Geometric multiplicity, invertible matrices Would appreciate help with the following problem:

Given an invertible $n \times n$ real matrix $A$ with eigenvalue $t$ of geometric multiplicity at least $2$, determine whether the following statement is true: "There is a real polynomial $Q(x)$ of degree smaller or equal to $n-2$ such that the inverse of $A$ equals $Q(A)$".

It's easy to prove a similar statement, that the inverse of A can be represented by an $n-1$ degree polynomial (if I am correct, this is true for any invertible matrix), but that's as far as I got. I was thinking maybe the Jordan form of $A$ can help me out, but I haven't gotten very far with that. Would appreciate hints!
 A: To expand on my comment: A typical way of showing that if $A$ is invertible then there is a polynomial $q(x)$ such that $A^{-1}=q(A)$ is to use Cayley-Hamilton: writing 
$$\chi(x) = x^n + a_{n-1}x^{n-1}+\cdots+a_0$$
for the characteristic polynomial of $A$, we know that $a_0$ is the product of the eigenvalues, and so $A$ is invertible if and only if $a_0\neq 0$. The Cayley-Hamilton Theorem tells us that $\chi(A)=0$, so
$$0 = A^n + a_{n-1}A^{n-1}+\cdots + a_1A + a_0I,$$
from which we get that
$$-a_0I = A^n+a_{n-1}A^{n-1}+\cdots + a_1A,$$
hence
$$I =-\frac{1}{a_0}A\left( A^{n-1} + a_{n-1}A^{n-2}+\cdots + a_1I\right).$$
Thus, putting $q(x) = -\frac{1}{a_0}\left(x^{n-1}+a_{n-1}x^{n-2}+\cdots + a_1\right)$ gives the desired polynomial.
Of course, you can play the exact same game with the minimal polynomial instead of the characteristic polynomial. The matrix $A$ is invertible if and only if $0$ is not an eigenvalue, if and only if $x$ does not divide the minimal polynomial. The geometric multiplicity of $\mathbf{t}$ tells you something about the degree of the minimal polynomial: since the highest degree of $x-\mathbf{t}$ that divides the minimal polynomial is the size of the largest Jordan block associated to $\mathbf{t}$, and since the geometric multiplicity equals the number of Jordan blocks, if the algebraic multiplicity of $\mathbf{t}$ is, say, $m$, having geometric multiplicity $2$ tells you that there are at least two Jordan blocks associated to $\mathbf{t}$, and so the largest block will have size at most $m-1$; which tells you that if the largest power of $x-\mathbf{t}$ that divides $\chi(x)$ is $m$, then the largest power of $x-\mathbf{t}$ that divides the minimal polynomial is at most $m-1$.
Can you leverage that into showing that a $Q(x)$ as you want must exist?
