Let $S$ be the set of 6x6 matrices with rational coefficients such that the characteristic polynomial is $x^6-x^2$ and the minimal polynomial is $x^5-x$. Problem: Given $A\in S$ what is the dimension of the null space of $(A^2+1)^2$.
Any two elements of $S$ have the same invariant factors so they're similar. The characteristic polynomial factors into $x^2(x-1)(x+1)(x^2+1)$ over the rationals. So $0,0,1,-1$ will be diagonal entries of $A$.
- What are the other diagonal entries?
- How does $(A^2+1)^2$ correspond to the minimal polynomial of $A$?
- How can I determine if there will be a 1 on the subdiagonal corresponding to the double root 0?
First of all, I'm not sure what you mean by "invariant factors". I haven't heard of any such thing that would be helpful in this context.
You can consider matrices in $S$ as matrices in the algebraic completion of $\Bbb Q$, so that the polynomial completely factors. So, we can write $$ x^6 - x^2 = x^2(x-1)(x+1)(x-i)(x+i) $$ where $i = \sqrt{-1}$.
Now, it is important to note that $A$ is diagonalizable. So, $A^2 + I$ is also be diagonalizable, as is $(A^2 + I)^2$. It follows that the dimension of the nullspace of $(A^2 + I)^2$ will simply be the algebraic multiplicity of the eigenvalue $0$.
Remember that for any polynomial $p(t)$: if $\lambda$ is an eigenvalue of $A$, then $p(\lambda)$ is an eigenvalue of $p(A)$.