# Transfer matrix

In applied mathematics, the transfer matrix is a formulation in terms of a block-Toeplitz matrix of the two-scale equation, which characterizes refinable functions. Refinable functions play an important role in wavelet theory and finite element theory.
For the mask $h$, which is a vector with component indexes from $a$ to $b$, the transfer matrix of $h$, we call it $T_{h}$ here, is defined as
    $(T_{h})_{j,k}=h_{2\cdot j-k}.$
More verbosely
    $T_{h}={\begin{pmatrix}h_{a}&&&&&\\h_{a+2}&h_{a+1}&h_{a}&&&\\h_{a+4}&h_{a+3}&h_{a+2}&h_{a+1}&h_{a}&\\\ddots &\ddots &\ddots &\ddots &\ddots &\ddots \\&h_{b}&h_{b-1}&h_{b-2}&h_{b-3}&h_{b-4}\\&&&h_{b}&h_{b-1}&h_{b-2}\\&&&&&h_{b}\end{pmatrix}}.$
The effect of $T_{h}$ can be expressed in terms of the downsampling operator "$\downarrow$":
    $T_{h}\cdot x=(h*x)\downarrow 2.$

## Properties

* $T_{h}\cdot x=T_{x}\cdot h$.
* If you drop the first and the last column and move the odd-indexed columns to the left and the even-indexed columns to the right, then you obtain a transposed Sylvester matrix.
* The determinant of a transfer matrix is essentially a resultant. More precisely: Let $h_{\mathrm {e} }$ be the even-indexed coefficients of $h$ ($(h_{\mathrm {e} })_{k}=h_{2k}$) and let $h_{\mathrm {o} }$ be the odd-indexed coefficients of $h$ ($(h_{\mathrm {o} })_{k}=h_{2k+1}$). Then $\det T_{h}=(-1)^{\lfloor {\frac {b-a+1}{4}}\rfloor }\cdot h_{a}\cdot h_{b}\cdot \mathrm {res} (h_{\mathrm {e} },h_{\mathrm {o} })$, where $\mathrm {res}$ is the resultant. This connection allows for fast computation using the Euclidean algorithm.
* For the trace of the transfer matrix of convolved masks holds $\mathrm {tr} ~T_{g*h}=\mathrm {tr} ~T_{g}\cdot \mathrm {tr} ~T_{h}$
* For the determinant of the transfer matrix of convolved mask holds $\det T_{g*h}=\det T_{g}\cdot \det T_{h}\cdot \mathrm {res} (g_{-},h)$ where $g_{-}$ denotes the mask with alternating signs, i.e. $(g_{-})_{k}=(-1)^{k}\cdot g_{k}$.
* If $T_{h}\cdot x=0$, then $T_{g*h}\cdot (g_{-}*x)=0$. This is a concretion of the determinant property above. From the determinant property one knows that $T_{g*h}$ is singular whenever $T_{h}$ is singular. This property also tells, how vectors from the null space of $T_{h}$ can be converted to null space vectors of $T_{g*h}$.
* If $x$ is an eigenvector of $T_{h}$ with respect to the eigenvalue $\lambda$, i.e. $T_{h}\cdot x=\lambda \cdot x$, then $x*(1,-1)$ is an eigenvector of $T_{h*(1,1)}$ with respect to the same eigenvalue, i.e. $T_{h*(1,1)}\cdot (x*(1,-1))=\lambda \cdot (x*(1,-1))$.
* Let $\lambda _{a},\dots ,\lambda _{b}$ be the eigenvalues of $T_{h}$, which implies $\lambda _{a}+\dots +\lambda _{b}=\mathrm {tr} ~T_{h}$ and more generally $\lambda _{a}^{n}+\dots +\lambda _{b}^{n}=\mathrm {tr} (T_{h}^{n})$. This sum is useful for estimating the spectral radius of $T_{h}$. There is an alternative possibility for computing the sum of eigenvalue powers, which is faster for small $n$. Let $C_{k}h$ be the periodization of $h$ with respect to period $2^{k}-1$. That is $C_{k}h$ is a circular filter, which means that the component indexes are residue classes with respect to the modulus $2^{k}-1$. Then with the upsampling operator $\uparrow$ it holds $\mathrm {tr} (T_{h}^{n})=\left(C_{k}h*(C_{k}h\uparrow 2)*(C_{k}h\uparrow 2^{2})*\cdots *(C_{k}h\uparrow 2^{n-1})\right)_{[0]_{2^{n}-1}}$ Actually not $n-2$ convolutions are necessary, but only $2\cdot \log _{2}n$ ones, when applying the strategy of efficient computation of powers. Even more the approach can be further sped up using the Fast Fourier transform.
* From the previous statement we can derive an estimate of the spectral radius of $\varrho (T_{h})$. It holds $\varrho (T_{h})\geq {\frac {a}{\sqrt {\#h}}}\geq {\frac {1}{\sqrt {3\cdot \#h}}}$ where $\#h$ is the size of the filter and if all eigenvalues are real, it is also true that $\varrho (T_{h})\leq a$, where $a=\Vert C_{2}h\Vert _{2}$.