given the minimal polynomial of matrix $A^2$ find minimal polynomials and Jordan forms of $A$

Question

I'm refreshing some linear algebra and came across this question: Consider a matrix $A \in \mathbb{C}^{4\times 4}$ and suppose the minimal polynomial of its square is $\phi_{A^2}=(x-1)^2$. Now we're asked to give all possible minimal polynomials of matrix $A$ with their respective Jordan normal form. My approach would be to just fill in $A^2$ in the minimal polynomial to get: $(A^2-I)^2=(A-I)^2(A+I)^2=0$. This would then lead to the Jordan form being: $J=\begin{pmatrix} 1&0&0&0\\ 1&1&0&0\\ 0&0&-1&0\\ 0&0&1&-1\\ \end{pmatrix}$. I suppose the minimal polynomial of $A$ could also be $(x-1)(x+1)^2$ or $(x-1)^2(x+1)$ or $(x-1)(x+1)$ each with their respective Jordan form but I don't see whether there's a thinking process behind it. like for example the minimal polynomial of $A^2$ limiting some of these from being possible but I don't see why. Any help or hint would be nice.

Answer 1

You know that $(A^2-I)^2 = 0$ but $A^2 - I \ne 0$. The minimal polynomial of $A$ divides $(x^2-1)^2 = (x-1)^2 (x+1)^2$ but does not divide $x^2 - 1 = (x+1)(x-1)$.
The possibilities are then $(x-1)^2 (x+1)^2$, $(x-1)^2 (x+1)$, $(x-1)^2$, $(x-1)(x+1)^2$ and $(x+1)^2$. For each of those it's easy to give an example.