Schur Leading Coefficient Lemma

Lemma

Let $\Delta [z]={\sum }_{i=0}^n{c}_i{z}^i$. If $\Delta$ is Schur, then $|c_n|>|c_0|$.

This is a necessary but not sufficient condition. So $|c_n|>|c_0|$ does not necessarily mean a polynomial is Schur.

Example 1:

$$z^3+z+3=\Delta [z]$$

where

• $c_n=c_3=1$
• $c_1=1$
• $c_0=3$

Then:

$$\begin{array}{r}|c_n|=|c_3|=1\\ |c_0|=3\end{array}⟹|c_n|<|c_0|⟹\Delta \text{ is NOT Schur!}$$

Example 2:

$$z^2+\frac{1}{2}z+\frac{1}{2}=\Delta [z]$$

Here, $|c_n|=1$ and $|c_0|=\frac{1}{2}$. Thus, $|c_n|>|c_0|$, so we cannot draw conclusions about whether $\Delta$ is Schur.

• However, if we factor out the characteristic polynomial, we can see that $\Delta z=(z+2)\left(z+\frac{1}{4}\right)$, so the polynomial is not Schur!
• This shows the limitations of the lemma – we still have to if $\Delta$ is Schur when $\left|c_n\right|>\left|c_0\right|$.

Proof

We have:

$$\Delta [z]=\sum _{i=1}^n{c}_i{z}^i=c_n\prod _{i=1}^n(z-\underset{\text{roots of }\Delta [z]}{\underbrace{p_i}})$$

Then, we have:

$$\Delta [0]=c_0=c_n\prod _{i=1}^n(-p_i)$$

Then:

$$\begin{array}{rl}⟹& \left|c_0\right|=\left|c_n\right|\prod _{i=1}^n\left|p_i\right|\\ ⟹& \frac{\left|c_0\right|}{\left|c_n\right|}=\prod _{i=1}^n\left|p_i\right|<1[\Delta [z]\text{ if Schur so all }{p}_i\in \mathbb{D}]\\ ⟹& \left|c_n\right|>\left|c_0\right|\end{array}$$