding to that eigenvalue for each $\ s=1,2,\dots,m\ $, then the rank of $\ (M-\lambda)^i\ $ is $\ n-\sum_\limits{s=1}^isk_s-i\sum_\limits{s=i+1}^mk_s\ $. The sum $\ \sum_\limits{s=1}^isk_s\ $ is the sum of the sizes of all the corresponding Jordan blocks of sizes not exceeding $\ i\ $, and the sum $\ \sum_\limits{s=i+1}^mk_s\ $ is the total number of corresponding Jordan blocks whose sizes do exceed $\ i\ $.
In your case of $\ A^2\ $, $\ m=2\ $ if $\ a\ne0\ $, or $\ m=1\ $ otherwise, and $\ k_1=b, k_2=a\ $. In either case, the above formula for the rank gives it as $\ n-2a-b\ $. Your equations $\ n-1=$$\,n-a-b\ $ and $\ n-2=$$\,n-2b-a\ $ then tell you that $\ a+b=1\ $ and $\ 2a+b=2\ $, or $\ a=1, b=0\ $, from which it also follows that $\ c=n-2\ $.