A matrix representation for the inverse matrix. I have a problem from "Methods of Algebriac Geometry in Control Theory by Peter Falb" textbook:

Show that if $A$ is $\,n\times n\,$ matrix, then $\displaystyle\,(zI-A)^{-1} = \sum_{j=1}^n \phi_j\left(z\right) A^{n-j} \big/ \det\left[zI-A\right]$. Compute $\phi_j\left(z\right)$.

I am using the following formula:
$$
{\left(zI-A\right)}^{-1} = 
\frac{1}{\det \left(zI-A\right)}
\sum_{s=0}^{n-1}\left(zI-A\right)^{s}
\sum_{k_1,k_2,\ldots ,k_{n-1}}
\prod_{l=1}^{n-1} \frac{\left(-1\right)^{k_l+1}}{l^{k_l}\,k_{l}!}\,
\left(\operatorname{tr}\left(zI-A\right)^l\right)^{k_l}
$$ 
where the second summation is over $\displaystyle\,j+\sum_{l=1}^{n-1}lk_l=n-1$.
But I don't see how to find $\,\phi_j\left(z\right)$.
Can anyone help me on this?
Thanks in advance.
P.S
In the book it's stated that $\phi_j(z)$ is a polynomial of degree $n-j$.
 A: This question is a bit unclear as written -- from the notation one would infer that $\phi_j(z)$ is supposed to be independent of the matrix $A$, but this cannot be true; consider for instance $n=2$; set $z=0, A = \left[\begin{array}{cc}a & b\\c & d\end{array}\right]$ to get that
$$\left[\begin{array}{cc}-d & b\\c & -a\end{array}\right] = \phi_1(0) \left[\begin{array}{cc}a & b\\c & d\end{array}\right] + \phi_2(0) I$$
and no constants $\phi_j(0)$ can exist for which this equality holds for all $A$.
If $\phi_j$ is not required to be independent of $A$, then you already have your answer in the formula you posted, after some reindexing:
\begin{align*}
\sum_{s=0}^{n-1} (zI-A)^s f(s) &= \sum_{s=0}^{n-1} \sum_{k=0}^s \binom{s}{k} (-1)^k A^k z^{s-k} f(s)\\
&= \sum_{k=0}^{n-1} \sum_{s=k}^{n-1} \binom{s}{k} (-1)^k A^k z^{s-k} f(s)\\
&= \sum_{j=1}^{n} \sum_{s=n-j}^{n-1} \binom{s}{n-j} (-1)^{n-j} A^{n-j} z^{s-n+j} f(s)
\end{align*}
and 
$$\phi_j(z) = \sum_{s=n-j}^{n-1} \binom{s}{n-j} (-1)^{n-j} z^{s-n+j} f(s)$$
where $f(s)$ is your
$$\sum_{k_1,k_2,\ldots ,k_{n-1}}
\prod_{l=1}^{n-1} \frac{\left(-1\right)^{k_l+1}}{l^{k_l}\,k_{l}!}\,
\left(\operatorname{tr}\left(zI-A\right)^l\right)^{k_l}.$$
