Transfer matrix

More precisely:

Let ${\displaystyle h_{\mathrm {e} }}$  be the even-indexed coefficients of ${\displaystyle h}$  (${\displaystyle (h_{\mathrm {e} })_{k}=h_{2k}}$ ) and let ${\displaystyle h_{\mathrm {o} }}$  be the odd-indexed coefficients of ${\displaystyle h}$  (${\displaystyle (h_{\mathrm {o} })_{k}=h_{2k+1}}$ ).

Then ${\displaystyle \det T_{h}=(-1)^{\lfloor {\frac {b-a+1}{4}}\rfloor }\cdot h_{a}\cdot h_{b}\cdot \mathrm {res} (h_{\mathrm {e} },h_{\mathrm {o} })}$ , where ${\displaystyle \mathrm {res} }$  is the resultant.

${\displaystyle \det T_{g*h}=\det T_{g}\cdot \det T_{h}\cdot \mathrm {res} (g_{-},h)}$ 

${\displaystyle T_{h}\cdot x=\lambda \cdot x}$ ,

then ${\displaystyle x*(1,-1)}$  is an eigenvector of ${\displaystyle T_{h*(1,1)}}$  with respect to the same eigenvalue, i.e.

Let ${\displaystyle C_{k}h}$  be the periodization of ${\displaystyle h}$  with respect to period ${\displaystyle 2^{k}-1}$ . That is ${\displaystyle C_{k}h}$  is a circular filter, which means that the component indexes are residue classes with respect to the modulus ${\displaystyle 2^{k}-1}$ . Then with the upsampling operator ${\displaystyle \uparrow }$  it holds

${\displaystyle \mathrm {tr} (T_{h}^{n})=\left(C_{k}h*(C_{k}h\uparrow 2)*(C_{k}h\uparrow 2^{2})*\cdots *(C_{k}h\uparrow 2^{n-1})\right)_{[0]_{2^{n}-1}}}$ 

${\displaystyle \varrho (T_{h})\geq {\frac {a}{\sqrt {\#h}}}\geq {\frac {1}{\sqrt {3\cdot \#h}}}}$ 

where ${\displaystyle \#h}$  is the size of the filter and if all eigenvalues are real, it is also true that

${\displaystyle \varrho (T_{h})\leq a}$ ,