Schur Checking Tutorial

Boundary Crossing Lemma

Let $\lambda \in [0,1]$ and $\Delta \lambda [z]={\Delta }_1[z]-\lambda {\Delta }_2[z]$. If ${\Delta }_1[z]$ is Schur and ${\Delta }_1[z]-{\Delta }_2[z]$ is not Schur (or vice versa), then $\exists {\lambda }^{\ast }\in [0,1]$ such that $\Delta {\lambda }^{\ast }[z]$ has a root on the unit circle.

Let $\Delta [z]={\sum }_{i=0}^n{c}_i{z}^i$. Let $Q[z]={\sum }_{i=0}^n{c}_{n-i}{z}^i$.

• Example: $\Delta [z]=5z^3+4z^2+1$. Then

$$Q[z]=z^3+0z^2+4z+5$$

Let

$$\begin{array}{rl}R[z]& =\frac{1}{z}\left(\Delta [z]-\frac{c_0}{c_n}Q[z]\right)\\ & =\frac{1}{z}\left(\sum _{i=0}^n{c}_i{z}^i-\frac{c_0}{c_n}\sum _{i=0}^n{c}_{n-i}{z}^i\right)\\ & =\frac{1}{z}\left(\sum _{i=0}^n\left(c_i-\frac{c_0}{c_n}{c}_{n-i}\right)z^i\right)\\ & =\frac{1}{z}\left(\sum _{i=1}^n\left(c_i-\frac{c_0}{c_n}{c}_{n-i}\right)z^i\right)\\ & =\frac{1}{z}\left(z\sum _{i=0}^{n-1}\left(c_{i+1}-\frac{c_0}{c_n}{c}_{n-i-1}\right)z^i\right)\\ & =\sum _{i=0}^{n-1}\left(c_{i+1}-\frac{c_0}{c_n}{c}_{n-i-1}\right)z^i\end{array}$$

• Line 3-4: For $i=0$ we get $c_0-\frac{c_0}{\cancel{c_n}}\cancel{c_n}=0$.

Lemma 2

Suppose $|c_n|>|c_0|$. Then $\Delta [z]$ is Schur $\iff$ $R[z]$ is Schur.

Define $R^{(0)}[z]:=\Delta [z]$:

$$\begin{array}{rl}& R^{(i+1)}[z]=\frac{1}{z}\left(R^{(i)}[z]-\frac{c_0}{c_n}{Q}^{(i)}[z]\right)\\ & R^{(0)}[z]⟶R^{(1)}[z]⟶R^{(2)}[z]⟶\dots ⟶R^{n-1}[z]\end{array}$$

• degree $n$, $n-1$, $n-2$, …, 1

For degree 1, we have $c_{1}z+c_0$, which has a root at $-\frac{c_0}{c_1}$. Then:

$$\frac{|c_0|}{|c_n|}<1(\text{aka Schur})⟺|c_n|>|c_0|$$