The formula for the $n$-th power of a square matrix and its inverse Let's assume we've found the formula for the nth power of a non-singular matrix $A$ for any positive integer $n$. Can we find the inverse of $A$ just by replacing $n$ with $-1$ in the formula of $A^n$?
I ask this question because in all the examples I've considered, doing this works.
EDITI'm not referring to a recursive formula. What I mean is that we've found (by any method) the nth term of the sequence $\{A^n\}$ as an expression of n. For example, for $$A=\begin{pmatrix}2&0\\1&1\end{pmatrix}$$ we have$$A^n=\begin{pmatrix}2^n&0\\-1+2^n&1\end{pmatrix}$$ where n is usually considered to be a positive integer. But we can see that if we replace n by -1 in that expression we get the inverse of A (and, hence, it's easy to see that the formula is true for any integer). In any other example I've considered (with A non singular), the same thing happens. If A is diagonalizable it's easy for me to prove this fact. My question is how to prove (or disprove) it if A is not diagonalizable.  
 A: Let $A\in GL_r(\mathbb{C})$ be s.t. $spectrum(A)=(\lambda_j)_{j\leq k}$, where the $(\lambda_j)$ are non-zero distinct complex numbers; we assume that $m$ is the maximum of the multiplicities of the $(\lambda_i)$ and the minimum polynomial of $A$ has degree $s$. Then, 
$\textbf{Proposition 1}$. There are functions $(p_i)_{0\leq i\leq s-1}$ s.t., for every positive integer $n$, 
i) $A^n=\sum_{i=0}^{s-1} p_i(n)A^i$
ii) each $p_i$ is in the form $p_i(n)=\sum_{1\leq j\leq k}\sum_{0\leq d<m} a_{i,j,d}({\lambda_j}^n)^{(d)}$ where the $a_{i,j,d}$ are complex constants (they don't depend on $n$) and $({\lambda_j}^n)^{(d)}$ is the $d^{th}$ derivative of ${\lambda_j}^n$ wrt. $\lambda_j$ (considered as a variable).
$\textbf{Proof}$. There is $P\in GL_r(\mathbb{C})$ s.t. $A=PDP^{-1}$ where $D$ is a block-diagonal matrix where each block is in the form $\Delta=\lambda I+J$ with $J$ a Jordan block of dimension say $u$. Then 
$(*)$ $\Delta^n=\lambda^nI+(\lambda^n)'J+1/2!(\lambda^n)''J^2+\cdots+\dfrac{1}{(u-1)!}(\lambda^n)^{(u-1)}J^{u-1}$ (binomial formula). 
Thus, each entry of $D^n$ is in the form of $p_i(n)$ and each entry of $A^n$ too.  $\square$
$\textbf{Proposition 2}$. There is a unique decomposition of the $(A^n)_{n\in \mathbb{Z}^+}$ of the previous type.
$\textbf{Proof}$. According to the unicity of the decomposition of the $(A^n)$ as linear combinations of $I,A,\cdots,A^{s-1}$, it suffices to show the result for $p_i$. Assume that $p_i(n)=0$ for $n=0,\cdots,mk-1$; we have a homogeneous linear system of $mk$ equations in the $mk$ unknowns $(a_{i,j,d})_{j,d}$. Its determinant, for $m=3$ and the distinct eigenvalues $x,y,z$ is the $9\times 9$ pseudo Vandermonde

$=8(x-y)^9(y-z)^9(z-x)^9\not= 0$; the generalization is clear. $\square$
$\textbf{Remark}$. There is not unicity of the above decomposition if we consider, more generally, analytic functions (for example, add $\sin( n\pi)$).
Assume that we have, in hand, the previous unique decomposition (valid for any positive integer $n$): $A^n=\sum_{i=0}^{s-1} p_i(n)A^i$.
$\textbf{Proposition 3}$. Then, for every positive integer $n$,  $A^{-n}=\sum_{i=0}^{s-1} p_i(-n)A^i$.
$\textbf{Proof}$. Recall that the eigenvalues of $A$ are non-zero. We consider the development in Taylor series of the $u\times u$ matrix $(\lambda I+J)^{-n}=\lambda^{-n}(I+1/\lambda J)^{-n}=$
$\lambda^{-n}(I-\dfrac{n}{\lambda} J+\dfrac{\binom{n}{2}}{\lambda^2} J^2+\cdots\pm \dfrac{\binom{n}{u-1}}{\lambda^{u-1}}J^{u-1})$.
We obtain exactly the formula $(*)$ in which we changed $n$ into $-n$. Thus if $\Delta^n=\phi(n)$, then $\Delta^{-n}=\phi(-n)$. We can use a similar trick for $D^n,D^{-n}$ and $A^n,A^{-n}$.  $\square$
EDIT. Answer to the OP. Yes, if we know that, for $n\in\mathbb{Z}^+$,  $A^n=\sum_{i=0}^{s-1} p_i(n)A^i$ where $p_i(n)$ is a linear combination of words $({\lambda_j}^n)^{(d)}$, with $0\notin spectrum(A)=(\lambda_j)_j$, then the formula is valid for $n\in\mathbb{Z}^-$. Yet, if a priori we do not know anything about the $(\lambda_j)$, then -for the unicity- we must prove that, necessarily, $spectrum(A)=(\lambda_j)_j$. I am sure that it is always true, but I did not proved it above. Of course, if you know explicitly $A$, then it is easy to verify.
On the other hand, it is very rare to have an explicit formula giving the $(A^n)$ because, in the case of a generic matrix, we cannot explicitly calculate its eigenvalues.
