## 1 Introduction
\[\mbox{A}_{1}=[\sqrt{k_{n}}\;]_{N+1}=[\sqrt{k_{m}}\;]_{N+1}={\rm diag }(\frac{1}{\sqrt[2]{2}},\,1,\,1,\,\ldots,\,1,\,\frac{1}{\sqrt[2]{2}}), \mbox{where }m,\,n=0,\,1,\,\ldots,\,N \tag{9}\] \[\mbox{A}_{2}=[\sqrt{k_{n}}\;]_{N}=[\sqrt{k_{m}}\;]_{N}={\rm diag }(\frac{1}{\sqrt[2]{2}},\,1,\,1,\,\ldots,\,1), \mbox{where }m,\,n=0,\,1,\,\ldots,\,N-1\] (10) \[\mbox{A}_{3}=[\sqrt{k_{n}}\;]_{N}=[\sqrt{k_{m}}\;]_{N}={\rm diag }(1,\,1,\,\ldots,\,1,\,\frac{1}{\sqrt[2]{2}}), \mbox{where }m,\,n=1,\,2,\,\ldots,\,N\] (11) \[\mbox{A}_{4}=[\sqrt{l_{n}}\;]_{N}=[\sqrt{l_{m}}\;]_{N}={\rm diag }(1,\,1,\,\ldots,\,1,\,\frac{1}{\sqrt[2]{2}}), \mbox{where }m,\,n=0,\,1,\,\ldots,\,N-1 \tag{12}\]  
\[k_{p}=1/2 p=0,\,N \tag{13}\] \[=1 p=1,\,2,\,\ldots,\,N-1\] (14) \[l_{p}=1 p=0,\,1,\,\ldots,\,N-2\] \[=1/2 p=N-1\]  
where diag(\(a_{11}\), \(a_{11}\), \(\ldots\), \(a_{NN}\)) implies a diagonal matrix with the diagonal elements as (\(a_{11}\), \(a_{11}\), \(\ldots\), \(a_{NN}\)). I\({}_{N}\) is the identity matrix of size \(N\times N\). Let matrices J, K, Q, and R be defined as:_Algorithms_**2009**, \(2\)  
\[J=\left( \begin{array}{c}I_{N}\\ O_{{}_{\rm lxN}}\end{array} \right) K=\left( \begin{array}{c}O_{{}_{\rm lxN}}\\ I_{N}\end{array} \right) Q=\left( \begin{array}{c}I_{N-1}\\ O_{{}_{\rm lx(N-1)}}\end{array} \right) R=\left( \begin{array}{c}O_{{}_{\rm lx(N-1)}}\\ I_{N-1}\end{array} \right) \tag{15}\] \[(N+1)\times N (N+1)\times N N \times(N-1) N\times(N-1) N\times(N-1)\]  
where \(O_{1\times N}\) is the zero row-vector, with \(N\) elements, whose entries are all zero. Multiplying one of those matrices with an input vector appends a single zero on the top of the first or under the last element of the vector. Multiplying the transpose of one of those matrices with an input vector discards the first or last element of the vector. **2. Symmetric-Convolution Multiplication Property**  
One of the forty cases of symmetric convolution is proven as an example. **Property**. y = A\({}_{1}\)\({}^{-1}\)C\({}_{IE}\)\({}^{-1}\) J \({\cal H}_{\rm C2\emph{e}}\) C\({}_{\rm IIE}\) x  
**Proof**. Equation (A.1) in the appendix of [2] can be rewritten as:  
\[C_{1\emph{e}}=2\ \hat{C}_{1\emph{e}}\ [k_{n}]_{N+1} \tag{16}\]  
where \(\hat{C}_{1\emph{e}}\) is the kernel. \[C_{1\emph{e}}=\sqrt{2/N}\ \ [\sqrt{k_{n}}\ ]_{N+1}\ \hat{C}_{1\emph{e}}\ [ \sqrt{k_{n}}\ ]_{N+1} \tag{17}\] \[([\sqrt{k_{n}}\ ]_{N+1})^{-1}=[\sqrt{1/k_{n}}\ ]_{N+1} \tag{18}\]  
From (16), (17) and (18):  
\[C_{1\emph{e}}=[\sqrt{2N/k_{n}}\ ]_{N+1}\ C_{1\emph{e}}\ [\sqrt{k_{n}}\ ]_{N+1} \tag{19}\] \[C_{1\emph{e}}{}^{-1}=[\sqrt{1/k_{n}}\ ]_{N+1}\ C_{1\emph{e}}{}^{-1}\ [ \sqrt{k_{n}/(2N)}\ ]_{N+1} \tag{20}\]  
From (1) and (10):  
\[C_{2\emph{e}}{}^{-1}=C_{\rm IIE}{}^{-1}\ [\sqrt{k_{m}/(2N)}\ ]_{N} \tag{21}\]  
Let x and w be input column vectors in time domain. Let y be an output column vector in time domain. Let \({\cal H}_{\rm C2\emph{e}}\) be a diagonal matrix defined as \({\cal H}_{\rm C2\emph{e}}={\rm diag}([C_{2\emph{e}}\ {\rm w}]^{T})\), where superscript \(T\) denotes the transpose operator. We rewrite the 8 th property in Table VI of [2] as:  
\[y = C_{1\emph{e}}{}^{-1}\ \{\ {\cal H}_{\rm C2\emph{e}}\ C_{2 \emph{e}}\ {\rm x}\} \tag{22}\] \[= C_{1\emph{e}}{}^{-1}\ \{\ {\cal H}_{\rm C2\emph{e}}\ [\sqrt{2N/k_{m}}\ ]_{N}\ C_{ \rm IIE}\ {\rm x}\}\] \[= [\sqrt{1/k_{n}}\ ]_{N+1}\ C_{1\emph{e}}{}^{-1}\ [\sqrt{k_{m}}\ ]_{N+1}\ J\ \{\ {\cal H}_{\rm C2\emph{e}}\ [\sqrt{1/k_{m}}\ ]_{N}\ C_{ \rm IIE}\ {\rm x}\ \}\]  
The matrix J shows up in the last line of (22) for zero padding. Since \({\cal H}_{\rm C2\emph{e}}\) and \([\sqrt{1/k_{m}}\ ]_{N}\) are diagonal matrices, they can commute. Thus:  
\[C_{1\emph{e}}=2\ \hat{C}_{1\emph{e}}\ [k_{n}]_{N+1} \tag{16}\]0 \[C_{1\emph{e}}=2\ \hat{C}_{1\emph{e}}\ [k_{n}]_{N+1} \tag{16}\]1  
Another example is shown in (11) of [6]. Equation (12) of [6] is expanded to full for our derivation and only results are listed in the first column of Tables 3 and 4 in the Appendix.