\msF(\na u)(\xxi)=\left(\msF(\p_1u)(\xxi),\msF(\p_2u)(\xxi)\right)=\sum_{j=1}^J\alpha_j\int_{\p\Om_j}e^{-2\pi i\xxi\cdot\x}\rd\nnu_j(\x).\end{aligned}$$ Let $\varphi\in\msS$, where $\msS$ is the space of Schwartz functions (e.g. [@Folland1999]). Then the direct computation shows that $\left(\msF(\na u)\ast\wh{\varphi}\right)(\xxi)=\left(\left(\msF(\p_1u)\ast\wh{\varphi}\right)(\xxi),\left(\msF(\p_2u)\ast\wh{\varphi}\right)(\xxi)\right)$ satisfies $$\begin{aligned}
\label{FouGraduConv}
\begin{split}
\left(\msF(\na u)\ast\wh{\varphi}\right)(\xxi)=\sum_{j=1}^J\alpha_j\int_{\p\Om_j}e^{-2\pi i\xxi\cdot\x}\varphi(\x)\rd\nnu_j(\x).
\end{split}\end{aligned}$$ From \[FouGraduConv\], it is easy to see that if $\Gamma\subseteq\left\{\x\in\R^2:\varphi(\x)=0\right\}$, then $\msF(\na u)\ast\wh{\varphi}=0$.

We further assume that there exists a finite (rectangular and symmetric) index set $\KK$ such that $$\begin{aligned}
\label{MajorAssumption}
\Gamma\subseteq\left\{\x\in\R^2:\varphi(\x)=0\right\}~~~~~\text{with}~~\varphi(\x)=\sum_{\bk\in\KK}\a[\bk]e^{2\pi i\bk\cdot\x/L}.\end{aligned}$$ For such $\varphi$, we have $$\begin{aligned}
\label{FouGradConvSemiDiscrete}
\left(\msF(\na u)\ast\wh{\varphi}\right)(\xxi)=\sum_{\bk\in\KK}\msF(\na u)(\xxi-L^{-1}\bk)\a[\bk]=0,~~~~~~~\xxi\in\R^2,\end{aligned}$$ which is referred to as the *linear annihilation relation* [@G.Ongie2015; @G.Ongie2016]. The trigonometric polynomial $\varphi$ in \[MajorAssumption\] is called the *annihilating polynomial*, $\a$ is called the *annihilating filter* with support $\KK$, and $\Gamma$ satisfying \[MajorAssumption\] is called the *trigonometric curves* [@G.Ongie2016]. It is proved in [@G.Ongie2016] that trigonometric curves can always be described as the zero set of a *real valued* trigonometric polynomial, using the tool of algebraic geometries.

In MRI, the Fourier transform of $u$ is sampled on the grid $L^{-1}\OO$ [@E.M.Haacke1999], so \[FouGradConvSemiDiscrete\] becomes the finite system of linear equations $$\begin{aligned}
\label{FouGradConvDiscrete}
\sum_{\bk\in\KK}\msF(\na u)(L^{-1}(\bsl+\bk))\a[-\bk]=0,~~~~~~~\bsl\in\OO:\KK.\end{aligned}$$ In the matrix-vector multiplication form, we have $$\begin{aligned}
\label{RankDeficientHankel}
\bmH\left(\msF(\na u)\big|_{L^{-1}\OO}\right)\a[-\cdot]=\0.\end{aligned}$$ Moreover, if $\varphi$ in \[MajorAssumption\] is the minimal polynomial with the support $\KK$ (i.e. $\KK$ has the smallest linear 