Difference Equations and displacement operator For a Prep exam
Exercise from the book: Numerical analysis of scientific computing. Section 1.3-3
Let $p$ be a polynomial of degree $m$, with $p(0) \neq 0$. If a sequence $x$ contains $m$ consecutive zeros and $p(E)x = 0$, then $x=0$
Where:
$E$ denotes the displacement operator, $Ex=[x_2, x_3, \dots]$ where $x = [x_1, x_2, \dots]$
$L: V \to V$
$L=P(E) = \sum_{i=0}^nc_i E^i$ where $P(\lambda) = \sum_{i=0}^m c_i \lambda^i$
I've tried:
I have studied the definitions of from Kincaid's book and I have found two theorems that says that: 
$\textbf{Theorem 1:}$ 
Let $p$ be a polynomial satisfying $p(0) \neq 0$. Then a basis for the null space of $p(E)$ is obtained as follows. With each zero $\lambda$ of $p$ having multiplicity $k$, associate the $k$ basic solutions $x(\lambda), x^{'}(\lambda), \dots, x^{k-1}(\lambda)$, where $x(\lambda) = [\lambda, \lambda^2, \lambda^3, \dots]$.
$\textbf{Theorem 2:}$ 
For a polynomial p satisfying $p(0) \neq 0$, these properties are equivalent:
(i) The difference equation $p(E)x = 0$ is stable
(ii) All zeros of $p$ satisfy $|z| \leq 1$, and all multiple zeros satisfy $|z|< 1$
but I don't know how to start, at the beginingI was thinking to show that $P(E)=\sum_{i=0}^n c_iE^i /neq 0$, but I am not sure if this is the way, any suggestions?
Thank u
 A: Let $p$ is a polynomial of degree $m$, $p(0)\neq 0$, and $x$ is a sequence with $m$ consecutive zeros. Suppose $p(E)(x)=0$. 
Well it is obvious that $x$ has length $\geq m$. When $x$ has exactly length $m$, then all the entries of the sequence is zero. Hence $x=0$.
Suppose $x$ has length greater than $m$. First we will begin with the assumption that $x$ has length $m+1$. Remember all the zero entries are consecutive. Hence we can assume, without loss of generality, than the first term is nonzero. i.e. Suppose $x=[x_1,0,\dots,0]$.
Let $p(b)=c_mb^m+c_{m-1}b^{m-1}+\dots+c_1b+c_0$. Note $p(0)\neq 0$ implies $c_0\neq 0$, and $p$ being a polynomial of degree $m$ implies $c_m\neq 0$. 
\begin{aligned}
p(E)(x) & =\left(\sum_{n=0}^{m} c_nE^n\right)x=\sum_{n=0}^{m}c_n(E^nx) \\
&= c_0E^0x+c_1E^1x+\dots+c_{m-1}E^{m-1}x+c_mE^mx \\
&=c_0\cdot[x_1,0,\dots,0]+c_1\cdot [0,x_1,0,\dots,0]+\dots+c_{m-1}\cdot [0,\dots,0,x_1,0,0]+c_m\cdot [0,\dots,0,x_1,0] \\
&=0 \text{ (by our assumption)}
\end{aligned}
Note that we have a system of $m+1$ equations.
[
\begin{cases}
c_0x_1=0 \\
c_1x_1=0 \\
\vdots
\end{cases}
]
By the first equation $x_1=0$ because $c_0\neq 0$. However, this is a contradiction because we assumed $x$ only has $m$ consecutive zeros. Hence $x$ must be of length $m$. 
We can extend this argument, for the remaining cases.
A: We have for the $k$'th element:
$$ (P(E)x)_k = c_0 x_k + \cdots + c_m x_{k+m}$$
Suppose $x_{j+1}=x_{j+2}=\cdots x_{j+m}=0$ for some $j\geq 0$.
Then $(P(E)x)_{j+1} = c_m x_{j+m+1} = 0$ implies $x_{j+m+1}=0$. By simple induction, shifting to the right, every $x_p=0$ for $p>j$.
When $j>0$: $ (P(E)x)_{j} = c_0 x_{j} = 0$ implies $x_j=0$. By simple induction, shifting to the left, every $x_p=0$ for every $p\leq j$.
