Let $R_{M\times N}$ be a space of size $M\times N$. Define the 2D Discrete Fourier Transform of $f\in R_{M\times N}$ to be
\begin{equation} \tilde{f}[m,n]=\sum_{p=0}^{M-1}\sum_{q=0}^{N-1}f[p,q]e^{\frac{-i 2\pi p m}{M} +\frac{-i2\pi q n}{N}},\ \ \ 0\leq m < M,\ \ 0 \leq n < N \end{equation}
If $f[m,n]$ and $g[m,n]$ are related by a translation i.e. $f[m, n] = g[m + u_0 (mod\ M), n + v_0 (mod\ N)]$ can I say anything special about the rank of the following matrix?
\begin{equation} H[m,n]= \frac{\tilde{f}[m,n]\tilde{g}[m,n]^*}{|\tilde{f}[m,n]\tilde{g}[m,n]^*|} \end{equation}
A bit more following AnonSubmitter85's hint: I work out the transform for the space $M$ with shift $u_0(mod\ M)$. We later generalize the result to $M\times N$.
\begin{align} \tilde{f}[m]&=\sum_{p=0}^{M-1}f[p]e^{\frac{-i 2\pi p m}{M}}\\ &=\sum_{p=0}^{M-1}g[p+u_0(mod\ M)]e^{\frac{-i 2\pi m(p + u_0(mod\ M))}{M}}e^{\frac{i 2\pi mu_0(mod\ M)}{M}}\\ &=\tilde{g}[m]e^{\frac{i 2\pi mu_0(mod\ M)}{M}} \end{align}
In general, we have $H[m,n]=e^{\frac{i 2\pi mu_0(mod\ M)}{M}}.e^{\frac{i 2\pi nv_0(mod\ N)}{N}}$ I know how to get the rank of matrix through Gaussian elimination but I've no idea how to do it for a general matrix like this. So what is the rank?
Choose any two rows, say $m = m_i$ and $m = m_j$. Their values are
$$ \begin{align} \mathrm{row} \, i & : & e^{j2\pi m_i u_0 / M }e^{j2\pi n v_0 / N } & , & n = 0,1,\dots,N-1 \\ \mathrm{row} \, j & : & e^{j2\pi m_j u_0 / M }e^{j2\pi n v_0 / N } & , & n = 0,1,\dots,N-1 \end{align} $$
If we multiply row $j$ by $e^{j2\pi (m_i-m_j)u_0/M}$, then the two rows will be equal. Thus, the RREF form of the matrix will contain only one non-zero row and the rank of the linear system is $1$.