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Transfer matrix

In wavelet theory

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Source updated February 26, 2026

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 " $\downarrow$ ":

$T_{h}\cdot x=(h*x)\downarrow 2.$

01Properties

$T_{h}\cdot x=T_{x}\cdot h$ .
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 $a=\Vert C_{2}h\Vert _{2}$ .

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Sources and credits

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Transfer matrix

01Properties

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