If I have a spatial domain signal $f$ $ = f(n,m)$, n $\in$ $(0,\ldots,(N-1))$, m $\in$ $(0,\ldots,(M-1))$. Then $F$, the fourier transform of $f$, is defined as
$$F(x,y) = \frac{1}{\sqrt{NM}} \sum_{n=0}^{N-1}\sum_{m=0}^{M-1} f(n,m)\cdot \exp{\left(i2\pi\left(\frac{nx}{N} + \frac{my}{M}\right)\right)}$$
Considering I want a low-pass filter (call it a transformation $L$) that sets to $0$ elements of $F$, is it possible to define a matrix $T$, such as $f^*= Tf$? ($f^*$ being the filtered signal).
The whole filtering process is shown below, intuition says each every step can be written as matrix multiplication, but the proof seems complicated.
$$f \rightarrow^{FT} F \rightarrow^{L} F^* \rightarrow^{FT^{-1}} f^* $$