Wiedemann for solving sparse linear equation

Question

I am new member. I am researching in Wiedemann algorithm to find solution $x$ of $$Ax=b$$ Firstly, I will show a Wiedemann's deterministic algorithm (Algorithm 2 in paper

1. Compute $A^ib$ for $i=0..2n-1$; n is szie of matrix A
2. Set k=0 and $g_0(z)=1$
3. Set $u_{k+1}$ to be $k+1$st unit vector
4. Extract from the result of step 1 the sequence $\{(u_{k+1},A^ib)\}^{2n-1}_{i=0}$, (,) is inner production
5. Apply $g_k(z)$ to this sequence
6. Set $f_{k+1}$ is minimal polynomial of that proceduce sequence in step 5
7. Set $g_{k+1}=f_{k+1}g_k$
8. Set $k=k+1$, and then if $deg(g_k)$

In which, the defination of Step 5 as following: Assume that, I have a polynominal $g(z)$ of degree $d$ and a sequence $\{s_i\}^l_{i=0}$. Let $$s(z)=\sum_{i=0}^l s_iz^i$$ and say that by definition the result of apply $g$ to the sequence is $$\{(g(z^{-1})s(z))[i] \}^{l-d}_{i=0}$$ Clearly, applying a polynomial to a sequence can be done by multiplying polynomials. It is definition of applying a polynominal to a sequence.

I am very clear step 1 to step 4. But I am confusing at step 5. I have two questions that need your help

1. What are coefficients of $g(z^{-1})$ given by $$g(z)=1z^2+1z^1+0$$

2. How to compute the applying of $g$ to the sequence $z$ from $0$ to $l-d$

Thanks in advance