Given the minimal polynomial of a matrix $A^2$, what could the minimal polynomial of $A$ be? It is given that the minimal polynomial of $A^2$ is $\phi_{A^2}(x) = (x-1)^2$, where $A$ is a complex $4\times4$ matrix. The question is, what are the possible minimal polynomials for $A$?
From the given that $\phi_{A^2}(x) = (x-1)^2$, I can derive several things. First off, I know that the characteristic polynomial $f_{A^2}$ of $A^2$ must have the same irreducible factors, which implies that the only eigenvalue of $A^2$ is $1$. Secondly, it is easily seen that $(A^2 -I)^2 = O$ and $A^2 - I \neq O$. Therefore, the matrix $A^2 - I$ is nilpotent with index 2. This implies that the invariant system is either $\{2,1,1\}$ or $\{2,2\}$. Putting this together, there are two possible Jordan forms of $A^2$:
$$ J_{A^2,1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} \quad\text{or}\quad J_{A^2,2} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix} $$
Great, all these things I know about $A^2$... but how does any of it lead to information about the minimal polynomial of $A$? Or just any information about $A$ at all? I'm kind of stuck on this.

HINT
You know a polynomial satisfied by $A$, namely $(A^2-1)^2$. So the minimal polynomial has to be a factor of $(x-1)^2(x+1)^2$.
Which of the possible factors could give the correct minimal polynomial for $A^2$?
If neither power of $(x-1)^i(x+1)^j$ is $2$ then we would have one of $A-I=0,A+I=0,(A+I)(A-I)=0$. In all these cases $A^2=I$, a contradiction. This cuts your work considerably.
Finally, to check the remaining possibilities easily, use simple 2x2 Jordan blocks e.g.
$A=\begin{pmatrix}1&1&0&0 \\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}$ has minimal polynomial $(x-1)^2$. What is the minimal polynomial for $A^2$?
